An object weighs 63.8 N in air. When it is suspended from a force scale and completely immersed in water the scale reads 16.8 N.
1. Determine the density of the object.
2. When the object is immersed in oil, the force scale reads 35.6 N. Calculate the density of the oil.

Answers

Answer 1

Answer:

The density of this object is approximately [tex]1.36\; {\rm kg \cdot L^{-1}}[/tex].

The density of the oil in this question is approximately [tex]0.600\; {\rm kg \cdot L^{-1}}[/tex].

(Assumption: the gravitational field strength is [tex]g =9.806\; {\rm N \cdot kg^{-1}}[/tex])

Explanation:

When the gravitational field strength is [tex]g[/tex], the weight [tex](\text{weight})[/tex] of an object of mass [tex]m[/tex] would be [tex]m\, g[/tex].

Conversely, if the weight of an object is [tex](\text{weight})[/tex] in a gravitational field of strength [tex]g[/tex], the mass [tex]m[/tex] of that object would be [tex]m = (\text{weight}) / g[/tex].

Assuming that [tex]g =9.806\; {\rm N \cdot kg^{-1}}[/tex]. The mass of this [tex]63.8\; {\rm N}[/tex]-object would be:

[tex]\begin{aligned} \text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{63.8\; {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 6.506\; {\rm kg}\end{aligned}[/tex].

When an object is immersed in a liquid, the buoyancy force on that object would be equal to the weight of the liquid that was displaced. For instance, since the object in this question was fully immersed in water, the volume of water displaced would be equal to the volume of this object.

When this object was suspended in water, the buoyancy force on this object was [tex](63.8\; {\rm N} - 16.8\; {\rm N}) = 47.0\; {\rm N}[/tex]. Hence, the weight of water that this object displaced would be [tex]47.0 \; {\rm N}[/tex].

The mass of water displaced would be:

[tex]\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{47.0\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 4.793\; {\rm kg}\end{aligned}[/tex].

The volume of that much water (which this object had displaced) would be:

[tex]\begin{aligned}\text{volume} &= \frac{\text{mass}}{\text{density}} \\ &\approx \frac{4.793\; {\rm kg}}{1.00\; {\rm kg \cdot L^{-1}}} \\ &\approx 4.793\; {\rm L}\end{aligned}[/tex].

Since this object was fully immersed in water, the volume of this object would be equal to the volume of water displaced. Hence, the volume of this object is approximately [tex]4.793\; {\rm L}[/tex].

The mass of this object is [tex]6.50\; {\rm kg}[/tex]. Hence, the density of this object would be:

[tex]\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{6.506\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 1.36\; {\rm kg \cdot L^{-1}} \end{aligned}[/tex].

(Rounded to [tex]\text{$3$ sig. fig.}[/tex])

Similarly, since this object was fully immersed in oil, the volume of oil displaced would be equal to the volume of this object: approximately [tex]4.793\; {\rm L}[/tex].

The weight of oil displaced would be equal to the magnitude of the buoyancy force: [tex]63.8\; {\rm N} - 35.6\; {\rm N} = 28.2\; {\rm N}[/tex].

The mass of that much oil would be:

[tex]\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{28.2\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 2.876\; {\rm kg}\end{aligned}[/tex].

Hence, the density of the oil in this question would be:

[tex]\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{2.876\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 0.600\; {\rm kg \cdot L^{-1}} \end{aligned}[/tex].

(Rounded to [tex]\text{$3$ sig. fig.}[/tex])


Related Questions

An 87-kg football player traveling 5.2 m/s is stopped in 1.0 s by a tackler.
What is the original kinetic energy of the player?
Express your answer to two significant figures and include the appropriate units.
What average power is required to stop him?
Express your answer to two significant figures and include the appropriate units.

Answers

The original kinetic energy of the player and the average power is required to stop him are 1200 J and 1200 W respectively.

What is Kinetic Energy ?

The energy possessed by a body in motion is known as Kinetic Energy. The S. I unit is Joule.

Given that an 87-kg football player traveling 5.2 m/s is stopped in 1.0 s by a tackler. The given parameters are;

Mass m = 87 KgVelocity v = 5.2 m/sTime t = 1 s

The original kinetic energy of the player can be calculated by using the formula K.E = 1/2m[tex]v^{2}[/tex]

Substitute all the parameters into the formula

K.E = 1/2 x 87 x [tex]5.2^{2}[/tex]

K.E = 1176.24

K.E = 1200 J

Power is the rate at which work is done.

