A small object with mass 3.95 kg moves counterclockwise with constant speed 1.65 rad/s in a circle of radius 2.95 m centered at the origin. It starts at the point with position vector 2.95î m. Then it undergoes an angular displacement of 9.10 rad.
(a) What is its new position vector? _____ m
(b) In what quadrant is the particle located and what angle does its position vector make with the positive x-axis? ______ quadrant at _______°
(c) What is its velocity? _____ m/s
(d) In what direction is it moving? _____ ° from the +x direction
(e) What is its acceleration? _____ m/s2
(f) Make a sketch of its position, velocity, and acceleration vectors.

Answer:

Explanation:

the value is -8 cm

The initial speed of the rock as it left the astronaut's hand is 168 m/s.

The speed of the satellite is 60.2 m/s.

g = GM/R²

where;

g = (6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(6.32 x 10⁴)²

g = 0.132 m/s²

v² = u² - 2gh

0 = u² - 2gh

u² = 2gh

u = √2gh

u = √(2 x 9.8 x 1440)

u = 168 m/s

v = √GM/r

v = √[(6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(1.45 x 10⁵)]

v = 60.2  m/s

Thus, the initial speed of the rock as it left the astronaut's hand is 168 m/s.

The speed of the satellite is 60.2 m/s.

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The tension in the cable is 380.55 N and the vertical component of the force exerted by the hi.nge on the beam is 11.45 N.

Apply the principle of moment and calculate the tension in the cable;

Clockwise torque = TL sinθ

Anticlockwise torque = ¹/₂WL

TL sinθ  =  ¹/₂WL

T sinθ  =  ¹/₂W

T = (W)/(2 sinθ)

T = (40 x 9.8)/(2 x sin31)

T = 380.55 N

T + F = W

F = W - T

F = (9.8 x 40) - 380.55

F = 11.45 N

Thus, the tension in the cable is 380.55 N and the vertical component of the force exerted by the hi.nge on the beam is 11.45 N.

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Answer:

Projectile motion formula or equations derived (In Tabular format)Motion Path equation:

y = (tanθ) x – (1/2) g . x2/(V0 cosθ)2

Explanation:

An object with a velocity (v) of 9 m/s and a linear momentum (p) of 72 kg.m/s, has a mass (m) of 8 kg.

In Newtonian mechanics, linear momentum, or simply momentum, is the product of the mass and velocity of an object.

It is a vector quantity, possessing a magnitude and a direction.

The mathematical expression for momentum is:

p = m . v

where,

An object has a velocity (v) of 9 m/s and its linear momentum (p) is 72 kg.m/s. We will use the definition of linear momentum to calculate the mass of the object.

p = m . v

m = p / v

m = (72 kg.m/s) / (9 m/s) = 8 kg

An object with a velocity (v) of 9 m/s and a linear momentum (p) of 72 kg.m/s, has a mass (m) of 8 kg.

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The magnitude of the magnetic field and the direction of the magnetic field at point P is mathematically given as

B=1.9*10^{-5}T

To determine the magnetic field direction, use the right-hand -rule on the page magnetic field is going.

Generally, the equation for Biot savant law is  mathematically given as

[tex]B=(\frac{u}{4\pi}*{\frac{Ilsin\theta}{r^2})[/tex]

Their net field is

Bn=2B

[tex]Bn=2* B=(\frac{u}{4\pi}*{\frac{Ilsin\theta}{r^2})[/tex]

Hence

[tex]B=(\frac{4*\p *10^{-7}}{4\pi}*{\frac{(30)(2*10^{-3}sin45)}{\sqrt{(3*10^{-2})^2+((3*10^{-2})^2)}/2})[/tex]

B=1.9*10^{-5}T

In conclusion, To determine the magnetic field direction, use the right-hand -rule on the page magnetic field is going.

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Answer:

The magnitude of the magnetic field at the point P is [tex]1.57*10^{-5}T[/tex] and the field is pointing into the page.

Explanation:

The general form of a similar question to this is:

[tex]\vec{B} = \frac{\mu _{0} }{4\pi } * \oint \frac{Id\vec{l} \times \hat{r}}{r^{2} }[/tex]

where [tex]\vec{B}[/tex] is the vector of the Magnetic Field, [tex]\mu _{0}[/tex] is the Free Space Permeability Constant (equal to [tex]4\pi * 10^{-7} \frac{N}{A^2}[/tex]), [tex]I[/tex] is the current, and [tex]r[/tex] is the distance from the segment to the point P. (I will get to the [tex]d\vec{l} \times \hat{r}[/tex] term in a bit)

This equation is fairly complicated. Luckily, it can be simplified by looking at the magnitude and direction separately.

