Answer:
A. True
Explanation:
Water exists in the air in its gaseous form. If this gaseous water is cooled sufficiently enough below its dew point, it will condense into liquid water. This is what happens to the body of a glass filled with cold water. The glass's temperature drops the temperature of the air immediately around it, cooling it below dew point temperature. This causes water to condense on the body of the glass.
Name at least three fundamental differences between the harmonic oscillator dynamics and the simple pendulum dynamics
Answer: please find the answer in the explanation.
Explanation:
Harmonic can be experienced by any body that repeats itself. The pattern can be sinusoidal, square, tooth etc.
The fundamental differences between the harmonic oscillator dynamics and the simple pendulum dynamics are:
1.) The harmonic oscillator dynamics can be sinusoidal or square wave so far the motion is periodic while the simple pendulum dynamics is always sinusoidal.
2.) In simple pendulum dynamics, the period of oscillation is independent of the amplitude. While the period in harmonic oscillator dynamics depends on the amplitude.
3.) Differential equation is only one method to analyze the simple pendulum dynamics where there are several methods to analyze the harmonic oscillator dynamics.
If we put negative charge between two similar positive charges then what is it's equilibrium? And how?
Your question has been heard loud and clear.
Well it depends on the magnitude of charges. Generally , when both positive charges have the same magnitude , their equilibrium point is towards the centre joining the two charges. But if magnitude of one positive charge is higher than the other , then the equilibrium point will be towards the charge having lesser magnitude.
Now , a negative charge is placed in between the two positive charges. So , if both positive charges have same magnitude , they both pull the negative charge towards each other with an equal force. Thus the equilibrium point will be where the negative charge is placed because , both forces are equal , and opposite , so they cancel out each other at the point where the negative charge is placed. However if they are of different magnitudes , then the equilibrium point will be shifted towards the positive charge having less magnitude.
Thank you
Answer:
it will be positive
Explanation:
Estimate the number of dollar bills (15.5 cm wide), placed end to end, that it would take to circle the Earth (radius = 6.40 × 103 km).
Answer:
#_bills = 2.59 10⁸ Bills
Explanation:
For this exercise we must calculate the length of the Earth's circle
L = 2π r
Let's reduce to the SI system
r = 6.40 10³ km (1000m / 1km) = 6.40 10⁶ m
d = 15.5 cm (1m / 100cm) = 0.155 m
L = 2π 6.40 10⁶
L = 40.21 10⁶ m
now we can use a direct rule of proportions (rule of three). If 1 bill has d = 0.155, how much bill will have L
#_bills = 1 L / d
#_bills = 1 40.21 10⁶6 / 0.155
#_bills = 2.59 10⁸ Bills
the total number of bills is 2.6 10⁸ bills
There are three long parallel wires arranged so that, in cross-section, they occupy the points of an equilateral triangle. Is there any way to arrange the currents so that all three wires attract each other?
Answer:
Yes, there is such a way.
Explanation:
If currents flow in the same direction in two or more long parallel wires, there will be an attractive force between the wires. If the current flows in different directions, there will be a repulsive force between the wires. In this case, these three parallel wires, can be be made to carry current in the same direction, creating an attractive force between all three wires.
Note that it is not possible to have at the least one of them carry current in the opposite direction and still have an attractive current between them.
A fish in an aquarium with flat sides looks out at a hungry cat. To the fish, the distance to the cat appears to be
Answer:
flat
Explanation:
ohms law? explanation.
Answer:
[tex] \boxed{ \bold{ \sf{see \: below}}}[/tex]
Explanation:
[tex] \boxed{ \bold{ \underline{ \huge{ \boxed{ \sf{ohm's \: law}}}}}}[/tex]
The relation between current through a metallic conductor and potential difference across its ends was studied systematically by a German physicist , George Simon Ohm in 1826 AD . This relation is now known as Ohm's law. It states that the electric current passing through a conductor is directly proportional to the potential difference across its two ends at a constant physical condition [ Temperature , cross - sectional area , length , nature of material etc ]
Hope I helped!
Best regards!!
