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Romashka [77]
3 years ago
14

Is it true that acceleration occurs when there is a change in speed

Physics
2 answers:
Amanda [17]3 years ago
4 0
Yes it does. But not always
Bess [88]3 years ago
4 0

Answer:

Acceleration is the rate of change of velocity.

Explanation: For example, acceleration is measured in m/2^2 or meters per second squared. That can also be split up into 2 components. M/s is how speed is measured, and since acceleration is the rate of change of velocity, m/(s*s). You could say that since acceleration is m/s^2, it can also be represented as how much the speed increases over 1 second. if there was an acceleration of 3 m/s^2, the rate of change of speed would be 3m/s every second, meaning that the object would speed up 3m/s every second that passes by.

Hope this helps!

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Guys help...A spring that obeys Hooke's law, with a spring constant k1, is cut into N identical springs, each with a spring cons
Tems11 [23]

Answer:

K2 = N*K1

Explanation:

The force you apply to each section is the same you apply to the whole spring, but the extension of each section is dX/N (if dX is the extension of the entire spring)

4 0
3 years ago
Convertir:<br> A. 3Km a m<br> B. 250 ma Km<br> C. 1000Cm a m<br> D. 10000 mm a Cm
Katen [24]

Answer:

A. 3,000,000 m

B. 0.25 km

C. 10 m

D. 1,000 cm

Explanation:

no hablo español, así que solo ingrese esto en el traductor de G*ogle

A. One kilometer equals 1000 meters, so

3,000*1,000 = 3,000,000 m

B. One meter equals 0.001 kilometer, so

250*0.001 = 0.25 km

C. One centimeter equals 0.01 meter

1,000*0.01 = 10 m

D. One milimeter equals 0.1 centimer, so

10,000*0.1 = 1,000

4 0
2 years ago
A car accelerates for 10 seconds. During this time, the angular
bagirrra123 [75]

Answer:

the angular acceleration of the car is 1.5 rad/s²

Explanation:

Given;

initial angular velocity, \omega_i = 10 rad/s

final angular velocity, \omega_f = 25 rad/s

time of motion, t = 10 s

The angular acceleration of the car is calculated as follows;

a_r = \frac{\omega_f - \omega_i }{t} \\\\a_r = \frac{25-10}{10} = 1.5 \ rad/s^2

Therefore, the angular acceleration of the car is 1.5 rad/s²

4 0
2 years ago
A train, traveling at a constant speed of 25.8 m/s, comes to an incline with a constant slope. While going up the incline, the t
zhannawk [14.2K]

The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

Answer:

Explanation:

So in this problem, the initial speed of the train is at 25.8 m/s before it comes to incline with constant slope. So the acceleration or the rate of change in velocity while moving on the incline is given as 1.66 m/s². So the final velocity need to be found after a time period of 8.3 s. According to the first equation of motion, v = u +at.

So we know the values for parameters u,a and t. Since, the train slows down on the slope, so the acceleration value will have negative sign with the magnitude of acceleration. Then

v = 25.8 + (-1.66×8.3)

v =12.022 m/s.

So the final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

8 0
2 years ago
A force of 10N is required to stretch a spring from 20cm to 25cm. What is the spring constant in N/m2 Be careful of unit
kupik [55]

Answer:

C) 40 N/m

Explanation:

If we ASSUME that the spring is un-stretched at the zero cm position

k = F/Δx = 10/0.25 = 40 N/m

5 0
2 years ago
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