A car slowing down for a stop sign is an example of.....?

Physics

2 answers:

Elza [17]3 years ago

7 0

D-negative Acceleration

maxonik [38]3 years ago

3 0

The answer would be D, negative acceleration

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the kinetic energy of an object with mass m moving with a velocity of 5 m/s is 25 j what will be its kinetic energy when its vel

Dmitriy789 [7]

Answer:

100 J, 225 J

Explanation:

The kinetic energy of an object is given by:

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the velocity of the object

In this problem, the initial kinetic energy of the object is

K = 25 J

Then, the velocity is doubled, which means

v' = 2v

Therefore, the new kinetic energy will be

$K'=\frac{1}{2}m(2v)^2 = 4(\frac{1}{2}mv^2)=4K$

Therefore, the kinetic energy has quadrupled:

$K' = 4(25)=100 J$

Later, the velocity is tripled, which means

v'' = 3v

Therefore, the new kinetic energy will be

$K''=\frac{1}{2}m(3v)^2 = 9(\frac{1}{2}mv^2)=9K$

Therefore, the kinetic energy has increased by a factor of 9:

$K' = 9(25)=225 J$

4 0

4 years ago

Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates

marta [7]

First, create an illustration of the motion of the two cars as shown in the attached picture. The essential equations used are:

For constant acceleration:
a = v,final - v,initial /t
d = v,initial*t + 1/2*at²

For constant velocity:
d = constant velocity*time

The solutions is as follows:

a = v,final - v,initial /t
3.8 = (v₁ - 0)/4.6 s
v₁ = 17.48 m/s

Total distance = d1 + d2 + d3
d1 = d = v,initial*t + 1/2*at²
d2 = constant velocity*time

Total distance = 0*(4.6) + 1/2*(3.8)(4.6)² + (17.48)(9.2) + d3= 257.71
d3 = 56.69 m