[9]
(b) A billiard ball B at rest is struck by an identical ball A moving with a speed of 10 m/s along
the positive y-axis. After the collision, the two balls move off in directions that are perpendicular
to each other, with ball A making 15° with the positive y-axis. Find the final velocities (Direction
and magnitude) of both balls. [16]
Answer:
9.66 m/s 15° with +y
2.59 m/s 75° with +y
Explanation:
Momentum is conserved in the y direction.
mu₁ + mu₂ = mv₁ + mv₂
u₁ + u₂ = v₁ + v₂
10 m/s + 0 m/s = v₁ cos 15° + v₂ cos 75°
10 = v₁ cos 15° + v₂ cos 75°
Momentum is conserved in the x direction.
mu₁ + mu₂ = mv₁ + mv₂
u₁ + u₂ = v₁ + v₂
0 m/s + 0 m/s = v₁ sin 15° − v₂ sin 75°
0 = v₁ sin 15° − v₂ sin 75°
v₁ sin 15° = v₂ sin 75°
v₂ = v₁ sin 15° / sin 75°
Substitute.
10 = v₁ cos 15° + (v₁ sin 15° / sin 75°) cos 75°
10 = v₁ cos 15° + v₁ sin 15° / tan 75°
10 = v₁ (cos 15° + sin 15° / tan 75°)
v₁ ≈ 9.66 m/s
v₂ ≈ 2.59 m/s
Sound waves reaching the olfactory mucosa do not elicit olfactory perception due to the ____________ not matching the type of stimulus of the receptor.
Answer:
Sound waves reaching the olfactory mucosa do not elicit olfactory perception due to the ___MODALITY_________ not matching the type of stimulus of the receptos
Explanation:
Because modality of sensation refers to what is perceived after the stimulus is effected eg the pressure modality is effected when the pressure receptors of the skin are stimulated so in this case sound will not elite olfactory sensation because the modality here is sound rather than smell
Since the noble gases are stable (VERY unreactive), oxygen would like to look like one in terms of the number and arrangement of its electrons. What would the atom of oxygen have to do to its electrons in order to look like neon?
Answer:
oxygen gains two electron to attain electron configuration like neon. Oe-²
Explanation:
The phase velocity of transverse waves in a crystal of atomic separation a is given byy = csin(ka/2) pka/2 1. What is the dispersion relation e(k)? 2. What is the group velocity as a function of k?
Answer:
a
e(k) = \frac{2a}{c} * sin (\frac{k*a}{2} )
b
G_{v} = \frac{d e(k ) }{dk } = \frac{a^2}{c} * cos (\frac{k* a}{2} )
Explanation:
From the question we are told that
The velocity of transverse waves in a crystal of atomic separation is
[tex]b_y = c \frac{sin (\frac{k*a}{2} )}{\frac{k*a}{2} }[/tex]
Generally the dispersion relation is mathematically represented as
[tex]e(k) = b_y * k[/tex]
=> [tex]e(k) = c \frac{sin(\frac{k*a}{2} ) }{ \frac{k*a}{2} } * k[/tex]
=> [tex]e(k) = c * \frac{sin (\frac{k_a}{2} )}{ \frac{a}{2} }[/tex]
=> [tex]e(k) = \frac{2a}{c} * sin (\frac{k*a}{2} )[/tex]
Generally the group velocity is mathematically represented as
[tex]G_{v} = \frac{d e(k ) }{dk } = \frac{a^2}{c} * cos (\frac{k* a}{2} ) [/tex]
IV) Fill in the blanks:
1) Equal volumes of different substances have
masses.
2) The more closely packed arrangement the particles of a substance have,
its density.
3) SI unit of power is
4) An iron nail sinks in water but can float on
5) Balloons used for advertisements are filled with_gas.
6)
is the primary mode of heat transfer in liquid and gases.
1.) equal volume of different substances have "different" masses.
2.)The more closely packed arrangement the particles of a substance have, "increases" its density.
3.)the SI unit of power is "Watts".
4.)an iron nail sinks in water but floats on " mercury ".
5.)balloons used for advertisements are filled with " helium" gas.
6.)"Conduction" is the primary mode of heat transfer in liquid and gases.
I hope this helps you...
If you run at a constant speed of 7.5 m/s, how long does it take for you to finish a 100 m race ? show your work
Answer:
13s
Explanation:
(7.5m/s)(t)=(100m)
t=(100/7.5)s=13s
14. A 5,000 kg train is traveling at a velocity of 100 m/s and hits another train. The two trains stick together, and the new velocity is 50 m/s. What is the mass of the second
train?