The first thing to simplify is the cross product. Due to the fact that a cross product can be simplified from [tex]\vec{x} \times \vec{y}[/tex] to [tex]xy\sin(\theta)[/tex], where [tex]\theta[/tex] is the angle between the 2 vectors, and [tex]\hat{r}[/tex] is the unit vector of [tex]r[/tex] (i.e. [tex]\hat{r} = \vec{r}/r[/tex]) we can simplify [tex]d\vec{l} \times \hat{r}[/tex] to just [tex]dl \sin(\theta)[/tex].

Next, we will look at the integral. In this scenario, everything will function as a constant, so we can essentially ignore it.

Finally, [tex]\frac{\mu_{0}}{4\pi}[/tex] simplifies down to [tex]10^{-7}[/tex].

This gives us our new equation for the Magnetic Field produced by a single segment at a point:

[tex]B = \frac{Il\sin\theta}{r^{2}}*10^{-7}[/tex]

Now we need to find values for [tex]r[/tex] and [tex]\theta[/tex]. Luckily, we are dealing with a 45-45-90 triangle with sides of [tex]1.5 cm[/tex]. This means the distance [tex]r[/tex] is [tex](1.5\sqrt2)cm[/tex]! Similarly, because it is a 45-45-90 triangle, our [tex]\theta[/tex] is [tex]45\textdegree[/tex]!

Now we can start plugging things in:

[tex]B = \frac{(25A)(2*10^{-3}m)\sin(45\textdegree)}{(1.5\sqrt2*10^{-2}m)^2}*10^{-7}\frac{N}{A^2}[/tex]

[tex]B = 7.86^{-6} \frac{N}{A}[/tex] or [tex]T[/tex]

This is the magnitude due to only one single segment. In order to find the total field, we need to know the direction of the field due to each segment.

Finding the direction is really easy. Just use the right hand rule. Point your thumb in the direction of the current and curl the rest of your fingers around an imaginary pole. The direction your fingers point is the direction of the field. In this case, the field lines due to the segments point into the page in the 4th quadrant (the origin is the bend). This means that at point P, both segments induce the same field in the same direction. Therefore, we can take our value from before and double it, giving us our final answer:

[tex]B = 1.57*10^{-5} T[/tex]; into the page.

The altitude of the pole vaulter as she crosses the bar is 5 m.

v² = u² - 2gh

where;

h = (u² - v²)/2g

h = (10² - 1.5²)/(2 x 9.8)

h = 5 m

Thus, the altitude of the pole vaulter as she crosses the bar is 5 m.

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The kinetic energy of the projectile at the highest point of its motion will be E/4.

When a projectile will be thrown obliquely near the surface of the earth, it travels a curved path under uniform acceleration directed toward the center of the Earth. The path of a particle is called a projectile while the motion of a projectile is projectile motion.

Given, the angle of projection  with horizontal, [tex]\theta = 60 ^o[/tex]

Consider that 'E' is the initial value of the kinetic energy of the projectile.

The equation for the initial kinetic energy is : [tex]E =\frac{1}{2}mu^2[/tex]

where m is the mass of the given projectile.

The component of the velocity of the projectile in the horizontal direction:

uₓ = u cosθ

uₓ = u cos 60°

uₓ = u/2

From the equation of motion: v = u +at

v = (u/2) + (0) t

v = u/2

The final kinetic energy of the projectile:

[tex]E_f = \frac{1}{2}mv^2[/tex]

[tex]E_f = \frac{1}{2}m(\frac{u}{2} )^2[/tex]

[tex]E_f = \frac{1}{4} (\frac{1}{2}mu^2 )[/tex]

[tex]E_f = \frac{E}{4}[/tex]

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The speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.

v = √GM/r

where;

v = √[(6.626 x 10⁻¹¹ x 5.97 x 10²⁴) / ((6371 + 4930) x 10³)]

v = 5,916.36 m/s

Thus, the speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.

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The percent difference between two numbers [tex]x[/tex] and [tex]y[/tex] is given by

[tex]\dfrac{|y-x|}x \times 100\%[/tex]

The absolute value is there because we only care about the absolute percent difference, and not taking into account whether we go from [tex]x[/tex] to [tex]y[/tex] or vice versa. If we remove them, we have two possible interpretations of percent difference.

