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Roman55 [17]
3 years ago
5

A car starts from rest and accelerates uniformly over a time of 18 seconds for a distance of 390 m. Determine the acceleration o

f the car.

Physics
1 answer:
Sergeeva-Olga [200]3 years ago
5 0

Answer:

a=2.4\ m/s^2

Explanation:

Given that,

The initial speed of a car, u = 0

Time, t = 18 s

Distance, d = 390 m

We need to find the acceleration of the car. Let it is a. Using the second equation of motion to find it.

d=ut+\dfrac{1}{2}at^2

or

d=\dfrac{1}{2}at^2\\\\a=\dfrac{2d}{t^2}\\\\a=\dfrac{2\times 390}{(18)^2}\\\\a=2.4\ m/s^2

So, the acceleration of the car is 2.4\ m/s^2.

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An object with a mass of 5.0 kg accelerates 8.0 m/s^2 when an unknown force is applied to it. What is the acceleration? (20 poin
Readme [11.4K]

Answer:

40N

Explanation:

F=ma

F= 5 × 8 = 40N

.......

4 0
3 years ago
Suggest a situation where we can obtain more than one shadow of an object at a time?
Murljashka [212]

Answer:

We can obtain more than one shadow of an object if light from more than one source falls on it. [For example during a match being played in a stadium multiple shadows of players are seen].

Explanation:

nah-

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3 years ago
As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arr
Nitella [24]

Answer:

  • <u><em>1. Part A: 648N</em></u>
  • <u><em></em></u>
  • <u><em>2. Part B: 324J</em></u>
  • <u><em></em></u>
  • <u><em>3. Part C: 1,296N</em></u>

Explanation:

<em><u>1. Part A:</u></em>

The magnitude of the force is calculated using the Hook's law:

          |F|=k\Delta x

You know \Delta x=0.250m but you do not have k.

You can calculate it using the equation for the work-energy for a spring.

The work done to compress the springs a distance \Delta x is:

          Work=\Delta PE=(1/2)k(\Delta x)^2

Where \Delta PE is the change in the elastic potential energy of the "spring".

Here you have two springs, but you can work as if they were one spring.

You know the work (81.0J) and the length the "spring" was compressed (0.250m). Thus, just substitute and solve for k:

             81.0J=(1/2)k(0.250m)^2\\\\k=2,592N/m

In reallity, the constant of each spring is half of that, but it is not relevant for the calculations and you are safe by assuming that it is just one spring with that constant.

Now calculate the magnitude of the force:

         |F|=k\Delta x=2,592N/m\times 0.250m=648N

<u><em></em></u>

<u><em>2. Part B. How much additional work must you do to move the platform a distance 0.250 m farther?</em></u>

<u><em></em></u>

The additional work will be the extra elastic potential energy that the springs earn.

You already know the elastic potential energy when Δx = 0.250m; now you must calculate the elastic potential energy when  Δx = 0.250m + 0.250m = 0.500m.

          \Delta E=(1/2)2,592n/m\times(0.500m)^2=324J

Therefore, you must do 324J of additional work to move the plattarform a distance 0.250 m farther.

<em><u></u></em>

<em><u>3. Part C</u></em>

<u><em></em></u>

<u><em>What maximum force must you apply to move the platform to the position in Part B?</em></u>

The maximum force is when the springs are compressed the maximum and that is 0.500m

Therefore, use Hook's law again, but now the compression length is Δx = 0.500m

           |F|=k\Delta x=2,592N/m\times 0.500m =1,296N

8 0
3 years ago
Please help me with this one
katen-ka-za [31]

Both of the pictures show how one picture is I believe in fall and the other one is in the forest and maybe in spring or summer. Also, the two pictures show 2 different animals. And maybe hunting for 2 different foods for them to survive.

* Hopefully this helps:) Mark me the brainliest:)

5 0
3 years ago
Aristotle believed which of the following to be true?
Bumek [7]

Answer:

number 2

Explanation:

5 0
4 years ago
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