Work done = energy

The average power is required to stop him can be calculated by using the formula P = E/t

Substitute all the parameters into the formula

P = 1200/1

P = 1200 W

Therefore, the original kinetic energy of the player and the average power is required to stop him are 1200 J and 1200 W respectively.

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A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
If the angle between the beam and the cable is θ = 57.0° what is the vertical component of the force exerted by the hi.nge on the beam?

Answers

26.10 N is the vertical component of the force.

Rx  represents the Horizontal component of force

Ry represents The Vertical component of force

According to the given diagram

Rx - Tcosθ = 0

Rx = Tcosθ

And,

Ry + Tsinθ = mg

Ry = mg - Tsinθ

The horizontal component of force =The Vertical component of force  

Rx = Ry

Tcosθ = mg - Tsinθ

T(cosθ + sinθ) = 29 × 9.8 = 284.2 N

T√2 cosθ = 284.2 N

T × √2 ×0.544 = 284.2 N

T × 0.769 = 284.2 N

T = 370 N (app)

So,

Ry = 284.2 - 370 (sin 57°)

    = 284.2 - 310.3 = -26.10 N

Hence, 26.10 N is the vertical component of the force exerted.

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Select the correct answer.
Using the statistical definition of entropy, what is the entropy of a system where W = 4?

Answers

Using the statistical definition of entropy, The entropy of a system W = 4 is 1.91×10²³.

What is entropy?

Entropy is typically referred to as a measurement of a system's randomness or disorder. In 1850, a German physicist named Rudolf Clausius first proposed this idea. Entropy is a thermodynamic property used to characterize a system's behavior in terms of temperature, pressure, entropy, and heat capacity. This thermodynamic explanation took the systems' equilibrium condition into account.

Entropy can be calculated using a mathematical expression.

Entropy = Total change of heat / thermodynamic temperature

S= KblnW

S is the statistical entropy

The value of Boltzmann's constant is 1.38×10⁻²³.

S=  1.38×10⁻²³ ln(4)

S = 1.91×10⁻²³

Using the statistical definition of entropy, The entropy of a system W = 4 is 1.91×10²³.

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Answer: C.

1.91 × 10⁻²³ joules/kelvin

Explanation: edmentum

The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down. Initial speed of the rock = 1.92×101 m/s. (Glob has no atmosphere, so no energy is lost to air friction. G = 6.67×10-11 Nm2/kg2.)
A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Calculate the speed of the satellite.

Answers

The orbiting speed of the satellite orbiting around the planet Glob is 60.8m/s.

To find the answer, we need to know about the orbital velocity a satellite.

What's the expression of orbital velocity of a satellite?Mathematically, orbital velocity= √(GM/r)G= gravitational constant= 6.67×10^(-11) Nm²/kg², M = mass of sun , r= radius of orbit

What's the orbital velocity of the satellite in a circular orbit with a radius of 1.45×10⁵ m around the planet Glob of mass 7.88×10¹⁸ kg?Here, M= 7.88×10¹⁸ kg, r= 1.45×10⁵ mOrbital velocity of the orbiting satellite = √(6.67×10^(-11)×7.88×10¹⁸/1.45×10⁵)

= 60.8m/s

Thus, we can conclude that the speed of the satellite orbiting the planet Glob is 60.8m/s.

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A constant force F is applied on a body for a time interval of delta t.This force changes the velocity of the body from V1 to V2.then change in momentum in time interval will be

Answers

The change in momentum in time interval, given the data will be F × Δt

What is momentum?

Momentum is defined as the product of mass and velocity. It is expressed as

Momentum = mass × velocity

What is impulse?

This is defined as the change in momentum of an object.

Impulse = change in momentum

But

Impulse = force × time

Therefore

Force × time = change in momentum

How to determine the change in momentumInitial velocity = v₁ Finalvelocity = v₂Force = FChange in time = ΔtChange in momentum = ?

Force × time = change in momentum

F × Δt = change in momentum

Change in momentum = F × Δt

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What is the x-component of a vector with a magnitude of 115 km at an angle of 22°?

