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Vika [28.1K]
3 years ago
14

Jennifer, who has a mass of 48.0 kg, is riding at 40.2 m/s in her red sports car when she must suddenly slam on the brakes. An a

ir bag deploys in the car. She strikes the air bag, bringing her body to a stop in 0.56 s. What average force does the air bag exert on her
Physics
2 answers:
andre [41]3 years ago
6 0

Answer:

F = 3,445.71 N

Explanation:

Finding the deceleration (a):

a = \frac{u - v}{t}

initial velocity = u = 40.2 m/s

final velocity = v = 0 m/s

a = 40.2/0.56 = 71.785 m/s^2

Now, Finding the force:

F = ma

mass = m = 48.0 kg

a = 71.785 m/s^2

the force is found to be 3,445.71 N

Bogdan [553]3 years ago
3 0

Answer:

F = 3445.714\,N

Explanation:

The system can be modelled appropriately by the use of the Principle of Momentum Conservation and Impact Theorem:

(48\,kg)\cdot (40.2\,\frac{m}{s} ) - F\cdot (0.56\,s) = 0\,\frac{kg\cdot m}{s}

The average force exerted on her:

F = 3445.714\,N

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What cause-and-effect relationship is described in this
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8 0
2 years ago
a bullet moving with a velocity of 100m/s pierce a block of wood and moves out with a velocityof 10 m/s.if the thickness of the
erma4kov [3.2K]

The emerging velocity of the bullet is <u>71 m/s.</u>

The bullet of mass <em>m</em> moving with a velocity <em>u</em>  has kinetic energy. When it pierces the block of wood, the block exerts a force of friction on the bullet. As the bullet passes through the block, work is done against the resistive forces exerted on the bullet by the block. This results in the reduction of the bullet's kinetic energy. The bullet has a speed <em>v</em> when it emerges from the block.

If the block exerts a resistive force <em>F</em> on the bullet and the thickness of the block is <em>x</em> then, the work done by the resistive force is given by,

W=Fx

This is equal to the change in the bullet's kinetic energy.

W=Fx=\frac{1}{2} m(u^2-v^2)......(1)

If the thickness of the block is reduced by one-half, the bullet emerges out with a velocity v<em>₁.</em>

Assuming the same resistive forces to act on the bullet,

F(\frac{x}{2} )=\frac{1}{2} m(u^2-v_1^2)......(2)

Divide equation (2) by equation (1) and simplify for v<em>₁.</em>

\frac{\frac{Fx}{2} }{Fx} =\frac{(u^2-v_1^2)}{(u^2-v^2)} \\\frac{100^2-v_1^2}{100^2-10^2} =\frac{1}{2} \\v_1^2=5050\\v_1=71.06 m/s

Thus the speed of the bullet is 71 m/s


3 0
3 years ago
A 350-g baseball is shot out of a small cannon with a velocity of 9.0 m/s. The baseball flies horizontally at a constant height
stira [4]

Answer:

9.5 kg m^2/s

Explanation:

The angular momentum of an object is given by:

L=mvr

where

m is the mass of the object

v is its velocity

r is the distance of the object from axis of rotation

Here we have:

m = 350 g = 0.35 kg is the mass of the ball

v = 9.0 m/s is the velocity

r = 3.0 m is the distance of the object from axis of rotation (if we take the ground as the centre of rotation)

Therefore, the angular momentum is:

L=(0.35)(9.0)(3.0)=9.5 kg m^2/s

4 0
3 years ago
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