For example, the (absolute) percent difference between 3 and 6 is

[tex]\dfrac{|6-3|}3 \times 100\% = 100\%[/tex]

In other words, we add 100% of 3 to 3 to end up with 6. This is the same as the percent difference going from 3 to 6. On the other hand, the percent difference going from 6 to 3 is

[tex]\dfrac{3-6}3\times100\%=-50\%[/tex]

which is to say, we take away 50% of 6 away from 6 to end up with 3.

"Make comparisons to object measurements" tells us that the differences should be computed relative to "measurements for object". In other words, take [tex]x[/tex] from the left column and [tex]y[/tex] from the right column.

[tex]\dfrac{|7.1-7.3|}{7.1} \times 100\% \approx 2.82\%[/tex]

[tex]\dfrac{|4.8-5.0|}{4.8} \times 100\% \approx 4.17\%[/tex]

[tex]\dfrac{|7.2-7.5|}{7.2} \times 100\% \approx 4.17\%[/tex]

The percentage difference are 2.82%, 4.17%, 4.17%.

The percent difference between two numbers a and b is given by formula:

[tex]\frac{y-x}{x}*100[/tex]

We are given absolute value as we are only concerned about the absolute percent difference.

Making comparisons to object measurements determines that the differences should be computed relative to object measurements.

Here take from the left column and  from the right column.

a) [tex]\frac{7.3-7.1}{7.1} *100=2.82[/tex]%

b) [tex]\frac{5.0-4.8}{4.8} *100=4.17[/tex]%

c) [tex]\frac{7.5-7.2}{7.2} *100=4.17[/tex]%

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The force F, necessary to support an object with a mass of 991 kg placed on piston 2. is 20.61J.

It should be noted that by Pascal Law, the pressure on piston 1 will have the same value as the pressure on piston 2.

This will be:

(991 × 10) /(π × 0.0946/2)²

= 9910/0.022

= 450454.6 Pa

F1 = A1 × 450454.6 = 3.14 × (0.0187/2)² × 450454.6

= 123.64

F = 123.64/6

F = 20.61

Therefore, the force F, necessary to support an object with a mass of 991 kg placed on piston 2. is 20.61J.

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The magnitude of the tension in the string marked A and B is mathematically given as

A = 52.5 N

Generally, the equation for is mathematically given as

tan=3/8

negative x

[tex]B= tan\phi \\tan \phi=5/4[/tex]

negative x

C= tan=1/6

Hence, considering the Horizontal components

74.9cos(9.46) = A*cos(20.6) + B*cos(51.3)

A = 78.9 - 0.668B

Vertical components

74.9*sin(9.46) + Asin(20.6) = Bsin(51.3)

40.07 = 1.015B

B = 39.5 N

In conclusion, Sub the value of B is the equation of A

A = 78.9 - 0.668B

Sub

A = 78.9 - 0.668( 39.5 N)

A = 52.5 N

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Answer: The rock

Explanation:

Answer:

80 kgm/s

Explanation:

Momentum = Mass x Velocity

It can be expressed as [tex]\displaystyle{p = mv}[/tex] where p is momentum, m is mass and v is velocity.

We know that mass is 10 kg and velocity is 8 m/s - therefore, substitute the given information in formula:

[tex]\displaystyle{p=10 \ \times \ 8}\\\\\displaystyle{p=80 \ \ kgm/s}[/tex]

Hence, the momentum is 80 kgm/s.

From the calculations, the power developed is 337.5 W.

First we must obtain the acceleration from;

v = u + at

u = 0 m/s because the motion started rom rest

v = at

a = v/t

a = 12 m/s/ 1.6 s

a= 7.5 m/s^2

The force is obtained from;

F = ma = 7.5 kg * 7.5 m/s^2 = 56.25 N

Now the distance covered is obtained from';

v^2 = u^2 + 2as

v^2 = 2as

s = v^2/2a

s = (12)^2/2 *  7.5

s = 9.6 m

Now;

Work = Fs = 56.25 N *  9.6 m = 540 J

Power expended =  540 J/ 1.6 s = 337.5 W

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5.77 ×[tex]10^1^4[/tex] Hz is the green photon's frequency .

The distance between similar points (adjacent crests) in adjacent cycles of a waveform signal that is propagated in space is known as the wavelength. A wave's wavelength is often measured in meters (m), centimeters (cm), or millimeters (mm) (mm). The relationship between frequency and wavelength is inverse.