Answers

The x-component of a vector are < 106.6, 43.07 >

Depending on the angle we are provided, the x-component of a vector can either be cos or sin. Cos always corresponds to the right triangle's side that contacts the specified angle.

If a vector v with magnitude ||v|| makes an angle θ with the positive x-axis then,

v = ||v|| cos θi + ||v|| sin θj

 =  < ||v|| cos θ , ||v|| sin θ >

Magnitude p = 115 km

Angle = 22°

p = ||p|| < cos θ, sin θ >

p = 115 < cos 22°, sin 22° >

p = 115 < 0.927, 0.3746 >

p = < 106.6, 43.07 >

Therefore,  the x-component of a vector are < 106.6, 43.07 >

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What is the approximate uncertainty in the area of a circle of radius 4.3×104cm ?

Answers

The approximate uncertainty in the area of a circle is 5.4%.

Area of the circle

A = πr²

where;

r is radius of the circle

A = π(4.3 x 10⁴)²

A = 5.81 x 10⁹ cm²

Let the error in measurement = 1 x 10⁴ cm

Error in Area measurement

A =  π(1 x 10⁴)² = 3.14 x 10⁸ cm²

Uncertainty in measurement

% = (error/actual area) x 100%

% = ( 3.14 x 10⁸) / (5.81 x 10⁹) x 100%

% = 5.4 %

Thus, the approximate uncertainty in the area of a circle is 5.4%.

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The Earth’s diameter is about 8,000 miles; our Moon’s diameter is about 2,000 miles; how
many Moon’s would fit inside of the volume of the Sun?

Answers

Three moons can fit inside the volume of the sun.

What is the moon?

The moon is a non luminous body found in the space. It could cause a solar eclipse when it comes between the sun and the earth.

Since the Earth’s diameter is about 8,000 miles and the Moon’s diameter is about 2,000 miles, to obtain the number of moons that could fit inside the sun we have;

8,000 miles/ 2,000 miles = 3

Hence, three moons can fit inside the volume of the sun.

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The Lunar Module could make a safe landing if its vertical velocity at impact is 2.8 m/s or less. Use conservation of energy to determine h in each case. The acceleration due to gravity at the surface of the Moon is 1.62 m/s2.
Suppose that you want to determine the greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is zero.
Express your answer to two significant figures and include the appropriate units.
Suppose that you want to determine the greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s downward.
Express your answer to two significant figures and include the appropriate units.
Suppose that you want to determine the greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s upward.
Express your answer to two significant figures and include the appropriate units.

Answers

(1) The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is zero is 0.4 m.

(2) The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s downward is 0.25 m.

(3) The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s upward is 0.25 m.

Maximum height

v² = u² - 2gh

where;

v is final velocityu is initial velocity

when the lander's velocity = 0

0 = u² - 2gh

u² = 2gh

h = u²/2g

h = (2.8²)/(2 x 9.8)

h = 0.4 m

when the velocity of the lander is 1.7 m/s downward

h = (u² - v²)/2g

h = (2.8² - 1.7²)/(2 x 9.8)

h = 0.25 m

when the velocity of the lander is 1.7 m/s upward

h = (u² - v²)/2g

h = (2.8² - 1.7²)/(2 x 9.8)

h = 0.25 m

Thus, the greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is zero is 0.4 m.

The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s downward is 0.25 m.

The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s upward is 0.25 m.

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A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (Figure 1). The field is changing with time, according to B(t)=(1.4T)e^−(0.057s^−1)t.
a) Find the emf induced in the loop as a function of time (assume t is in seconds).
b) When is the induced emf equal to 110 of its initial value?
c) Find the direction of the current induced in the loop, as viewed from above the loop.

Answers

The solution to the questions are given as

[tex]t=40.39 \mathrm{sec}[/tex][tex]\varepsilon &=(0.12v)e^{0.057t}[/tex]the direction of induced current will be Counterclock vise.

What is the direction of the current induced in the loop, as viewed from above the loop.?

Given, $B(t)=(1.4 T) e^{-0.057 t}$

[tex]$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}[/tex]

[tex]\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$[/tex]

[tex]\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}[/tex]

[tex]\varepsilon &=(0.12v)e^{0.057t}[/tex]

(b) [tex]Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$[/tex]

[tex]\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}[/tex]

c)

In conclusion, the direction of the induced current will be Counterclockwise.