A. 15.000 kg
B. 8.000 kg
C. 5,000 kg
D. 10.000 kg
Please help ASAP
Answer:
C
Explanation:
Momentum of first train is = m * v= (5000 kg)·(100 m/s)
= 500,000 kg·m/s.
Momentum after collision is = (5000+M) kg)·(50 m/s) And using conservation of momentum5000+M) kg)·(50 m/s) = 500,000 kg·m/s
Dividing by 50 m/s and subtracting 5000 kg, we have ...
(5000 +M) kg = 10,000 kg
M kg = 5000 kg
Which scenario is an example of a physical change?
Wood is burned in a campfire and is turned into ash and smoke.
A piece of metal is left in the rain and forms rust.
Lead is melted into a liquid to form pellets.
Yeast turns sugar into energy, water, and carbon dioxide gas.
Answer:
Lead is melted into a liquid to form pellets.
Answer:
C. lead is melted into a liquid to form pellets
Explanation:
7. When initially-unpolarized light passes through three polarizing filters, each oriented at 45 degree angles from the preceding filter, _____ of the initial light passes through the last filter.
Answer:
Explanation:
Let the intensity of unpolarised light be I₀ . After passing through the first polarising filter , the intensity is I₀ / 2 .
After second filter , the intensity will be I₀ / 2 x cos²45 = I₀ / 4
After third filter , the intensity will be I₀ / 4 x cos²45 = I₀ / 8 .
So,
1 / 8 the of initial light passes through the last filter .
__________ have large spaces between high energy particles.
Ions
Liquids
Gases
Solids
Answer:
the answer should be gasses
a planet of mass 2*10^24 kg is at location <2*10^11,-4*10^11,0>m. A star of mass 5*10^30 kg is at location <-5*10^11,6*10^11,0>m. The planet is in a circular orbit around the star, moving at a constant speed. What is the gravitational force vector on the planet due to the star
Answer:
(-2.57*10^20, 3.67*10^20, 0)
Explanation:
Refer to the attachment.
I hope that helps
Q.Solve the following circuit find total resistance RT. Also find value of voltage across resister RC.
Answer:
14.57 ohms
Explanation:
Here in the figure ,Rb & R₄are in series & also Rc & R₅ are in series. As they are in series , ( Rb + R₄ ) & (Rc & R₅) are in parallel . So the equivalent resistance in that branch = ( 2 + 18 ) ║ ( 3 + 12 )
= 20 ║ 15
= (20×15) / (20 + 15)
= 8.57 ohms
Also Ra ( 6 ohm ) is in series with that branch ,. So the equivalent resistance of the whole circuit = 8.57 + 6 = 14.57 ohms.
total resistance is 26 ohm and voltage across [tex]R_C[/tex] resister is 1.1554 Ampere.
According to the circuit,
[tex]R_A[/tex] = 6 Ω.
[tex]R_B[/tex] = 2 Ω.
[tex]R_C[/tex] = 3 Ω.
R₄ = 18 Ω.
R₅ = 12 Ω.
V = 10 V.
In the circuit, the series combination of [tex]R_C[/tex] and R₅ is parallel connected with series combination of [tex]R_B[/tex] and R₄. And this network is connected with [tex]R_A[/tex] in series combination.
So, Equivalent resistance of the circuit is; R = [tex]R_A[/tex] + ( [tex]R_C[/tex] + R₅) II ([tex]R_B[/tex] +R₄)
= 18 Ω + ( 3 Ω + 12 Ω)II(2 Ω+ 18 Ω)
= 18 Ω + 15ΩII20Ω
= 18 Ω + (15×20)/(15+20)Ω
= 18 Ω + 8.57 Ω
= 26 Ω
So, current flowing through the circuit: I = V/ R = 10V / 26Ω = 0.38 Amp.
Voltage drop across the parallel network; V₁ = V - I[tex]R_A[/tex]
= 10V - 0.38×6 V
= 7.72 V.
So, voltage across the resister [tex]R_C[/tex] = V₁[tex]R_C[/tex]/[tex](R_C[/tex] + R₅)
= 7.72 × 3/(3 + 12) Amp
= 1.1544 Amp.
Hence, total resistance is 26 ohm and voltage across specific resister is 1.1554 Ampere.
Learn more about circuit here:
https://brainly.com/question/21505732
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hydraulic lift is to be used to lift a 2100-kg weight by putting a weight of 25 kg on a piston with a diameter of 10 cm. Determine the diameter of the piston on which the weight is to be placed.
