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Aleks04 [339]
3 years ago
14

How far will a body move in 4 seconds if uniformly accelerated from rest at rate of 2m\sm\s​

Physics
1 answer:
Lubov Fominskaja [6]3 years ago
3 0

Answer:

16 m

Explanation:

For an object in uniformly accelerated motion (=with constant acceleration), the distance covered by the body can be found by using the following suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the body in this problem,

u = 0 (it starts from rest)

a=2 m/s^2 is the acceleration

Substituting t = 4 s, we find the distance covered:

s=0+\frac{1}{2}(2)(4)^2=16 m

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You walk 12.0 m West and then 4.00 m south. What is your displacement?​
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Answer:

12.6 (3 sig. fig.)

Explanation:

(12^2 + 4^2)^1/2 = 12.6 (3 sig. fig.)

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Plastics are used in millions of objects we use every day. Why are plastics NOT considered natural resources?
jeka57 [31]

Answer:

B

Explanation:

Yes plastic is artficial

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suppose a car manufacturer tested its cars for front end collsion by hauling them up on a crane and dropping them from a certain
IRINA_888 [86]

Initial height: 66.5 m

Explanation:

The problem can be solved by using the principle of conservation of energy.

If we neglect air resistance, the total mechanical energy of the car is conserved during the fall, therefore we can write:

K_i + U_i = K_f + U_f

where :

K_i = 0 is the kinetic energy of the car at the top (it starts from rest)

U_i = mgh is the gravitational potential energy of the car at the top, with:

m = the mass of the car

g = the acceleration of gravity

h = the heigth of the car

K_f = \frac{1}{2}mv^2 is the kinetic energy of the car just before hitting the ground, with

v = 130 km/h final speed of the car

U_f = 0 is the gravitational potential energy of the car at the bottom

Re-arranging the equation,  we find

mgh=\frac{1}{2}mv^2

and we have:

g=9.8 m/s^2\\v = 130 km/h = 36.1 m/s

Solving for h, we find the initial height of the car:

h=\frac{v^2}{2g}=\frac{36.1^2}{2(9.8)}=66.5 m

Learn more about kinetic energy and potential energy:

brainly.com/question/6536722

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5 0
3 years ago
What is 6.9 × 107 written in standard form?
n200080 [17]
Answer: 6.9x 107 in standard form is 69,000,000

7 0
2 years ago
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A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an
3241004551 [841]

Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

F_s = am = 1.31*16 = 20.92 N

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N

So the maximum acceleration on the block is

a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2

4)As the box slides, it is now subjected to kinetic friction, which is

F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

3 0
3 years ago
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