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Figure 21.62 shows a box on whose surfaces the electric field is measured to be horizontal and to the right. On the left face (3 cm by 2 cm) the magnitude of the electric field is 400 V/m, and on the right face the magnitude of the electric field is 1000 V/m. On the other faces only the direction is known (horizontal). Calculate the electric flux on every face of the box, the total flux, and the total amount of charge that is inside the box.

Answers

(a) The electric flux on left face is 0.24 Vm and on the right face is 1 Vm.

(b) The total flux is 1.24 Vm.

(c) The total amount of charge that is inside the box is  1.1 x 10⁻¹¹ C.

Area of the left face

The area of the left face is calculated as follows;

A1 = 0.03 m x 0.02 m = 0.0006 m²

Electric flux on the left face

Ф1 = EA1

Ф1 = (400 V/m)( 0.0006 m²) = 0.24 Vm

Let the dimension of the right face = 5 cm by 2 cm

Area of the right face

A2 = 0.05 m x 0.02 m = 0.001 m²

Electric flux on the right face

Ф2 = EA2

Ф2 = (1000 V/m)( 0.001 m²) = 1 Vm

Total flux

Ф = Ф1 + Ф2

Ф = 0.24 Vm + 1 Vm = 1.24 Vm

Total charge inside the box

Ф = Q/ε

Q = εФ

Q = (8.85 x 10⁻¹²)(1.24)

Q = 1.1 x 10⁻¹¹ C

Thus, the electric flux on left face is 0.24 Vm and on the right face is 1 Vm.

The total flux is 1.24 Vm and the total amount of charge that is inside the box is  1.1 x 10⁻¹¹ C.

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A 1100-N crate rests on the floor.
1) How much work is required to move it at constant speed 5.0 m along the floor against a friction force of 310 N .
Express your answer to two significant figures and include the appropriate units.
2)
How much work is required to move it at constant speed 5.0 m vertically.
Express your answer to two significant figures and include the appropriate units.

Answers

(a) The work done in moving the crate along the floor is 7,050 J.

(b) The work done in moving the crate vertically is 0 J.

Work done in moving the crate horizontally

W = Fdcosθ

where;

F is total force to be appliedd is displacement of the crateθ is the horizontal angle = 0

W = (1100 + 310) x 5

W = 7,050 J

Work done in moving the crate vertically

W  = Fd cos(90)

W = 0

Thus, the work done in moving the crate along the floor is 7,050 J.

the work done in moving the crate vertically is 0 J.

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The small amount of missing mass that occurs when _____ atoms fuse to form helium
atoms is converted into energy in the form of heat and light.

nitrogen

hydrogen

barium

Ochromium

Answers

Answer:

hydrogen

Explanation:

Atomic fusion occurs when 2 or more atoms combine to create a new element.

Helium

Before figuring out what elements create helium, we need to understand its properties. Helium is the second element on the periodic table. It has 2 protons and usually 2 electrons in the base state.

There are technically ions but base helium will have 2 electrons, especially because it is a noble gas.

Fusion

In most cases, including this one, when 2 atoms fuse together you can add their protons together to find the atomic number of the new element. Since helium only has 2 protons, the elements fused to form helium must both have 1 proton.

The element with 1 proton is hydrogen. So, 2 hydrogen atoms fuse to form helium.

Real World Applications

Fusion can be seen across the universe. One of the most common examples of hydrogen fusing to become helium is in stars. Due to the heat within stars, 2 hydrogen atoms can fuse together to create helium. This process also creates energy in the form of heat and light that the star is made of.

Glucose solution is administered to a patient in a hospital. The density of the solution is 1.308 kg/l. If the blood pressure in the vein is 35.7 mmHg, then what is the minimum necessary height of the IV bag above the position of the needle?

Answers

The minimum  necessary height of the IV bag above the position of the needle is 0.37 m.

we know that,

P = ρgh

where,

P = 35.7mmHg

  = 4759.609 Pa

g = 9.8[tex]\frac{m}{s^{2} }[/tex]

ρ = 1.308 kg / m^3

now, substituting all the values, we get,

4759.609 = 1.308 × 9.8 × h

h = 0.37 m

The minimum  necessary height of the IV bag above the position of the needle = 0.37 m.

what is an IV bag ?

