Ask question

Ask question

All categories

Zielflug [23.3K]

2 years ago

10

During a test crash an airbag inflates to stop a dummy's forward motion. the dummy's mass is 75kg. If the net force on the dummy

is 825n toward the rear of the car , what is the dummy's deceleration.

1 answer:

Black_prince [1.1K]2 years ago

4 0

F=ma
a=F/m
=825N/75kg
=825kg*m/75kg*s^2
=11m/s^2 in the direction of the force (ans)

You might be interested in

A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule

ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

$p_i = 0$

The final total momentum is instead:

$p_f = m_a v_a + m_c v_c$

where

$m_a = 125 kg$ is the mass of the astronaut

$v_a = 2.50 m/s$ is the velocity of the astronaut

$m_c = 1900 kg$ is the mass of the capsule

$v_c$ is the velocity of the capsule

Since the total momentum must be conserved, we have

$p_i = p_f = 0$

so

$m_a v_a + m_c v_c=0$

Solving the equation for $v_c$ , we find

$v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s$

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

$F \Delta t = \Delta p$

The change in momentum of the astronaut is

$\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s$

And the duration of the push is

$\Delta t = 0.600 s$

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

$F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N$

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

$K=\frac{1}{2}mv^2$

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

$K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J$

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

$K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J$

3 0

3 years ago

On a 10 kg cart (shown below), the cart is brought up to speed with 50N of force for 7m, horizontally. At this point (A), the ca

pav-90 [236]

Answer: Letter B! Is your answer

3 0

2 years ago

Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference

strojnjashka [21]

Answer:

1.7323

Explanation:

To develop this problem, it is necessary to apply the concepts related to refractive indices and Snell's law.

From the data given we have to:

$n_{air}=1$

$\theta_{liquid} = 19.38\°$

$\theta_{air}35.09\°$

Where n means the index of refraction.

We need to calculate the index of refraction of the liquid, then applying Snell's law we have:

$n_1sin\theta_1 = n_2sin\theta_2$

$n_{air}sin\theta_{air} = n_{liquid}sin\theta_{liquid}$

$n_{liquid} = \frac{n_{air}sin\theta_{air}}{sin\theta_{liquid}}$

Replacing the values we have:

$n_{liquid}=\frac{(1)sin(35.09)}{sin(19.38)}$

$n_liquid = 1.7323$

Therefore the refractive index for the liquid is 1.7323

6 0

3 years ago

What causes glowsticks to give off light?

iren2701 [21]

Phenyl oxalate ester is responsible for the luminescence in aglow stick<span>. The reaction with hydrogen peroxide </span>causes<span> the liquid inside a </span><span>glow stick to glow</span>

3 0

2 years ago

What is the SI unit of acceleration and work?

Lerok [7]

A=m/s^2(meter per second square)
Work=joule

6 0

2 years ago