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VladimirAG [237]
3 years ago
11

What happens to the velocity and the acceleration of the car when the car increases in speed?

Physics
1 answer:
saul85 [17]3 years ago
3 0

Answer:

it goes faster

Explanation:

acceleration = +

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Which one is itttttttttttt
Vlada [557]

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Red

Explanation:

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3 years ago
A mass executes SHM at the end of a light spring. (a) What fraction of the total energy of the system is potential and what frac
PolarNik [594]

Answer:

Explained

Explanation:

A) The total energy of the system is defined by the energy at maximum amplitude, which we'll call A. At that point, the energy of the system is

E = 1/2×m×A^2;

since energy is conserved, this is also the total amount of energy that the system ever has.

So at x=1/2A,

the potential energy of the system is 1/8×m×A^2

which is one-fourth of the system's total energy. Therefore, the remaining three-fourths is kinetic.

B) (i) Doubling the maximum amplitude will quadruple the total energy:

E= \frac{1}{2}m(2A)^2

(ii) Doubling the maximum amplitude will double the maximum velocity

\frac{1}{2}m(2A)^2= \frac{1}{2}mV^2

(iii) Doubling the maximum amplitude will double the maximum acceleration: m×a = -k(2A)

(iv) Doubling the maximum amplitude leaves the period unchanged:

T= 2\pi\sqrt{\frac{m}{k} }

(neither m nor k has changed).

6 0
3 years ago
You push a cart with mass 30 kg forward, giving it an acceleration of 5 m/s2How much force did you apply? O A. 0.17 N O B. 35 N
umka21 [38]

Answer:

150N

Explanation:

To find the force applied remember that force equals to Mass multiply by the Acceleration.

F=Ma

=30*5

=150N

3 0
3 years ago
Which of the following are correct statements about the way an atom is put
Ann [662]

Answer:

valenc e shell

Explanation:

4 0
3 years ago
Sam heaves a 16lb shot straight upward, giving it a constant upward acceleration from rest of 35 m/s^2 for 64.0 cm. He releases
makvit [3.9K]

Answer:

6.69 m/s

4.483 m

1.42s

Explanation:

Given that:

Initial Velocity, u = 0

Final velocity, v =?

Acceleration, a = 35m/s²

1.) using the relation :

v² = u² + 2as

v² = 0 + 2(35) * 64*10^-2m

v² = 70 * 0.64

v = sqrt(44.8)

v = 6.693

v = 6.69 m/s

B.) height from the ground, h0 = 2.2

How high ball went , h:

Using :

v² = u² + 2as

Upward motion, g = - ve

0 = 6.69² + 2(-9.8)*(h - 2.2)

0= 6.69² - 19.6(h - 2.2)

44.7561 + 43.12 - 19.6h = 0

19.6h = 44.7561 - 43.12

h = 87.8761 / 19.6

h = 4.483 m

C.)

vt - 0.5gt² = h - h0

6.69t - 0.5(9.8)t²

6.69t - 4.9t² = 1.83 - 2.2

-4.9t² + 6.69t + 0.37 = 0

Using the quadratic equation solver :

Taking the positive root:

1.4185 = 1.42s

5 0
3 years ago
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