Answer:
840 cm
Explanation:
Note: A hydraulic press operate based on pascal's principle.
From pascal's principle
W₁/d₁ = W₂/d₂...................... Equation 1
Where W₁ and W₂ are the first and second weight, and d₁ and d₂ are the first and second diameter of the piston.
make d₁ the subject of the equation
d₁ = W₁×d₂/W₂................ Equation 2
Given: W₁ = 2100 kg, W₂ = 25 kg, d₂ = 10 cm = 0.1 m.
Substitute these values into equation 2
d₁ = 2100(0.1)/25
d₁ = 8.4 m
d₁ = 840 cm
A child of mass 22.0 kg is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force must she experience to stay on if she is 1.25 m from the center
Answer:
Centripetal Force = 483.3 N
Explanation:
A centripetal force is the force that tends to keep a mocing object along a curved path and it is directed towards the centre of the rotatio, while centrifugal force is an apparent force that tends to force a rotating object away from the center of the rotation.
The formula for centripetal force is given by:
[tex]F_c = \frac{mv^2}{r} \\where:\\F_C = centripetal\ force\\m = mass\ = 22kg\\\omega =angular\ velocity = 40.0\ rev/min[/tex]
Let us work on the angular velocity (ω), by converting to radians/ seconds
ω = 40 rev/min,
1 rev = 2π rad
∴ 40 rev = 2π × 40 rad = 80π rad
1 min = 60 seconds
[tex]\therefore\ 40\ rev \slash min = \frac{80\ \times\ \pi\ rad}{60\ seconds} \\40\ rev \slash min = 4.189\ rad \slash sec[/tex]
Next let us find the velocity (v) from the angular velocity. Velocity (v) and angulsr velocity (ω) are related by the equation:
v = ω × r (m/s)
v = 4.189 × 1.25
v = 5.24 m/s
Finally, the centripetal force is calculated thus:
[tex]F_c = \frac{mv^2}{r} \\\\F_c = \frac{22 \times (5.24)^2 }{1.25} \\\\F_c = \frac{604.07}{1.25}\\ F_c = 483.3N[/tex]
Two positive point charges, each with charge q, separated by a distance d, repel each other with a force of magnitude 20 N. What is the magnitude of the force between two positive point charges of magnitude 11.11 q, separated by a distance 2.5 d in units of N
Answer:
The value is [tex]F_2 = 395 \ N[/tex]
Explanation:
From the question we are told that
The magnitude of the charge of each positive charge for the first case is [tex]q_1 = q_2 = q[/tex]
The distance between the charges for the first case is [tex]d[/tex]
The force between the charges for the first case is [tex]F = 20 \ N[/tex]
The magnitude of the charge of each positive charge for the second case is [tex]q_1 = q_2=11.11q[/tex]
The distance between the charge for the second case is [tex]2.5d[/tex]
Generally for the first case the force between the charge is mathematically represented as
[tex]F_1 = \frac{k * q^2 }{d^2}[/tex]
Where k is the Coulomb constant with value [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
So
[tex]20= \frac{k * q^2 }{d^2}[/tex]
Generally for the second case the force between the charge is mathematically represented as
[tex]F_2 = \frac{k * (11.11q)^2}{(2.5d)^2}[/tex]
[tex]F_2 = \frac{k * 11.11^2 *q^2}{2.5^2d^2}[/tex]
=> [tex]F_2 = \frac{11.11^2}{2.5^2} F_1[/tex]
=> [tex]F_2 = \frac{11.11^2}{2.5^2} * 20[/tex]
=> [tex]F_2 = 395 \ N[/tex]
A rectangular wooden block, 22 cm x 13.2 cm x 4.4 cm, has a mass of 1562.0 g. The density of the wood in kg/m 3 is:
Answer:
1222.45 Kg/m³
Explanation:
The following data were obtained from the question:
Dimension of rectangular block = 22 cm x 13.2 cm x 4.4 cm
Mass (m) = 1562.0 g
Density (D) =?
Next, we shall determine the volume of the rectangular block.
This is illustrated below:
Volume = length x breadth x height
V = 22 cm x 13.2 cm x 4.4 cm
V = 1277.76 cm³
Next, we shall convert 1277.76 cm³ to m³.