A reagent, also called as an analytical reagent, is a substance or compound that is added to a system in chemistry to bring about a chemical reaction or examine to see if one happens. Even though the terms "reagent" and "reactant" are frequently used synonymously, "reactant" refers to a substance that is consumed during a chemical reaction.

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Radiation from the sun hits Earth unequally and is absorbed by different materials in varying amounts. This is called
A. the Coriolis Effect
B. a thermocline
OC. El Nino Southern Oscillation
OD. differential heating

Answers

Radiation from the sun hits earth unequally and is absorbed by different materials in varying amounts. This is called differential heating.

What is differential heating?

Differential heating is the property that causes different surfaces to heat up and cool down at different rates. The earth's surface receives different magnitudes of solar radiation and also the earth's surfaces absorb thermal energy in different magnitudes.

The color, shape, texture, surface and presence of constructions can influence the heating or cooling of the earth.

The earth in the equator line is heated more by solar action than that of the poles, since it receives more amount of radiation per unit area.

In general, dry surfaces heat up and cool down faster than wet ones.

Therefore, we can confirm that when radiation from the sun hits earth unequally and is absorbed by different materials in varying amounts is called differential heating.

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Which of the following changes would increase the force between two
charged particles to 9 times the original force?
A. Decreasing the distance between the particles by a factor of 3
B. Decreasing the amount of charge on one of the particles by a
factor of 9
C. Increasing the distance between the particles by a factor of 3
D. Increasing the amount of charge on each particle by a factor of 9

Answers

The answer is A. Decreasing the distance between the particles by a factor of 3.

The Universal Law of Gravitation is :

F = Gm₁m₂ / r² (where 'r' is the distance between them)

Since force is inversely proportional to the square of the distance between them, distance has to be decreased by a factor of 3 to increase the force to 9 times the original force.

Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a spitting cobra rears up to a height of 0.420 m above the ground and launches venom at 2.50 m/s, directed 35.0° above the horizon. Neglecting air resistance, find the horizontal distance (in m) traveled by the venom before it hits the ground.

Answers

The horizontal distance traveled by the venom before it hits the ground is 0.6 m

What is Projectile ?

A stone or ball or anything projected is known as a projectile.

Given that  a spitting cobra rears up to a height of 0.420 m above the ground and launches venom at 2.50 m/s, directed 35.0° above the horizon.

Assuming air resistance is neglected, the parameters to be considered are;

Height h = 0.42 mVelocity v = 2.5 m/sAngle Ф = 35°

The ball we reach the ground at the same time when it is dropped vertically or horizontally.

In a vertical direction,

h = 1/2g[tex]t^{2}[/tex]

Substitute all the parameters into the formula

0.42 = 1/2 x 9.8 x [tex]t^{2}[/tex]

0.42 = 4.9[tex]t^{2}[/tex]

[tex]t^{2}[/tex] = 0.42/4.9

[tex]t^{2}[/tex] = 0.0857

t = [tex]\sqrt{0.0857}[/tex]

t = 0.29 s

The horizontal distance traveled by the venom before it hits the ground will be

Distance = ucosФ x t

Distance = 2.5cos35 x 0.29

Distance = 0.599 m

Distance = 0.6 m

Therefore,  the horizontal distance (in m) traveled by the venom before it hits the ground is 0.6 m

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Two builders carry a sheet of drywall up a ramp. Assume that W = 1.80 m, L = 3.30 m, θ = 24.0°, and that the lead builder carries a (vertical) weight of 147.0 N (33.0 lb).
1. What is the (vertical) weight carried by the builder at the rear?
2. The builder at the rear gets tired and suggest that the drywall should be held by its narrow side. What is the weight (in N) he must now carry?

Answers

The vertical weight carried by the builder at the rear is 240.89N. The weight he must now carry is 352.26N

1. How to solve for the vertical weight

We have w = 1.8

Then we have L as 3.30

θ = 24.0

FC = 147

We have to find FB

147 (3.3 + 1.8 tan24)/(3.3 - 1.8 tan24)

= 240.896

The vertical weight carried by the builder is 240.896

2. 240.896 + 147

= 387.896

387.896/[1 + (1.8 + 3.3 tan24) /(1.8 - 3.3 tan24)]

= 387.896/10.885

= 35.64

387.896 - 35.64

= 352.26N

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A moonshiner makes the error of filling a glass jar to the brim and capping it tightly. The moonshine expands more than the glass when it warms up, in such a way that the volume increases by 0.6% (that is, ΔV/V0 = 6 ✕ 10-3) relative to the space available. Calculate the force exerted by the moonshine per square centimeter if the bulk modulus is 2.1 ✕ 109 N/m2, assuming the jar does not break.

