Ask question

Ask question

All categories

3 years ago

6

During a crash test, two identical cars crash into two different barriers. Both cars are initially traveling at the same constan

t velocity. Car A crashes into a solid brick wall and decelerates to v = 0.0m/s over a distance dA = 10.0cm. Car B crashes into a row of shrubs and decelerates to v = 0.0m/s over a distance dB = 10.0m. During the collision Car A experiences an average force FA and Car B experiences and average force FB. What is the ratio FA/FB?

1 answer:

Sergio039 [100]3 years ago

6 0

Answer:

FA / FB = 100

Explanation:

given two identical cars

let their initial speed be U m/s

for car 1 let it deaccelerates with a1

0 = U² - 2*a1*(10cm)

a1 = U² / (2*(0.1))

car2 deaccelerates with a2

0 = U² - 2*a2*(10)

a2 = U² / (2*(10))

FA = m*a1

FB = m*a2

FA/FB = a1 / a2

= 10/0.1

= 100

FA/FB = 100

You might be interested in

if the Horse and Rider have a combined mass of 572 kg what force would be required to accelerate them 5 kph per second

Vlad [161]

Answer:

Force required to accelerate = 794.44 N

Explanation:

Force required = Mass of horse x Acceleration of horse

Mass of horse and rider, m= 572 kg

Acceleration of horse and rider, a = 5 kph per second

$=\frac{5*1000}{60*60} =1.39 m/s^2$

Force required = ma

= 572 x 1.39 = 794.44 N

Force required to accelerate = 794.44 N

8 0

3 years ago

The core of a certain reflected reactor consist of a cylinder 10 ft high 10 ft in diameter The measured maximum-to-average flux

Westkost [7]

Answer:

The maximum power density in the reactor is 37.562 KW/L.

Explanation:

Given that,

Height = 10 ft = 3.048 m

Diameter = 10 ft = 3.048 m

Flux = 1.5

Power = 835 MW

We need to calculate the volume of cylinder

Using formula of volume

$V =\pi r^2 h$

Put the value into the formula

$V=\pi\times(1.524)^2\times 3.048$

$V= 22.23\m^3$

$V = 22.23\times10^{3}\ Liter$

We need to calculate the maximum power density in the reactor

Using formula of power density

$P=\dfrac{E}{V}$

Where, P = power density

E = energy

V = volume

Put the value into the formula

$P=\dfrac{835\times10^{6}}{22.23\times10^{3}}$

$P=37561.85 = 37.562\times KW/L$

Hence, The maximum power density in the reactor is 37.562 KW/L.

6 0

3 years ago

Crank

$\\ \rm\Rrightarrow \dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$

$\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{1}{-10}+\dfrac{1}{38}$

$\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{-19+5}{190}$

$\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{-14}{190}$

$\\ \rm\Rrightarrow u=\dfrac{190}{-14}$

$\\ \rm\Rrightarrow u=13.6cm$

Real

5 0

2 years ago

How the sun causes the weather?

Lerok [7]

If the sunlight is covered up by clouds or other things the temperature will be a little cooler, because the sun rays arent hitting things and making them warmer<span />

6 0

3 years ago

Read 2 more answers

A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 59° w

frosja888 [35]

Answer:

(a) $n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

(b) $n_{2} = 1.349$

(c) $v_{1} = 2.04\times 10^{8}\ m/s$

(d) $v_{2} = 2.22\times 10^{8}\ m/s$

Solution:

As per the question:

Refractive index of medium 1, $n_{1} = 1.47$

Angle of refraction for medium 1, $\theta_{1} = 59^{\circ}$

Angle of refraction for medium 2, $\theta_{1} = 69^{\circ}$

Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

$n_{1}sin\theta_{1} = n_{2}sin\theta_{2}$

where

$n_{2}$ = Refractive Index of medium 2

Now,

$n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:

$n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}$

$n_{2} = 1.349$

(c) To calculate the velocity of light in medium 1:

We know that:

$Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}$

Thus for medium 1

$n_{1} = \frac{c}{v_{1}$

$v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s$

(d) To calculate the velocity of light in medium 2:

For medium 2:

$n_{2} = \frac{c}{v_{2}$

$v_{2} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.349} = 2.22\times 10^{8}\ m/s$

5 0

3 years ago

Read 2 more answers