This can be obtained as follow:
1 cm³ = 1×10¯⁶ m³
Therefore,
1277.76 cm³ = 1277.76 × 1×10¯⁶
1277.76 cm³ = 1.27776×10¯³ m³
Therefore, 1277.76 cm³ is equivalent to 1.27776×10¯³ m³.
Next, we shall convert 1562.0 g to kg. This can be obtained as follow:
1000 g = 1 kg
Therefore,
1562.0 g = 1562/1000 = 1.562 kg
Therefore, 1562.0 g is equivalent to 1.562 Kg.
Finally, we shall determine the density of rectangular block as follow:
Volume = 1.27776×10¯³ m³.
Mass = 1.562 Kg.
Density =?
Density = mass /volume
Density = 1.562 / 1.27776×10¯³
Density = 1222.45 Kg/m³
Therefore, the density of rectangular block is 1222.45 Kg/m³
Calculate the reactance of a 0.5 F capacitor that is connected to a battery with peak voltage 2 V and angular frequency 200 radians/s.
Answer:
Reactance of the capacitor is 0.01 ohms.
Explanation:
It is given that,
Capacitance of the capacitor is 0.5 F
Peak voltage of the battery is 2 V
Angular frequency is 200 rad/s
We need to find the capacitive reactance. The capacitive reactance is due to the capacitor in the circuit. It is given by :
[tex]X_c=\dfrac{1}{\omega C}\\\\X_c=\dfrac{1}{200\times 0.5}\\\\X_c=0.01\ \Omega[/tex]
So, the reactance is 0.01 ohms.
What is the direction of the force? Answer with a positive angle measured counterclockwise from the positive x axis.
Answer:
hello the diagram related to your question is missing attached below is the missing diagram
answer : 107.07⁰
Explanation:
The direction of the force can be calculated as below
angle = 180 - ∝ - ∅ ------ 1
∅ = [tex]tan^{-1} b/a[/tex] = tan^-1 ( 0.429 )
∅ = 23.2⁰
∝ can be determined by applying the parallelogram law
∝ = [tex]tan^{-1} ( \frac{R24sin56.4}{f13 + R24cos56.4} )[/tex]
= tan^-1 ( 21.173 / 17.935 )
∝ = 49.73⁰
back to equation 1 direction of force with positive angle
= 180 - 49.73 - 23.2 = 107.07⁰
Find the magnitude of u × v and the unit vector parrallel to u × v in the direction of u × v. u = 2i + 2j - k, v = -i + k
Answer:
magnitude = 3
unit vector = [tex]\frac{2i}{3} - \frac{j}{3} - \frac{2k}{3}[/tex]
Explanation:
Given vectors:
u = 2i + 2j - k
v = -i + k = -i + 0j + k
(a) u x v is the cross product of u and v, and is given by;
[tex]u X v = \left[\begin{array}{ccc}i&j&k\\2&2&-1\\-1&0&1\end{array}\right][/tex]
u x v = i(2+0) - j(2 - 1) + k(0 - 2)
u x v = 2i - j - 2k
Now the magnitude of u x v is calculated as follows:
| u x v | = [tex]\sqrt{2^2 + (-1)^2 + (-2)^2}[/tex]
| u x v | = [tex]\sqrt{4 + 1 + 4}[/tex]
| u x v | = [tex]\sqrt{9}[/tex]
| u x v | = 3
Therefore, the magnitude of u x v is 3
(b) The unit vector û parallel to u x v in the direction of u x v is given by the ratio of u x v and the magnitude of u x v. i.e
û = [tex]\frac{u X v}{|u X v|}[/tex]
u x v = 2i - j - 2k [calculated in (a) above]
|u x v| = 3 [calculated in (a) above]
∴ û = [tex]\frac{2i - j - 2k}{3}[/tex]
∴ û = [tex]\frac{2i}{3} - \frac{j}{3} - \frac{2k}{3}[/tex]
Which parenting style is thought to produce children who grow up to be emotionally withdrawn,
distant, and lacking in curiosity?
Answer:
Uninvolved parenting style
Explanation:
Uninvolved parenting style is the parenting style which is thought to produce children who grow up to be emotionally withdrawn, distant, and lacking in curiosity.
This is usually because there is little or no contact between the parents and children which leaves a wide gap and the children start to become emotionally withdrawn as a result. They also aren’t curious to learn new things as a result of the absence of parental care.
how does the earth release energy back into the atmosphere?