Answers

The force exerted per square centimeter is 126 N/cm².

What is pressure?

Pressure is the force acting per unit area.

Pressure = force/area

Based on the data given:

volume increase, ΔV/V0 = 6 * 10⁻³

Bulk Modulus, B = 2.1 * 10⁹ N/m²

Bulk modulus B of a material is ratio of change in pressure and change in volume as given below:

B = ΔP/ [(ΔV/V)]

Solving for ΔP;

ΔP = B * [(ΔV/V)]

ΔP = (2.1 * 10⁹ N/m²) * (6 * 10⁻³)

ΔP = 1.26 * 10⁶ N/m²

Converting to per square centimeter

ΔP = (1.26 * 10⁶ N/m²)/10⁴

ΔP = 126 N/cm²

In conclusion, the force exerted per square centimeter is a measure of the pressure.

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The mass of Saturn is 5.68 x 1026 kg, and use an orbital radius of 3.00 x 105 km. (G = 6.67 × 10-11 N ∙ m2/kg2). Find the orbital speed of an ice cube in the rings of Saturn.

Answers

The orbital speed of an ice cube in the rings of Saturn is determined as  355,366.5 m/s.

What is orbital speed?

The orbital speed of an astronomical body or object is the speed at which it orbits around the center of mass of the most massive body.

Orbital speed of ice cube in the rings of Saturn

The orbital speed of ice cube in the rings of Saturn is calculated as follows;

v = √GM/r

where;

G is universal gravitation constantM is mass of Saturnr is the distance of the ice cube

v = √(6.67 x 10⁻¹¹ x 5.68 x 10²⁶)/(3 x 10⁵)

v = 355,366.5 m/s

Thus, the orbital speed of an ice cube in the rings of Saturn is determined as  355,366.5 m/s.

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What is formed when 2 or more elements combine chemically

Answers

Answer:

compound

Explanation:

a substance made from two or more different elements that have been chemically joined. Examples of compounds include water (H2O), which is made from the elements hydrogen and oxygen, and table salt (NaCl), which is made from the elements sodium and chloride.

6.
A swimmer bounces straight up from a diving board and falls feet first into a pool. She
starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool. (3pt)
a) How long are her feet in the air?
b) What is her highest point above the board?
c) What is her velocity when her feet hit the water?

Answers

a) Her feet are in the air for 0.73+0.41 = 1.14 seconds

b) Her highest height above the board is 0.82 m

c) Her velocity when her feet hit the water is 7.16 m/s

Given,

t = Time taken

u = Initial velocity = 4 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

a) Her feet are in the air for 0.73+0.41 = 1.14 seconds

s = ut + [tex]\frac{1}{2}[/tex]at²

2.62 = 0t + [tex]\frac{1}{2}[/tex] ₓ 9.8  ₓ [tex]t^{2}[/tex]

t = 0.73 s

b) Her highest height above the board is 0.82 m

The total height she would fall is 0.82+1.8 = 2.62 m

v = u + at

0 = 4 ₋ 9.8 ₓ t

t = 0.41 s

s = ut +[tex]\frac{1}{2}[/tex] at²

s = 4 ₓ 0.41 ₊ [tex]\frac{1}{2}[/tex] ₓ ₋9.8 ₓ 0.41 [tex]t^{2}[/tex]

c) Her velocity when her feet hit the water is 7.16 m/s

[tex]v = u + at \\v = 0 + 9.8[/tex] ₓ [tex]0.73[/tex]

v = 7.16 m/s

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A coil 3.55 cm in radius, containing 470 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.20×10^−2 T/s )t+( 2.85×10^−5 T/s4 )t^4. The coil is connected to a 610 Ω resistor, and its plane is perpendicular to the magnetic field. You can ignore the resistance of the coil.
a) Find the magnitude of the induced emf in the coil as a function of time.
b) What is the current in the resistor at time t0 = 5.20 s ?

