Ask question

Ask question

All categories

2 years ago

15

A 61 kg person is in a head-on collision. The car's speed at impact is 14 m/s. Estimate the net force on the person if he or she

is wearing a seat belt and if the air bag deploys.

1 answer:

forsale [732]2 years ago

8 0

Explanation:

It is known that the relation between force and acceleration is as follows.

F = -ma ............ (1)

Also, acceleration (a) = $\frac{v^{2} - u^{2}}{2s}$ ......... (2)

The given data is as follows.

v = 0, u = 14 m/s, s = 1

Therefore, putting the values of expression (2) into expression (1) as follows.

F = $-m \times \frac{v^{2} - u^{2}}{2s}$

= $-61 kg \times \frac{(0)^{2} - (14)^{2}}{2 \times 1}$

= 5978 N

thus, we can conclude that net force on the person if he or she is wearing a seat belt and if the air bag deploys is 5978 N.

You might be interested in

A force off 700 newtons is applied to a 600 kg bowling ball. What is the acceleration of the bowling ball once the force is appl

Readme [11.4K]

Answer:

<h2>1.17 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question we have

$a = \frac{700}{600} = \frac{7}{6} \\ = 1.1666666...$

We have the final answer as

<h3>1.17 m/s²</h3>

Hope this helps you

6 0

2 years ago

If a transformer has a primary with 100 V and fifty coils, and a secondary that yields 20 V, how many coils are on the secondary

melomori [17]

Ideally, if all the magnetic of one winding cuts the other winding, and there isn't any loss in the transformer core or the resistance of the wire, then the voltage across each winding is proportional to the number of turns in its coil.

If you apply 100 V to a winding of 50 turns, then a winding that yields 20 volts
must be wound with

(20/100) of 50 turns = 10 turns

7 0

2 years ago

A spring stretches 0.2 cm per newton of applied force. An object is suspended from the spring and a deflection of 3 cm is observ

siniylev [52]

Answer:

$m=1.53kg$

Explanation:

To solve this problem we use the Hooke's Law:

$F=k*\Delta x$ (1)

F is the Force needed to expand or compress the spring by a distance Δx.

The spring stretches 0.2cm per Newton, in other words:

1N=k*0.2cm ⇒ k=1N/0.2cm=5N/cm

The force applied is due to the weight

$F=mg$

We replace in (1):

$mg=k*\Delta x$

We solve the equation for m:

$m=k*\Delta x/g=5*3/9.81=1.53kg$

7 0

2 years ago

How can you increase the potential energy of a bouncing ball

ad-work [718]

Answer:

When you lift the ball, you are doing work to increase its gravitational potential energy. When you then release the ball, gravitational energy is transformed into kinetic energy as the ball falls. When the ball hits the floor, the ball's shape changes as it flattens against the floor.

Explanation:thats should be the way^^ in explaining

7 0

2 years ago

Two circular holes, one larger than the other, are cut in the side of a large water tank whose top is open to the atmosphere. Ho

lidiya [134]

Answer: $\frac{r_1}{r_2}=1.565$

Explanation:

Given

two holes are made with different sizes

Hole 1 is large in size and hole 2 is small

If the volume flow rate of water is same for both the hole then small hole must be below the large hole because for same flow rate, velocity of water is large while cross-sectional area is small so it compensate to give same flow for both the holes.

Now for radius apply Bernoulli's theorem at hole 1 and 2

$P_1+\rho gh_1=P_{atm}+\frac{1}{2}\rho v_1^2$

$P_2+\rho g6h_2=P_{atm}+\frac{1}{2}\rho v_2^2$

if hole 1 is h distance below water surface then $h_2=6h$

and $P_1=P_2=P_{atm}$

Also $v_1=\sqrt{2gh}$

$v_2=\sqrt{2g(6h)}$

and $Q=A_1v_1=A_2v_2$

$A=\pi r^2$

thus $\dfrac{r_1}{r_2}=\sqrt{\dfrac{v_2}{v_1}}$

$\dfrac{r_1}{r_2}=\sqrt{\dfrac{\sqrt{6h}}{\sqrt{h}}}$

$\frac{r_1}{r_2}=1.565$

5 0

2 years ago