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Sav [38]
3 years ago
12

In a stunt, three people jump off a platform and fall 8.5 m onto a large air bag. A fourth person at the other end of the air ba

g, the "flier," is launched 16 m vertically into the air. Assume all four people have a mass of 74 kg. What is the momentum of the flier just after launch? Let upward be the positive direction.
Physics
1 answer:
docker41 [41]3 years ago
8 0

Answer:

They Died Period NO MORE

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A red rubber ball rolls down a hill from rest with an acceleration of 7.8 m/s 2 . How fast is it moving after it has traveled 5
masya89 [10]

Answer:

39 m/s

Explanation:

7 0
3 years ago
Vision is blurred if the head is vibrated at 29 Hz because the vibrations are resonant with the natural frequency of the eyeball
Likurg_2 [28]

Answer:

k = 4422.35  KN/m

Explanation:

Given that

Frequency ,f= 29 Hz

m = 7.5 g

Natural frequency ω

ω = 2 π f

We also know that for spring mass system

ω ² m =k        

k=Spring constant

So we can say that

( 2 π f)² =  m k

By putting the values

(2 x π x 29)² = 7.5 x 10⁻³  k

33167.69 = 7.5 x 10⁻³  k

k=4422.35 x  10³ N/m

k = 4422.35  KN/m

Therefore spring constant will be 4422.35  KN/m

4 0
3 years ago
If you were to stand in the exact center of a rotating disc, you would only have what kind of
Dmitriy789 [7]

Answer:

Tangential speed or Rotational speed

3 0
3 years ago
A 30-kg child rides a 20-kg bicycle together,the child and the bicycle have a momentum of 110 kg•m/s. What is the velocity of th
Korolek [52]
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3 0
3 years ago
An astronaut goes out for a space walk. Her mass (including space suit, oxygen tank, etc.) is 100 kg. Suddenly, disaster strikes
Marina CMI [18]

Answer:

<u>Part A:</u>

Unknown variables:

velocity of the astronaut after throwing the tank.

maximum distance the astronaut can be away from the spacecraft to make it back before she runs out of oxygen.

Known variables:

velocity and mass of the tank.

mass of the astronaut after and before throwing the tank.

maximum time it can take the astronaut to return to the spacecraft.

<u>Part B: </u>

To obtain the velocity of the astronaut we use this equation:

-(momentum of the oxygen tank) = momentum of the astronaut

-mt · vt = ma · vt

Where:

mt = mass of the tank

vt = velocity of the tank

ma = mass of the astronaut

va = velocity of the astronaut

To obtain the maximum distance the astronaut can be away from the spacecraft we use this equation:

x = x0 + v · t

Where:

x = position of the astronaut at time t.

x0 = initial position.

v = velocity.

t = time.

<u>Part C:</u>

The maximum distance the astronaut can be away from the spacecraft is 162 m.

Explanation:

Hi there!

Due to conservation of momentum, the momentum of the oxygen tank when it is thrown away must be equal to the momentum of the astronaut but in opposite direction. In other words, the momentum of the system astronaut-oxygen tank is the same before and after throwing the tank.

The momentum of the system before throwing the tank is zero because the astronaut is at rest:

Initial momentum = m · v

Where m is the mass of the astronaut plus the equipment (100 kg) and v is its velocity (0 m/s).

Then:

initial momentum = 0

After throwing the tank, the momentum of the system is the sum of the momentums of the astronaut plus the momentum of the tank.

final momentum = mt · vt + ma · va

Where:

mt = mass of the tank

vt = velocity of the tank

ma = mass of the astronaut

va = velocity of the astronaut

Since the initial momentum is equal to final momentum:

initial momentum = final momentum

0 = mt · vt + ma · va

- mt · vt = ma · va

Now, we have proved that the momentum of the tank must be equal to the momentum of the astronaut but in opposite direction.

Solving that equation for the velocity of the astronaut (va):

- (mt · vt)/ma = va

mt = 15 kg

vt = 10 m/s

ma = 100 kg - 15 kg = 85 kg

-(15 kg · 10 m/s)/ 85 kg = -1.8 m/s

The velocity of the astronaut is 1.8 m/s in direction to the spacecraft.

Let´s place the origin of the frame of reference at the spacecraft. The equation of position for an object moving in a straight line at constant velocity is the following:

x = x0 + v · t

where:

x = position of the object at time t.

x0 = initial position.

v = velocity.

t = time.

Initially, the astronaut is at a distance x away from the spacecraft so that

the initial position of the astronaut, x0, is equal to x.

Since the origin of the frame of reference is located at the spacecraft, the position of the spacecraft will be 0 m.

The velocity of the astronaut is directed towards the spacecraft (the origin of the frame of reference), then, v = -1.8 m/s

The maximum time it can take the astronaut to reach the position of the spacecraft is 1.5 min = 90 s.

Then:

x = x0 + v · t

0 m = x - 1.8 m/s · 90 s

Solving for x:

1.8 m/s · 90 s = x

x = 162 m

The maximum distance the astronaut can be away from the spacecraft is 162 m.

6 0
3 years ago
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