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german
2 years ago
15

A 69.5-kg person throws a 0.0475-kg snowball forward with a ground speed of 31.5 m/s. A second person, with a mass of 57.5 kg, c

atches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.35 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged
Physics
1 answer:
Leno4ka [110]2 years ago
3 0

Answer:

- After throwing the snow, velocity of the thrower is 2.33 m/s

- the velocity of the receiver is 0.026 m/s

Explanation:

Given the data in the question;

Using conservation of momentum,

Initial thrower has a momentum of mv; m_{totalv

(69.5 kg + 0.0475 kg) × 2.35 m/s = 163.4366 kg.m/s

Now, When he throws it at 31.5 m/s, these constitutes a momentum of;

(0.0475 kg )(31.5 m/s) = 1.49625 kg.m/s

hence his momentum now is: 163.4366 - 1.49625 = 161.94035 kg.m/s

To get his velocity, we say;

161.94035 = mv

{ he lost weight of the snow ball so, m = 69.5 kg )

161.94035 = 69.5 × v

v = 161.94035 / 69.5

v = 2.33 m/s

Therefore, After throwing the snow, velocity of the thrower is 2.33 m/s

Next is the Receiver;

the receiver will gain momentum of 1.49625 kg.m/s

he has no momentum initially and after he catches the snow ball;

1.49625 kg.m/s = mv

1.49625 kg.m/s = ( 57.5 kg +  0.0475 kg ) × v

1.49625 kg.m/s = 57.5475 kg × v

v = ( 1.49625 kg.m/s ) / 57.5475 kg

v = 0.026 m/s

Therefore, the velocity of the receiver is 0.026 m/s

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A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr
KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s

Time the parachutist falls without friction is 3.19 seconds

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2

Now the initial velocity of the last half height will be the final velocity of the first half height.

v=u+at\\\Rightarrow v=at

Since the height are equal

\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0

t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2

Magnitude of acceleration is -43200 ft/s²

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You are traveling 70 mph (31.3 m/s) and slam on the brakes to avoid hitting another car. How far do you travel if it takes you 8
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You traveled a distance of 620.075 meters if it takes you 8.5 seconds to stop.

<u>Given the following data:</u>

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  • Time, t = 8.5 seconds.

We know that acceleration due to gravity (a) for an object is equal to 9.8 meter per seconds square.

To find the distance traveled, we would use the second equation of motion:

Mathematically, the second equation of motion is given by the formula;

S = ut + \frac{1}{2} at^2

Where:

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Substituting the parameters into the formula, we have;

S = 31.3(8.5) + \frac{x}{y} (9.8)(8.5^2)\\\\S = 266.05 + 4.9(72.25)\\\\S = 266.05 + 354.025

<em>Distance, S</em><em> = </em><em>620.075 meters.</em>

Therefore, you traveled a distance of 620.075 meters if it takes you 8.5 seconds to stop.

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