Wavelength of green light = 520 nm

f = c / λ

where, f = Frequency

            c = Speed of light = 3 × [tex]10^8[/tex] m/s

            λ = Wavelength of light

∴ f = c / λ

  f = [tex]\frac{3*10^8}{520 * 10^-^9}[/tex]

    = 5.77 ×[tex]10^1^4[/tex] Hz

Therefore,  5.77 ×[tex]10^1^4[/tex] Hz is the green photon's frequency .

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Answer:

VR = Velocity of effort / Velocity of Load

Explanation:

The ratio of the out-lever to the in-lever (length of the resistance arm to effort arm) is known as the velocity ratio (VR). It indicates that the distance covered by effort is four times that covered by the load.

The load-to-effort ratio of a machine, or alternatively, the output-to-input ratio of a machine, is known as its mechanical advantage. Another definition of velocity ratio is the ratio of the velocity of effort to the velocity of load.

When a machine has a force ratio of 4 and a velocity ratio of 4, this indicates that the weight moved is multiplied by 4 and the distance moved by the effort is multiplied by 4 at the same time.

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"V = IR" is the equation that relates current to voltage.

Relationship between current and voltage:

                                           V = IR or,

                                           I = V/R,

        the amount of current (I) flowing through a circuit is inversely    

        related to the amount of resistance (r) and directly proportional to  

        the voltage (v).

Therefore the correct answer is option C i.e., "V = IR".

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(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 24.67 N, opposite as the motion of the crate.

(b) The net work done on the crate while it is on the rough surface is -16.04 J.

(c) The speed of the crate when it reaches the end of the rough surface is 0.65 m/s.

F(net) = F - μFf

F(net) = 299 - 0.351(92 x 9.8)

F(net) = -24.67 N

W = F(net) x L

W = -24.67 x 0.65

W = - 16.04 J

a = F(net)/m

a = -24.67/92

a = - 0.268 m/s²

v² = u² + 2as

v² = 0.88² + 2(-0.268)(0.65)

v² = 0.426

v = √0.426

v = 0.65 m/s

Thus, the magnitude and direction of the net force on the crate while it is on the rough surface is 24.67 N, opposite as the motion of the crate.

The net work done on the crate while it is on the rough surface is -16.04 J.

The speed of the crate when it reaches the end of the rough surface is 0.65 m/s.

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The minimum work needed to pus the cart up the inclined plane is 960000 J.

The work done on a inclined plane is given below as:

Distance = 700 m

The force on an inclined plane, F = mgsinθ

where;

m is mass in kg

g = 9.81 m/s²

θ = 8.5°

Work done = 950 * 9.81 * sin 8.5 * 700

Work done = 960000 J

Therefore, minimum work needed is 960000 J.

In conclusion, the work done is a product of force and distance.

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The acceleration due to gravity near the surface of the planet is 27.38 m/s².

g = GM/R²

where;

g = (6.626 x 10⁻¹¹ x 2.81 x 5.97 x 10²⁴) / (6371 x 10³)²

g = 27.38 m/s²

Thus, the acceleration due to gravity near the surface of the planet is 27.38 m/s².

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The minimum diameter of the steel wire is 7.58 mm.

Young's modulus = stress/strain

strain = e/l = (7.32 x 10⁻³ m) / (19.3 m) = 3.79 x 10⁻⁴

stress = Young's modulus  x  strain

stress = 200 x 10⁹ N/m²   x 3.79 x 10⁻⁴  = 7.59 x 10⁷ N/m²

stress = Force/Area

Area = Force/stress

Area = mg/stress

Area = (350 x 9.8) / (7.59 x 10⁷)

Area = 4.519 x 10⁻⁵ m²

Area = πd²/₄

πd²/₄ = 4.519 x 10⁻⁵ m²

πd² = 4(4.519 x 10⁻⁵)

d² = (4 x 4.519 x 10⁻⁵)/π

d² = 5.75 x 10⁻⁵

d = √(5.75 x 10⁻⁵)

d = 7.58 x 10⁻³ m

d = 7.58 mm

Thus, the minimum diameter of the steel wire is 7.58 mm.

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The horizontal distance XQ traveled by the bomb is 250 m.

