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rjkz [21]
3 years ago
11

Two people are standing on a 3.12-m-long platform, one at each end. The platform floats parallel to the ground on a cushion of a

ir, like a hovercraft. The two people, the platform and a 3.36-kg ball are all initially at rest. One person throws the 3.36-kg ball to the other, who catches it. The ball travels nearly horizontally. Excluding the ball, the total mass of the platform and people is 119 kg. Because of the throw, this 119-kg mass recoils. How far does the platform move before coming to rest again
Physics
2 answers:
Ann [662]3 years ago
7 0

Answer:

The magnitude of displacement is 0.082m

Explanation:

While the ball is in motion,we have MV + mv= 0 ...eq1

Where M = combined mass of the platform and the two people.

V = velocity of the platform

m = mass of the ball

v = velocity of the ball

The distance that the platform moves is given by:

X = Vt ...eq2

Where t is the time that the ball is in the air.

The time the ball is in the air is given by:

L/(v-V) ...eq3

Where L is the length of the platform

The quantity(v-V) = velocity of the ball relative to the platform.

Combining eq2 and eq3

X = (V/(v - V))L

From eq1 , the ratios of the velocities is V/v = -m/M

X = (V/v)L / (1 - (V/v) = (-m/M) L /(1+ (m/M))

X = -mL/(M + m)

X = - (3.36kg × 3.12m) /( 119kg + 3.36kg)

X = - 10.48/ 122.36

X = -0.082m

The minus sign implies that the displacement of the platform is in the opposite direction to the displacement of the ball.

Therefore, the distance moved by the platform is the magnitude of this displacement 0.082m.

GarryVolchara [31]3 years ago
6 0

Answer:

Displacement = 0.086m

Explanation:

Given

M = Combined mass of the two people and the platform = 119kg

m = mass of the ball = 3.36kg

v = velocity of the ball

V = Recoil velocity of the platform

t = time spent by the ball in the air

L = length of the platform = 3.12m

MV + mv = 0

The distance moved by the platform is given by; x = Vt

Where t = L/(v - V)

v - V represents the velocity of the ball relative to the platform.

Substitute L/(v - V) for t

so, x = VL/(v-V)

From (MV + mv = 0), we have

MV = -mv

V/v = -m/M

So, x = VL/(v-V) becomes

x = ((V/v)L)/(1 - V/v)

x = ((-m/M)L)/(1 + m/M)

x = -mL/(M+m)

x = -(3.36 * 3.12)/(3.36 + 119)

x = -0.08567505720823798627002

x = -0.086m

The negative sign implies that the displacement of the platform is in the opposite direction to the displacement of the ball.

The distance moved by the platform before coming to rest is 0.086m

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Answer:

Length, l = 33.4 m

Explanation:

Given that,

Electrical field, E=3\times 10^6\ V/m

Let the electrical potential is, V=10^8\ V

We need to find the length of a thundercloud lightning bolt. The relation between electric field and the electric potential is given by :

V=E\times d\\\\d=\dfrac{V}{E}\\\\d=\dfrac{10^8}{3\times 10^6}\\\\d=33.4\ m

So, the length of a thundercloud lightning bolt is 33.4 meters. Hence, this is the required solution.

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3 years ago
Sam receives the kicked football on the 3 yd line and runs straight ahead toward the goal line before cutting to the right at th
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Answer:

Distance: 21 yd, displacement: 15 yd, gain in the play: 12 yd

Explanation:

The distance travelled by Sam is just the sum of the length of each part of Sam's motion, regardless of the direction. Initially, Sam run from the 3 yd line to the 15 yd line, so (15-3)=12 yd. Then, he run also 9 yd to the right. Therefore, the total distance is

d = 12 + 9 = 21 yd

The displacement instead is a vector connecting the starting point with the final point of the motion. Sam run first 12 yd straight ahead and then 9 yd to the right; these two motions are perpendicular to each other, so we can find the displacement simply by using Pythagorean's theorem:

d=\sqrt{12^2+9^2}=15 yd

Finally, the yards gained by Sam in the play are simply given by the distance covered along the forward-backward direction only. Since Sam only run from the 3 yd line to the 15 yd line along this direction, then the gain in this play was

d = 15 - 3 = 12 yd

7 0
3 years ago
4. A ball is thrown vertically upward from the ground with a velocity of 30m/s. (a) how long will it take to rise to the highest
yarga [219]

All the answers are:

a) The time that will it take to rise to the highest point is 3.06 seconds.

b) The ball will rise to a height of 45.87 meters.

c) The time at which the ball will have a velocity of 10 m/s upward is 2.04 seconds.

The time when the ball has 10 m/s downward is 1.02 seconds.

d) The displacement of the ball will be zero at 6.12 seconds.

e) The time when the magnitude of the ball's velocity is equal to half its velocity of projection is 1.53 seconds.

f) The ball's displacement is equal to half the maximum height to which it rises after 0.90 seconds.

g) In each moment (upward and downward) the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.

Let's calculate the values for each case.

a) At the highest point, the final velocity is 0, so we can use the following equation.  

v_{f}=v_{i}-gt (1)

Where:

  • v(i) is the initial velocity
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  • g is the acceleration due to gravity (9.81 m/s²)

We know that v(i) = 30 m/s.

0=30-9.81t

Solve it for t:

t=3.06\: s

Hence, the time is 3.06 s.

b) At the highest point, the final velocity is 0, so we can use the following equation.  

v_{f}^{2}=v_{i}^{2}-2gh (2)

0=v_{i}^{2}-2gh

We know that the initial velocity is 30 m/s.

0=30^{2}-2gh

Solving it for h we have:  

h=\frac{30^{2}}{2*9.81}

h=45.87 \: m

Then, the height is 45.87 m.

c) Using equation (1) we can find the time (t).

10=30-(9.81t)

So, the time elapsed to get 10 m/s is:

t_{upward}=2.04\: s

We know the upward time is equal to the downward time. So the time from v=10 m/s to v=0 m/s will be.

t_{upward}=2.04+t  

t=1.02\: s

This is the time when the ball has 10 m/s downward.          

Therefore, the time upward is 2.04 s, and the time downward is 1.02 s.

d) It will be when the ball returns to the ground.

t=2t_{upward}

t=2*3.06      

t=6.12\: s

The displacement will be zero after 6.12 s.  

e) Here we need to find the time when v(f) is 15 m/s

15=30-gt

t=\frac{15}{9.81}  

t=1.53\: s

The time when the v(f) is 15 m/s is 1.53 s.

f) Here, we need to find t when h = 45.87/2 m = 22.94 m

We can use the next equation:

[tex]h=v_{i}t-0.5gt^{2}/tex]

[tex]22.94=30t-0.5*9.81*t^{2}/tex]

Solving this quadratic equation, t will be:

[tex]t=0.90\: s/tex]

Hence, the ball's displacement is equal to half the maximum h, at 0.90 s.

g) In each moment the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.

Learn more about vertical motion here:

brainly.com/question/13966860

I hope it helps you!

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Answer:

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C = 1.85*10^{-5}F

Therefore the capacitance of the capacitor is 1.85*10^{-5}F

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