Ask question

Ask question

All categories

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

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

You might be interested in

How can tectonic activity change the climate of the earth?

soldi70 [24.7K]

If you really want to find out the answer just pray about it and it will come to you..

3 0

3 years ago

for every 120 joules of energy input a car waste 85 joules, find the useful energy output of the car?

harina [27]

Answer:

Useful Output = 35 J

Explanation:

The useful energy output of the car must be equal to the difference between the total input energy supplied to the car and the energy wasted by the car:

Useful Output = Total Input - Waste

where,

Total Input = 120 J

Waste = 85 J

Therefore,

Useful Output = 120 J - 85 J

<u>Useful Output = 35 J</u>

7 0

3 years ago

A block of ice is sliding down a ramp of slope 40 to the horizontal. What is the rate of acceleration of the block? Assume the f

uysha [10]

The given problem can be exemplified in the following diagram:

Since there is no friction or any other external force, the only force acting in the direction of the movement is the component of the weight of the block, therefore, applying Newton's second law:

$\Sigma F=ma$

Replacing the values:

$mg\sin 40=ma$

We may cancel out the mass:

$g\sin 40=a$

Using the gravity constant as 9.8 meters per square second:

$9.8\frac{m}{s^2}\sin 40=a$

Solving the operations:

$6.3\frac{m}{s^2}=a$

Therefore, the acceleration is 6.3 meters per square second.

8 0

1 year ago

A basketball center holds a basketball straight out, 2.0 m above the floor, and releases it. It bounces off the floor and rises

atroni [7]

Answer:

a) The velocity of the ball before it hits the floor is -6.3 m/s

b) The velocity of the ball after it hits the floor is 3.1 m/s

c) The magnitude of the average acceleration is 470 m/s². The direction is upward at an angle of 90º with the ground.

Explanation:

First, let´s calcualte how much time it takes the ball to hit the floor:

The equation for the position of the ball is:

y = y0 + v0 * t + 1/2 g * t²

Where:

y = position at time t

y0 = initial position

v0 = initial velocity

t = time

g = acceleration due to gravity

We take the ground as the origin of the reference system.

a) Since the ball is realesed and not thrown, the initial velocity v0 is 0. The direction of the acceleration is downward, towards the origin, then "g" will be negative. When the ball hits the ground its position will be 0. Then:

0 = 2.0 m + 0 m/s *t - 1/2 * 9.8 m/s² * t²

-2.0 m = -4.9 m/s² * t²

t² = -2.0 m / - 4.9 m/s²

t = 0.64 s

The equation for the velocity of a falling object is:

v = v0 + g * t where "v" is the velocity

since v0= 0:

v = g * t = -9.8 m/s² * 0.64 s = -6.3 m/s

b) Now, we know that the velocity of the ball when it reaches the max height must be 0. We can obtain the time it takes the ball to reach that height from the equation for velocity and then use that time in the equation for position to obtain the initial velocity:

v = v0 + g * t

0 = v0 + g * t

-v0/g = t

now we replace t in the equation for position, since we know that the maximum height is 1.5 m:

y = y0 + v0 * t + 1/2* g * t² y = 1.5 m y0 = 0 m t = -v0/g

1.5 m = v0 * (-v0/g) + 1/2 * g (-v0/g)²

1.5 m = - v0²/g - 1/2 * v0²/g

1.5 m = -3/2 v0²/g

1.5 m * (-2/3) * g = v0²

1.5 m * (-2/3) * (-9.8 m/s²) = v0²

v0 = 3.1 m/s

c) The average acceleration will be:

a = final velocity - initial velocity / time

a = 3.1 m/s - (-6.3 m/s) / 0.02 s = 470 m/s²

the direction of the acceleration is upward perpendicular to the ground.

The vector average acceleration will be:

a = (0, 470 m/s²) or (470 m/s² * cos 90º, 470 m/s² * sin 90º)

4 0

3 years ago

At a certain place, Earth's magnetic field has magnitude B =0.703 gauss and is inclined downward at an angle of 75.4° to the hor

Irina18 [472]

Answer:

The charge flows in coulombs is

$dq=1.843x10^{-5}C$

Explanation:

The current magnitude of current is given by the resistance and the induced Emf as:

$I=N*\frac{dF}{Rdt}$

$\frac{dq}{dt}=\frac{dF}{Rdt}=dq=N*\frac{dF}{R}$

$dq=\frac{N*\beta*A*(Cos(\alpha_f)-Cos(\alpha_i}{R}$

$N=1300$ , $\beta=0.703$ , $A=\pi*r^2=\pi*0.10^2=0.01\pi m^2$ , $R=99.4+202=301.4$ Ω

$\alpha_f=14.6$ , $\alpha_i=165.4$

Replacing :

$dq=\frac{1300*0.703x10^{-4}*0.01\pi*(0.9667-(-0.9667))}{202+99.4}$

$dq=1.843x10^{-5}C$

5 0

3 years ago