Let the XQ be the horizontal distance traveled by the bomb.

h = vt + ¹/₂gt

Let the vertical velocity = 0

h = 0 + ¹/₂gt²

h = ¹/₂gt²

2h = gt²

t² = 2h/g

t = √(2h/g)

t = √(2 x 100 / 9.8)

t = 4.5 s

XQ = vx(t)

where;

vx is horizontal speed, = 200 km/hr = 55.56 m/s

XQ = 55.56 x 4.5

XQ = 250 m

Thus, the horizontal distance XQ traveled by the bomb is 250 m.

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The angle between the two beams inside the glass is mathematically given as

<(BR)= 0.563°

Generally, the equation for an angle for blue is  mathematically given as

[tex]< B= arcsin( sin\theta / i )[/tex]

Therefore

<B= arcsin( sin30 / 1.660 )

<B= 17.53°

For angle for red

<R= arcsin(sin30 / 1.610)

<R= 18.09°

For the angle in between

<(BR)= 18.093 - 17.530

<(BR)= 0.563°

In conclusion,  the angle between the two beams inside the glass

<(BR)= 0.563°

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The initial speed of the sled at the given height is 4.85 m/s.

Apply the principle of conservation of energy;

K.E = P.E

¹/₂mv² = mgh

v² = 2gh

v = √2gh

where;

v = √(2 x 9.8 x 1.2)

v = 4.85 m/s

Thus, the initial speed of the sled at the given height is 4.85 m/s.

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(A) The spring constant of the spring is 4.94 N.

(B) The speed of the cart after pushing it is 0.47 m/s.

(C) The force applied to the cart is 0.75 N.

ω = √k/m

where;

0.5 rev/s = 0.5(2π) rad/s = π rad/s = 3.142 rad/s

ω² = k/m

k = mω²

k = 0.5 x (3.142)²

k = 4.94 N/m

E = ¹/₂kA²

where;

A is amplitude

E = ¹/₂(4.94)(0.15)²

E = 0.056 J

E = ¹/₂mv²

2E = mv²

v² = 2E/m

v² = (2 x 0.056)/(0.5)

v² = 0.224

v = √0.224

v = 0.47 m/s

E = ¹/₂FA

2E = FA

F = 2E/A

F = (2 x 0.056)/(0.15)

F = 0.75 N

Thus, the spring constant of the spring is 4.94 N. The speed of the cart after pushing it is 0.47 m/s. The force applied to the cart is 0.75 N.

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The behavior of water when subjected to temperature change is that the volume of water will increase if cooled from 3° to 2°C.

The chemical compound water, which can exist in the gaseous, liquid, and solid phases, is made up of the elements hydrogen and oxygen in the ratio 2: 1 i.e. 2 atoms of hydrogen and 1 atom of oxygen.

In general

Volume of water depends on the temperature and is directly proportional to it.

Thus, as the temperature rises, the molecules of water gain energy and move more quickly, which causes the molecules to spread apart and increase the volume of the liquid.

When water cools, it initially contracts (decreases in volume) until a temperature of about four degrees Celsius (4°C).

But at temperatures below 4.0° C, water undergoes an abnormal expansion that causes its volume to start to rise.

This ability is related to the formation of hexagonal structures, which take up a lot of room and increase the volume of the water, as a result of strong hydrogen bonding between water molecules at a lower temperature.

Hence, the volume of water will increase if cooled from 3° to 2°C

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The amount of heat needed to raise the temperature of 6.00 g of copper by 15.0°C is 34.65J (option A).

The amount of heat absorbed or released by a substance can be calculated using the following formula:

Q = mc∆T

Where;

Q = 6 × 0.385 × 15

Q = 90 × 0.385

Q = 34.65J

Therefore, the amount of heat needed to raise the temperature of 6.00 g of copper by 15.0°C is 34.65J.

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The minimum value of the coefficient of static friction is equal to 0.92.

First of all, we would derive an expression for the horizontal and vertical component of forces acting acting on the car.

For the vertical component, we have:

∑Fy = 0

Fn + Fg = 0

Fn - mg = 0

Fn = mg     .....equation 1.

For the horizontal component, we have:

∑Fx = mAc

uFn = m(V²/r)    .....equation 2.

Substituting eqn. 1 into eqn. 2, we have:

umg = m(V²/r)

u = 1/g(V²/r)

u = (V²/gr)

u = (30²/9.8 × 100)

u = 900/980

u = 0.92.

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Since upward forces must be equal to the downward forces, the tension in the cable is 225.3 N

The two conditions are;

The given parameters are;

Tsinθ = mg

Tsin 66 = 21 x 9.8

T = 205.8 / 0.914

T = 225.3 N

Therefore, the tension in the cable is 225.3 N

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