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Lynna [10]
4 years ago
7

How does uniform motion relate to velocity and acceleration?

Physics
1 answer:
nekit [7.7K]4 years ago
8 0

-- Velocity is the speed and direction of motion.

-- Acceleration is any change of velocity ...
. . . speeding up, slowing down, or changing direction.

-- Uniform motion means no change ...
. . . straight line motion at constant speed
. . . constant velocity and zero acceleration.
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As you slide a heavy box across the floor, friction applies a force of -100 N
nirvana33 [79]

Answer:

A -500J

Explanation:

because W=Fs

100 × 5 = 500

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3 years ago
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What is quantity?explain.
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Answer:

the amount or number of a material or immaterial thing not usually estimated by spatial measurement

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How much potential energy does a 40-N medicine ball gain when it is lifted 5 m?
777dan777 [17]
We know, Potential Energy = Force * Height
Here, F = 40 N
h = 5 m

Substitute their values, 
U = 40 * 5
U = 200 J

In short, Your Answer would be Option A

Hope this helps!
6 0
3 years ago
A particle moves along the x axis. It is initially at the position 0.180 m, moving with velocity 0.060 m/s and acceleration -0.3
shusha [124]

Answer:

a

  x_2 = -2.3356

b

 v = -1.384 \ m/s

Explanation:

From the question we are told that

  The initial position of the particle is  x_1 = 0.180 \ m

  The initial  velocity of the particle is  u = 0.060  \  m/s

  The acceleration is   a = -0.380 \  m/s^2

   The time duration is  t = 3.80  \ s

Generally from kinematic equation

    v = u + at

=>  v = 0.060 + (-0.380 * 3.80)

=>  v = -1.384 \ m/s

Generally from kinematic equation

   v^2 = u^2 + 2as

Here s is the distance covered by the particle, so

   (-1.384)^2 = (0.060)^2 + 2(-0.380)* s

=>  s = -2.5156 \ m

Generally the final position of the particle is  

    x_2 = x_1 + s

=>   x_2 = 0.180 + (-2.5156)

=>   x_2 = -2.3356

6 0
3 years ago
A cart on an air track is moving at 0.5m/s when the air issuddenly turned off. THe cart comes to rest in 1m. The experimentis re
Licemer1 [7]

Answer:

D = 4 m

Explanation:

Speed of cart in air track v₁ = 0.5 m/s

Speed of cart moved when air is turned off v₂= 1 m/s

The distance travelled by the cart is d₁ = 1 m

Work done (W) = F x d

Work done is equal to the kinetic energy

F x d = 1/2mv²

velocity is directly proportional to distance

therefore,

v₁²/ v₂²  = d₁ / d₂

d₂ = d₁v₂² / v₁²

= 1 m x (1 m /s)² / (0.5 m/s)²

= 4 m

5 0
3 years ago
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