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tamaranim1 [39]
3 years ago
7

An object, initially at rest, is subject to an acceleration of 34 m/s^2. How long will it take for that object to reach 3400m ?

Physics
1 answer:
Norma-Jean [14]3 years ago
6 0
Vf^2 = Vi^2 + 2ad
a= 34 m/s^2
Vi = 0 m/s
d = 3400m

Vf = 480.83 m/s

a=v/t
t=v/a
t=480.83/34
t=14.142 s
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6.38 x 10^4 J

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The space shuttle usually orbited Earth at altitudes of around 300.0 km. 1) Determine the time for one orbit of the shuttle abou
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You stand on top a building 44 m tall with a water balloon. You drop the water balloon from rest. How fast is the balloon moving
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3 years ago
A 150kg motorcycle starts from rest and accelerates at a constant rate along a distance of 350m. The applied force is 250N and t
notsponge [240]

A) The net force on the motorbike is 205.9 N

B) The acceleration of the motorbike is 1.37 m/s^2

C) The final speed is 5.2 m/s

D) The elapsed time is 3.80 s

Explanation:

A)

There are two forces acting on the motorbike:

- The applied force, F = 250 N, forward

- The frictional force, F_f, backward

The frictional force can be written as

F_f = \mu mg

where

\mu=0.03 is the coefficient of kinetic friction

m=150 kg is the mass of the motorbike

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Therefore the net force is given by

\sum F = F - F_f = F - \mu mg

And substituting, we find

\sum F=250 - (0.03)(150)(9.8)=205.9 N

2)

The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:

\sum F = ma

where

m is the mass

a is the acceleration

In this problem, we have

\sum F = 205.9 N is the net force

m = 150 kg is the mass

Solving for a, we find the acceleration:

a=\frac{\sum F}{m}=\frac{205.9}{150}=1.37 m/s^2

C)

Since the motion of the motorbike is a uniformly accelerated motion, we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For this motorbike, we have:

u = 0 (it starts from rest)

a=1.37 m/s^2

s = 350 m

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(1.37)(9.8)}=5.2 m/s

4)

For this part of the problem, we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

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Here we have:

v = 5.2 m/s

u = 0

a=1.37 m/s^2

Solving for t, we find

t=\frac{v-u}{a}=\frac{5.2-0}{1.37}=3.80 s

Learn more about Newton's laws of motion and accelerated motion:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

6 0
3 years ago
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