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dlinn [17]
1 year ago
8

What constant acceleration is required to increase the speed of a car from 27 mi/h to 53 mi/h in 3s?

Physics
1 answer:
mario62 [17]1 year ago
7 0

Answer:

8.87 m/s^2  

Explanation:

Acceleration is change in velocity over a change in time

27 to 53 = 26 m/s   change in velocity

26 m/s / 3 s = 8.87 m /s^2

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An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at 12.8 m/s2 . At t1 the rocket e
Simora [160]

Answer:4.39 s

Explanation:

Given

initial velocity u=0

acceleration a=12.8 m/s^2

velocity acquired by sled in t_1 time

v=0+at

v=12.8t_1

distance traveled by sled in t_1 s

v^2-u^2=2as

(12.8t_1)^2-0=2\times 12.8\times s_1

s_1=6.4\cdot t_1^2

distance traveled in t_2 time with velocity v=12.8t_1

s_2=v\times t_2

s_2=12.8\times t_1\times t_2

s_2=12.8\cdot t_1\cdot t_2

s_1+s_2=5.37\times 10^3

6.4t_1^2+12.8t_1t_2=5370----1

t_1+t_2=97.7 s

t_2=97.7-t_1

substitute the value of t_2 in 1

we get

6.4t_1^2-1250.56t_1+5370=0

thus t_1=\frac{1250.56-1194.33}{12.8}=4.39 s

t_1=4.39 s

5 0
3 years ago
What is the relationship between the angle of an incline and the acceleration of an object moving down the incline? How would yo
iren2701 [21]

Answer:

See Explanation

Explanation:

The relationship between angle of an incline and the acceleration of an object moving down the incline.

As the angle of an incline increases, so does the acceleration of the body moving down the incline increases, resolving the force acting on an inclined object

Parallel force = mgsin, perpendicular = mgcosΘ

With th weigh component 'mg' of the parallel force accounting for the acceleration of the body down the incline.

mgsinΘ = ma

Fnet = ma

B.) From Fnet = ma

Fnet = ma

a = Fnet / m

Where Fnet = Net force = mgsinΘ, a = acceleration

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Minchanka [31]
C is the answer because it mix and water combining together
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A 4000kg truck has a head-on inelastic collision with a 2500kg truck.
iogann1982 [59]

Answer:it could be B

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im not sure

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