Answers

The current is 1.13* 10^{-4}A

Given that r = radius of the coil = 4 cm = 0.04 m

Area of coil is given as

A = πr²

A = (3.14) (0.04)² = 0.005024 m²

N = Number of turns = 500

R = Resistance = 600 Ω

B = magnetic field = (0.0120)t + (3 x 10⁻⁵) t⁴

Taking derivative at both the side

[tex] \frac{dB}{dt} = (0.120 + (12 \times 10^-5)t^3)[/tex]

Induced current is given as

[tex]i= (\frac{NA}{R} )( \frac{db}{dt} )[/tex]

[tex]i \: = (\frac{NA}{R})(12 \times 10 ^{-5} )t^3[/tex]

substituting the value t = 5

[tex]i = ( \frac{(500)(0.005024)}{600}) (12 \times 10 ^{-5} )5^3[/tex]

[tex]i = 1.13 \times 10 ^{ - 4} A[/tex]

Hence the current is

[tex]1.13 \times 10^-4A[/tex]

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The Electric current is 1.11* 10^{-4}A


Given that the coil's radius is 3.55 cm (0.35 m),

The formula for the coil's area is A = r2 A = (3.14) (0.35)2 = 0.005024 m2.

R = Resistance = 600 N = Number of spins = 500 B = Magnetic field = (0.0120)

t + (3 x 10⁻⁵) t⁴

The number t = 5 is substituted for taking the derivative at both the induced current and the electric current.

The Electric current is therefore 1.11* 10^{-4}A
Electric current - The rate of electron passage in a conductor is known as electric current. The ampere is the electric current's SI unit. Electrons are little particles that are part of a substance's molecular structure. These electrons can be held loosely or securely depending on the situation.

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The Venn diagram compares protons with electrons. Which shared property belongs in the region marked "B”?

Answers

The shared property of electron and proton that belongs in the region marked "B” is the presence of charge on both of them.

What charge is present on proton and electron?

We know that proton is positively charge particle while on the other hand, electron is negatively charge particle. Due to opposite charges, these particles attract each other.

So we can conclude that the shared property of electron and proton that belongs in the region marked "B” is the presence of charge on both of them.

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Answer:

electrically charged

Explanation:

got it right

The light beam shown in the figure below makes an angle of = 24.8° with the normal line NN' in the linseed oil. Determine the angles and '. (The refractive index for linseed oil is 1.48.)

Answers

The angle of refraction of the light beam is determined as  16.46 ⁰.

Angle of refraction of the light beam

n = sin i / sin r

where;'

n is refractive indexi is angle of incidencer is angle of refraction

Angle between the ray and the normal = incident angle = 24.8⁰

1.48 = sin (24.8) / (sin r)

sin r = sin (24.8) / (1.48)

sin r = 0.283

r = sin ⁻¹(0.283)

r = 16.46 ⁰

Thus, the angle of refraction of the light beam is determined as  16.46 ⁰.

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A 2.30 mH toroidal solenoid has an average radius of 6.20 cm and a cross-sectional area of 2.80 cm2.
a) How many coils does it have? In calculating the flux, assume that B is uniform across a cross section, neglect the variation of B with distance from the toroidal axis.
b) At what rate must the current through it change so that a potential difference of 2.60 V is developed across its ends?

Answers

(a) The number of turns of the coil is determined as 1,596 turns.

(b) The rate of change of current is determined as 1,130.43 A/s.

Number of turns of the solenoid

L = N²μA/l

where;

L is inductance N is number of turnsA is areal is average length = 2πr

N²μA = LI

N² = LI/μA

N² = (2.3 x 10⁻³ x 2π x 0.062)/(4π x 10⁻⁷ x 2.8 x 10⁻⁴)

N² = 2,546,428.6

N = √2,546,428.6

N ≈ 1,596 turns

Rate of current change

L = (emf)/I

I = (emf)/L

I = (2.6)/(2.3 x 10⁻³)

I = 1,130.43 A/s

Thus, the number of turns of the coil is determined as 1,596 turns.

The rate of change of current is determined as 1,130.43 A/s.

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Checking the what will reduce the possibility of having to rebuild or replaced the engine?

Answers

Checking the oil will reduce the possibility of having to rebuild or replaced the engine.

what is an engine ?

