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
FromTheMoon [43]
2 years ago
13

Write a collision scenario here. If you choose your own collision, you can have neither, one, or both of the objects break. Be s

ure to describe which object (if any) breaks.
Physics
1 answer:
OleMash [197]2 years ago
5 0

Answer:

My scenario would be A Car vs. a guard rail on a road.  You have a car that is coming down a Highway at a speed of 43 Mph Miles per hour (69.2018 Kmh)

And it hits a steel guardrail and the car smashes in at the front and the guardrail is only bent while the car has the bumper and the hood along with the headlights and windshield along with the passenger side window break.

Explanation:

This is caused by so much force reacting from one object to another but also depends on molecular density.

You might be interested in
Which of the following statements are true about magnets?
timama [110]
The correct answer is 3
4 0
3 years ago
What is matter, atom, molecules?
bagirrra123 [75]

matter is a solid liquid or gas , a atom is a basic unit of a chemical element and a molecule is a group of atoms together

6 0
3 years ago
One terminal of a car battery is said to be connected to “ground.” since it is not really connected to the ground, what is meant
GalinKa [24]
This expression means that the negative terminal (-) is connected to the metal chassis or engine, which means that all voltages used for the electrical devices in the car are measured with respect to the car's chassis or engine. 
Today's vehicles have a negative ground system, which means that the vehicle's steel frame or chassis is directly connected to the negative side of the battery via the negative battery cable. 
3 0
3 years ago
A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive
Ivahew [28]

Answer:

a

The  x- and y-components of the total force exerted is

           F_{31 +32} =  (8.64i - 5.52 j) *10^{-5}

b

 The magnitude of the force is  

            |F_{31 +32}| = 10.25 *10^{-5} N

   The direction of the force is  

         \theta =327.43 ^o   Clockwise from x-axis

Explanation:

From the question we are told that

    The magnitude of the first charge is q_1 = +5.00nC = 5.00*10^{-9}C

      The magnitude of the second charge is q_2 = -2.00nC = -2.00*10^{-9}C

        The position of the second charge  from the first one is  d_{12} = 4.00i \  cm = \frac{4.00i}{100} = 4.00i *10^{-2} m

        The  magnitude of the third charge is q_3 = +6.00nC = 6.00*10^{-9}C

       The position of the third charge from the first one is  \= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} =  (4i + 3j) *10^{-2}m

                |d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m

                |d_{31}| =5 *10^{-2} m

        The position of the third charge from the second  one is

                \= d_{32} = 3j cm = 3j *10^{-2}m

               |d_{32}| =(\sqrt{ 3^2}) *10^{-2} m

               |d_{32}| =3 *10^{-2} m

The force acting on the third charge due to the first and second charge is mathematically represented as

           F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}

 Substituting values

          F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3}  * (4i + 3j ) *10^{-2}  \\ \ +  \ \ \ \ \ \ \ \ \   \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3}  * (4i + 3j ) *10^{-2}

            F_{31 +32} = 2.16 *10^{-5} (4i + 3j)  - 12*10^{-5} j

            F_{31 +32} =  (8.64i - 5.52 j) *10^{-5}

The magnitude of     F_{31 +32}  is mathematically evaluated as

            |F_{31 +32}| = \sqrt{(8.64^2 + 5.52 ^2) } *10^{-5}

             |F_{31 +32}| = 10.25 *10^{-5} N

The direction is obtained as

            tan \theta = \frac{-5.52 *10^{-5}}{8.64 *10^{-5}}

              \theta = tan ^{-1} [-0.63889]

             \theta = - 32.57 ^o

             \theta = 360 - 32.57

            \theta =327.43 ^o

               

                         

5 0
3 years ago
a spring is stretched 20 cm by a 30.0 n force. determine the work done in stretching the spring from 0 to 40 cm?
Alekssandra [29.7K]

The work done when a spring is stretched from 0 to 40cm is 4J.

What is work done?

Work done is the magnitude of force multiplied by displacement of an object. It is also the amount of energy transferred to an object when work is done on that.

The work done on the spring to stretch to 40cm is,

F = kx

where F is force, k is force constant.

k = F / x = 10 N / 20 * 10^-2 m = 50 N/m

W = 0.5 * k * (x)^2

where W = work done, k = force constant.

W = 0.5 x 50 x (40 x 10^-2)^2 = 4 J.

Therefore, the work done on the spring when it is stretched to 40cm is 4J.

To learn more about work done click on the given link brainly.com/question/25573309

#SPJ4

3 0
1 year ago
Other questions:
  • The deflection of alpha particles in rutherford’s gold foil experiments resulted in what change to the atomic model? the additio
    11·2 answers
  • Two formula one racing cars are negotiating a circular turn, and they have the same centripetal acceleration. however, the path
    11·1 answer
  • Dos masas están conectadas por una cuerda ligera que pasa por una polea sin rozamiento. Determine la aceleración de las masas y
    7·1 answer
  • What happens if a species is unable to adapt to a changing environment
    8·2 answers
  • A 0.25-kg ball sits on the roof of a building that is 10 meters tall. Find the GPE. (Gravity = 9.8 on Earth)
    10·1 answer
  • Why are vectors so important?
    13·1 answer
  • A typical ceiling fan running at high speed has an airflow of about 1.85 ✕ 103 ft3/min, meaning that about 1.85 ✕ 103 cubic feet
    5·1 answer
  • Group elements number 11 to 20 as either metallic, non metallic or metalloid.​
    5·2 answers
  • IS THIS CORRECT???...........................
    11·1 answer
  • An object has a mass of 7 kg and is accelerating at 4 m/s2. How far would the object move if it took 168 J of work to move it?​
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!