1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Artemon [7]
1 year ago
5

A car is being tested for safety by colliding it with a brick wall.The 1300 kg car is initially driving towards the wall at a sp

eedof 15 m/s, and after colliding with the wall the car movesaway from the wall at 2 m/s. If the car is in contact with thewall for 0.5 s, calculate the average force exerted on the carby the wall.
Physics
1 answer:
serg [7]1 year ago
4 0

Answer:

44200 N

Explanation:

To calculate the average force exerted on the car, we will use the following equation

\begin{gathered} Ft=\Delta p \\ Ft=m(v_f-v_i) \end{gathered}

Where F is the average force, t is the time, m is the mass, vf is the final velocity and vi is the initial velocity of the car.

Replacing t = 0.5s, m = 1300 kg, vf = -2 m/s, and vi = 15 m/s and solving for F, we get

\begin{gathered} F(0.5s)=(1300\text{ kg\rparen\lparen-2 m/s - 15 m/s\rparen} \\ F(0.5s)=(1300\text{ kg\rparen\lparen-17 m/s\rparen} \\ F(0.5s)=-22100\text{ kg m/s} \\ F=\frac{-22100\text{ kg m/s}}{0.5\text{ s}} \\ F=-44200\text{ N} \end{gathered}

Therefore, the average force exerted on the car by the wall was 44200 N

You might be interested in
An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e
White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}

with this value you can compute the frequency:

a)

f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz

b)

the mass of the block is given by the formula:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s

Finally, the amplitude is:

x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m

5 0
2 years ago
An object travels 20 m in 10s What is its speed?
ella [17]

speed=distance/time

distance=20m

time=10s

speed=?

speed=20×10

speed=200m/s

3 0
2 years ago
100 °C is a greater temperature than which of the following?
ohaa [14]
When somebody hands you a Celsius°, it's easy to find the equivalent Fahrenheit°.

Fahrenheit° = (1.8 · Celsius°) + 32° .

So 100°C works out to 212°F.  

It's also easy to find the equivalent Kelvin.  Just add 273.15 to the Celsius.

So now you can see that  100°C  is equal to  A  and  D,
and it's less than  B .

The only one it's greater than is  C .
6 0
2 years ago
If wheel turning at a constant rate completes 100 revolutions in 10 s its angular speed is:
GenaCL600 [577]
The frequency of the wheel is given by:
f= \frac{N}{t}
where N is the number of revolutions and t is the time taken. By using N=100 and t=10 s, we find the frequency of the wheel:
f= \frac{100}{10 s}=10 s^{-1}

And now we can find the angular speed of the wheel, which is related to the frequency by:
\omega=2 \pi f=2 \pi (10 s^{-1})=62.8 s^{-1}
6 0
3 years ago
A strong man and a weak man are trying to carry a ladder . How should they carry it in such a way that the weak person feels les
dsp73

Answer:

The strong person should carry the ladder at the  front end and the weak person should carry it at the back end.

Explanation:

this is because in such a case the strong person has to pull the ladder whereas the weak person at the back end have to push the ladder. In such case it is easier to push because the weak person can use the force of gravity of his own body for pushing the ladde.

However in case of pulling the ladder one has to overcome his own gravity to pull the heavy object

<u>                                                                                                                                 </u>

<h2><em>I Hope it help you </em></h2>
7 0
2 years ago
Read 2 more answers
Other questions:
  • What information does the atomic mass of an element provide?
    9·1 answer
  • The ability to store electrical energy is called<br><br> Voltage <br> Capacitance
    6·1 answer
  • The Kinetic Molecular Theory helps explain relationships between:
    10·2 answers
  • Kepler's second law: as a planet moves around its orbit, it sweeps out ______areas in ______ times.
    5·1 answer
  • Assume we’re able to travel to your planet and decide to take some fireworks with us to celebrate our journey.
    11·1 answer
  • A toroidal solenoid has an inner radius of 12.0 cm and an outer radius of 15.0 cm . It carries a current of 1.50 A . Part A How
    15·1 answer
  • After the car with the dead battery is running, the cables can be disconnected in the reverse order that they were connected. wh
    9·1 answer
  • A body is moving with an acceleration of 60 m/s^2(square).If the force applied to the body is 4200N.Calculate its mass.​
    10·1 answer
  • Can someone help me?
    15·1 answer
  • Hi..I'm new here..Can some one help mi with my school work plz​
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!