Answer:
Explanation:
The earth-atmosphere energy balance is the balance between incoming energy from the Sun and outgoing energy from the Earth. Energy released from the Sun is emitted as shortwave light and ultraviolet energy. When it reaches the Earth, some is reflected back to space by clouds, some is absorbed by the atmosphere, and some is absorbed at the Earth's surface.
Hope this helped you!
You hike two thirds of the way to the top of a hill at a speed of 2.9 mi/h and run the final third at a speed of 5.6 mi/h. What was your average speed?
Answer:
The average speed is 3.5 mi/h
Explanation:
Average speed is given by
[tex]Average speed = \frac{Total distance}{Total time}[/tex]
If the total distance covered is [tex]x[/tex] mi,
Then [tex]\frac{2}{3}x[/tex] mi was covered while hiking and
[tex]\frac{1}{3}x[/tex] mi was covered while running.
Now, we will find the time taken while hiking and the time taken while running
[tex]Speed = \frac{Distance}{ Time}\\ Time = \frac{Distance}{Speed}[/tex]
For the time taken while hikingSpeed = 2.9 mi/h
Distance = [tex]\frac{2}{3}x[/tex] mi
From,
[tex]Time = \frac{Distance}{Speed}[/tex]
[tex]Time = \frac{\frac{2}{3}x }{2.9}[/tex]
Time = [tex]0.2299x[/tex] h
Time taken while hiking is 0.2299 h
For the time taken while runningSpeed = 5.6 mi/h
Distance = [tex]\frac{1}{3}x[/tex] mi
[tex]Time = \frac{Distance}{Speed}[/tex]
[tex]Time = \frac{\frac{1}{3}x }{5.6}[/tex]
Time = [tex]0.05952x[/tex] h
Now, for the average speed
[tex]Average speed = \frac{Total distance}{Total time}[/tex]
Total distance = [tex]\frac{2}{3}x[/tex] mi + [tex]\frac{1}{3}x[/tex] mi = [tex]x[/tex] mi
Total time = [tex]0.2299x[/tex] + [tex]0.05952x[/tex] = [tex]0.28942x[/tex] h
∴ [tex]Average speed = \frac{x}{0.28942x}[/tex]
Average speed = 3.4552 mi/h
Average speed ≅ 3.5 mi/h
An apple is thrown at an angle of 30° above the horizontal from the top of a building 20 m high. Its initial speed is 40 ms-1. Calculate
a)the time taken for the apple to strikes the ground
b) the distance from the foot of the building will it strikes the ground
c) the maximun height reached by the apple from the ground
Answer:
(a) 4.83 seconds
(b) 167.3m
(c) 40m
Explanation:
This is a two-dimensional motion. Therefore, the components of the initial velocity - [tex]u_X[/tex] and [tex]u_Y[/tex] - in the x and y directions are given as follows:
[tex]u_X[/tex] = u cos θ --------------(*)
[tex]u_Y[/tex] = u sin θ -------------(**)
Where
θ = angle of projection
(a) To calculate the time taken for the apple to strike the ground.
For simplicity, let's first calculate the maximum height H reached by the apple.
Using one of the equations of motion as follows, we can find H:
v² = u² + 2as ---------------(i)
Where;
v = velocity at maximum height = 0 [at maximum height, velocity is 0]
u = initial vertical velocity of the apple = [tex]u_Y[/tex]
=> u = [tex]u_Y[/tex] = u sin θ
=> [tex]u_Y[/tex] = 40 sin 30°
=> [tex]u_Y[/tex] = 40 x 0.5
=> [tex]u_Y[/tex] = 20 m/s
a = acceleration due to gravity = -g [apple moves upwards against gravity]
a = -10m/s²
s = H = maximum height reached from the top of the building
Substitute these values into equation (i) above to have;
0² = [tex]u_Y[/tex]² + 2aH
0² = (20)² + 2(-10)H
0 = 400 - 20H
20H = 400
H = 20m
The total time taken to strike the ground is the sum of the time taken to reach maximum height and the time taken to strike the ground from maximum height.
=>Calculate time t₁ to reach maximum height.
Using one of the equations of motion, we can calculate t₁ as follows;
v = [tex]u_Y[/tex] + at ---------------(ii)
Where;
v = velocity at maximum height = 0
u = initial vertical velocity of the apple = [tex]u_Y[/tex]
a = -g = -10m/s² [acceleration due to gravity is negative since the apple is thrown upwards to reach maximum height]
t = t₁ = time taken to reach maximum height.
