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Marrrta [24]
3 years ago
6

A car traveling at 25 meters per second is brought to stop by a force over a 10 second timer interval. What was the acceleration

of the car during the 10 second time interval.
Physics
1 answer:
nordsb [41]3 years ago
6 0

Answer:

–2.5 m/s²

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 25 m/s

Final velocity (v) = 0 m/s

Time (t) = 10 s

Acceleration (a) =?

Acceleration can simply be defined as the rate of change of velocity with time. Mathematically, it can be expressed as:

a = (v – u) /t

Where:

a => is the acceleration

v => is the final velocity

u => is the initial velocity

t => is the time.

With the above formula, we can obtain the acceleration of the car as follow:

Initial velocity (u) = 25 m/s

Final velocity (v) = 0 m/s

Time (t) = 10 s

Acceleration (a) =?

a = (v – u) /t

a = (0 – 25) /10

a = –25 / 10

a = –2.5 m/s²

The acceleration of the car is –2.5 m/s². The negative sign indicate that the car is decelerating i.e coming to rest.

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Doug rubs a piece of fur on a hard rubber rod, giving the rod a negative charge. Which of the following statements best describe
Leto [7]

Answer:

B is correct. Electrons are added to the rod.

Explanation:

Because the fur lose electrons to the rode and because positively charged while the rod because negative

7 0
3 years ago
Tom has built a large slingshot, but it is not working quite right. He thinks he can model the slingshot like an ideal spring wi
Amiraneli [1.4K]

Answer:

8.9

Explanation:

We can start by calculating the initial elastic potential energy of the spring. This is given by:

U=\frac{1}{2}kx^2 (1)

where

k = 35.0 N/m is the initial spring constant

x = 0.375 m is the compression of the spring

Solving the equation,

U=\frac{1}{2}(35.0)(0.375)^2=2.5 J

Later, the professor told the student that he needs an elastic potential energy of

U' = 22.0 J

to achieve his goal. Assuming that the compression of the spring will remain the same, this means that we can calculate the new spring constant that is needed to achieve this energy, by solving eq.(1) for k:

k'=\frac{2U'}{x^2}=\frac{2(22.0)}{0.375^2}=313 N/m

Therefore, Tom needs to increase the spring constant by a factor:

\frac{k'}{k}=\frac{313}{35}=8.9

7 0
4 years ago
Electromagnets are created by
Scrat [10]
I electromagnets, the magnetic field is turned off when there is no induced current. So, electromagnets act as magnets only when current is induced in an insulated wire wrapped around a material mage of ferromagnetic material such as iron core rotating in a magnetic field.

So, option D is the correct one.



If you wanna know more about electromagnets, write down in comments {:
6 0
3 years ago
Read 2 more answers
A lamp is 10% efficient. How much electrical energy must be supplied to the lamp each second if it produces 20 J of light energy
Leto [7]

Answer:

200J

Explanation:

200J is 100%, 10% of 200J would be 20J

3 0
3 years ago
Work and Energy
Sergeeva-Olga [200]

a) W_1 = 2332 J, W_2= 2332 J

The work done by the student in each trial is equal to the gravitational potential energy gained by the student:

W=mg\Delta h

where

m = 68 kg is the mass of the student

g = 9.8 m/s^2 is the acceleration of gravity

\Delta h is the gain in height of the student

For the first student, \Delta h = 3.5 m, so the work done is

W_1 = (68)(9.8)(3.5)=2332 J

The second student runs up to the same height (3.5 m), so the work done by the second student is the same:

W_2 = (68)(9.8)(3.5)=2332 J

2) P_1 = 204.6 W, P_2 = 274.4 W

The power exerted by each student is given by

P=\frac{W}{t}

where

W is the work done

t is the time taken

For the first student, W_1 = 2332 J and t=11.4 s, so the power exerted is

P_1 = \frac{W_1}{t_1}=\frac{2332 J}{11.4 s}=204.6 W

For the second student, W_2 = 2332 J and t=8.5 s, so the power exerted is

P_2 = \frac{W_2}{t_2}=\frac{2332 J}{8.5 s}=274.4 W

5 0
3 years ago
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