A device created to transform one or more sources of energy into mechanical energy is known as an engine or motor. Potential energy, heat energy, chemical energy, electric potential, and nuclear energy are all forms of energy that are readily available.

The lubricating function of engine oil is crucial. It shields and stops all the moving parts from rubbing against one another. Metal-on-metal wear would quickly kill your engine without lubrication.

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9.80 1. The density of mercury is 13600 kg/m³. What is this value in g/cm³? 2. Find the mass of water which will fit in a large tank measuring 2 m x 1 m x 20 cm. Density of water is 1000 kg/m³ or 1.0 g/cm³. 3. Find the volume of a lump of softwood whose mass is 120 g. Density of softwood is 0.6 gcm-3 or 600 kgm-³. -3​

Answers

Answer:

1. 13..6 grams per centimeters cubed.

2. Mass = 400kg

3. Volume = 200cm = 2m

Explanation:

1. The conversion for kg/m^3 to g/cm^3 is divide by 1000.

2. [tex]density=\frac{mass}{volume}[/tex]

[tex]1000=\frac{mass}{2 * 1 * 0.2}[/tex]

[tex]1000*0.4=mass[/tex]

[tex]400kg = mass[/tex]

3. [tex]density=\frac{mass}{volume}[/tex]

[tex]0.6=\frac{120}{volume}[/tex]

[tex]volume=\frac{120}{0.6}[/tex]

[tex]volume= 200cm[/tex]

Calculate the de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s (about 60 mi/h)

Answers

The de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s is 4.55×10⁻³⁵ m.

De Broglie wavelength:

The wavelength that is incorporated with the moving object and it has the relation with the momentum of that object and mass of that object. It is inversely proportional to the momentum of that moving object.

λ=h/p

Where, λ is the de Broglie wavelength, h is the Plank constant, p is the momentum of the moving object.

Whereas, p=mv, m is the mass of the object and v is the velocity of the moving object.

Therefore, λ=h/(mv)

λ=(6.63×10⁻³⁴)/(0.56×26)

λ=4.55×10⁻³⁵ m.

The de Broglie wavelength associated with the object weight 0.56 kg moving with the velocity of 26 m/s is λ=4.55×10⁻³⁵ m.

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In an experiment replicating Millikan’s oil drop experiment, a pair of parallel plates are placed 0.0200 m apart and the top plate is positive. When the potential difference across the plates is 240.0 V, an oil drop of mass 2.0 × 10-11 kg gets suspended between the plates. (e = 1.6 × 10-19 C)

a) Draw a free-body diagram for the charge.

b) What is the charge on the oil drop?

c) Is there an excess or deficit of electrons on the oil drop? How many electrons are in excess or deficit?

Answers

Answer: See below

Explanation:

Given:

The potential between plates, V = 240 V

Distance between plates, d = 0.02 m

The mass of drop, m = 2x10^-11

Charge on electron, e = 1.6x10^-19

Part (a)

The free-body diagram is attached below

Part (b)

The electric field is given by,

[tex]E=\frac{V}{d}[/tex]

On applying force balance, the force on oil drop is equal to the weight of the oil,

[tex]$$\begin{aligned}F_{E} &=m g \\q E &=m g \\q \frac{V}{d} &=m g \\q &=\frac{m g d}{V}\end{aligned}$$[/tex]

Substituting the given values in the above equation,

[tex]\begin{aligned}&q=\frac{2 \times 10^{-11} \mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{s}^{2} \times \frac{1 \mathrm{~N}}{1 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}} \times 0.02 \mathrm{~m}}{240 \mathrm{~V} \times \frac{1 \mathrm{~N} \cdot \mathrm{m} / \mathrm{C}}{1 \mathrm{~V}}} \\&q=1.63 \times 10^{-14} \mathrm{C}\end{aligned}[/tex]

Therefore, the charge on the oil drop is 1.63x10^-14 C

Part (c)

There will be an excess of electrons on the oil drop.

The number of electrons on oil drop can be calculated as,

[tex]\begin{aligned}q &=n e \\1.63 \times 10^{-14} \mathrm{C} &=n \times 1.6 \times 10^{-19} \mathrm{C} \\n &=1.01 \times 10^{5}\end{aligned}[/tex]

Therefore, the number of excess electrons is 1.01x10^5

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