Equation (ii) then becomes;
0 = 20 + (-10)t₁
10t₁ = 20
t₁ = 2 seconds
=>Calculate time t₂ to strike the ground from maximum height.
Now, using one of the equations of motion, we can calculate the time taken as follows;
Δy = [tex]u_Y[/tex] t + [tex]\frac{1}{2}[/tex]at ---------------(iii)
Where;
Δy = displacement from maximum height to the ground = maximum height from top of building + height of building = 20 + 20 = 40m
a = g = 10m/s² [acceleration due to gravity is positive since the apple is now coming downwards from maximum height]
t = t₂ [time taken to strike the ground from maximum height]
[tex]u_Y[/tex] = initial vertical velocity from maximum height = 0
[tex]u_Y[/tex] = 0
Equation (iii) then becomes
40 = 0t + [tex]\frac{1}{2}[/tex](10)t₂²
40 = 5t₂² [divide through by 5]
8 = t₂²
t₂ = ±[tex]\sqrt{8}[/tex]
t₂ = ±[tex]2\sqrt{2}[/tex]
t₂ = +2[tex]\sqrt{2}[/tex] or -2[tex]\sqrt{2}[/tex]
since time cannot be negative,
t₂ = 2[tex]\sqrt{2}[/tex] = 2.83 seconds
Therefore, the time taken for the apple to strike the ground is;
t₁ + t₂ = 2 + 2.83 = 4.83 seconds
(b) The distance from the foot of the building where the apple will strike the ground
Since this is the horizontal distance, we use the horizontal version of equation (iii) as follows;
Δx = [tex]u_X[/tex] t + [tex]\frac{1}{2}[/tex]at -----------(v)
Where
Δx = distance from the foot of the building to where the apple strikes the ground.
[tex]u_X[/tex] = initial horizontal velocity of the apple as expressed in equation (*)
[tex]u_X[/tex] = 40 cos 30
[tex]u_X[/tex] = 34.64 m/s
t = time taken for the motion of the apple = 4.83 seconds [calculated above]
a = acceleration due to gravity in the horizontal direction = 0. [For a projectile, there is no acceleration in the horizontal direction since velocity is constant]
Substitute these values into equation (v) as follows;
Δx = [tex]u_X[/tex] t + [tex]\frac{1}{2}[/tex](0)t
Δx = 34.64 x 4.83
Δx = 167.3 m
Therefore, the distance from the foot of the building is 167.3m
(c) The maximum height reached by the apple from the ground
This is the sum of the height reached from the top of the building (20m which has been calculated in (a) above) and the height of the building.
= 20m + 20m = 40m
A U-tube is open to the atmosphere at both ends. Water is poured into the tube until the water column on the vertical sides of the U is more than 10 cm deep. Then, oil with a density of 950 kg/m3 is poured into one side, until the column of oil is 6.0 cm tall. How much higher is the top surface of the oil on that side of the tube compared with the surface of the water on the other side of the tube?
Answer:
The value is [tex]\Delta h = 0.003 \ m[/tex]
Explanation:
From the question we are told that
The height of the water is [tex]h_1 = 10 \ cm = 0.10 \ m[/tex]
The density of oil is [tex]\rho_o = 950 \ kg/m^3[/tex]
The height of oil is [tex]h_2 = 6 \ cm = 0.06 \ m[/tex]
Given that both arms of the tube are open then the pressure on both side is the same
So
[tex]P_a = P_b[/tex]
=> Here
[tex]P_a = P_z + \rho_w * g * h[/tex]
where [tex]\rho_w[/tex] is the density of water with value [tex]\rho_w = 1000 \ kg/m^3[/tex]
and [tex]P_z[/tex] is the atmospheric pressure
and
[tex]P_b = P_z + \rho_o * g * h_2[/tex]
=> [tex]P_z + \rho_w * g * h = P_z + \rho_o * g * h_2[/tex]
=> [tex]\rho_w * h = \rho_o * h_2[/tex]
=> [tex]h = \frac{950 * 0.06 }{1000}[/tex]
=> [tex]h = 0.057 \ m[/tex]
The difference in height is evaluated as
[tex]\Delta h = 0.06 - 0.057[/tex]
[tex]\Delta h = 0.003 \ m[/tex]