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2 years ago

Activity A:

1 answer:

6 0

The crumple zone help protect a passenger from injury by converting some of the kinetic energy possessed by the car into a controlled deformation, or crumpling when there is a crash.

<h3>What is impact?</h3>

The term impact refers to a large force that acts for a short time. Passengers in a vehicles sustain injuries after an accident due to the high impact of the crash.

As such, the crumple zone help protect a passenger from injury by converting some of the kinetic energy possessed by the car into a controlled deformation, or crumpling when there is a crash.

Learn more about impact in physics: brainly.com/question/10952916

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How to change V=72km/hr to m/s

lapo4ka [179]

Multiply by (1000 meters / 1 km).
Then multiply by (1 hour / 3600 seconds).

Both of those fractions are equal to ' 1 ', because the top
and bottom numbers are equal, so the multiplications
won't change the VALUE of the 72 km/hr. They'll only
change the units.

(72 km/hour) · (1000 meters / 1 km) · (1 hour / 3600 seconds)

= (72 · 1000 / 3600) (km·meter·hour / hour·km·second)

= 20 meter/second

7 0

3 years ago

What is the momentum of a man of mass 10kg when he walks with a uniform velocity of 2m/s.

Stolb23 [73]

Answer:

$Momentum(p) = 20kgm/s$

Explanation:

Given

Mass = 10kg

Velocity = 2m/s

Required

Calculate the momentum of the man

Momentum is calculated as thus

$Momentum(p) = Mass(m) * Velocity(v)$

$p = mv$

So; to solve this question; we simply substitute 10kg for mass and 2m/s for velocity in the above formula;

The formula becomes

$Momentum(p) = 10kg * 2m/s$

$Momentum(p) = 10 * 2 * kg * m/s$

$Momentum(p) = 20kgm/s$

Hence, the momentum of the man is $20kgm/s$

5 0

3 years ago

One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i

kati45 [8]

Answer:

The force exerted on an electron is $7.2\times10^{-18}\ N$

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

$E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

$E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}$

$E_{1}=1.0183\times10^{3}\ N/C$

Using formula of electric field again

$E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}$

$E_{2}=-1.064\times10^{3}\ N/C$

We need to calculate the resultant electric field

Using formula of electric field

$E=E_{1}+E_{2}$

Put the value into the formula

$E=1.0183\times10^{3}-1.064\times10^{3}$

$E=-0.045\times10^{3}\ N/C$

We need to calculate the force exerted on an electron

Using formula of electric field

$E = \dfrac{F}{q}$

$F=E\times q$

Put the value into the formula

$F=-0.045\times10^{3}\times(-1.6\times10^{-19})$

$F=7.2\times10^{-18}\ N$

Hence, The force exerted on an electron is $7.2\times10^{-18}\ N$

8 0

3 years ago

A 0.095 kg tennis ball is traveling to the right at 40 m/s when it bounces of a wall and travels in the opposite direction it ca

Mashutka [201]

given that

mass of ball = 0.095 kg

initial velocity of ball towards the wall = 40 m/s

final velocity of the ball after it rebound = 30 m/s

now change in momentum is given as

$\Delta P = m(v_f - v_i)$

$\Delta P = 0.095(30 - (-40))$

$\Delta P = 6.65 kg m/s$

So change in momentum will be 6.65 kg m/s

3 0

3 years ago

How would the seasons change if the earth were tilted at 90 degrees instead of 23.5?

PIT_PIT [208]

<span>The 23.5 degree tilt is responsible for the seasons. If the earth had no tilt there would not be seasons. If the earth was tilted by 90 degrees the seasonal changes would be at the most extreme. The Earth's pole would point directly at the sun at a point on the track around the sun. As the Earth revolves around the Sun the pole would alternate twice each year between pointing directly at the sun and being perpendicular to the sun.

I hope this helps you!
xo, Leafling</span>

6 0

3 years ago