What condition is necessary for an object to make a good reference point?

if a car has a kinetic energy of 400 J at 20m/s what.is the cars kinetic energy in kJ at a speed of 5 m/s

Nutka1998 [239]

Kinetic energy is an expression of the fact that a moving object can do work on anything it hits; it quantifies the amount of work the object could do as a result of its motion. It can be calculated as:

KE = mv^2 /2

We calculate the problem as follows:

KE = mv^2 /2

400 = m(20)^2 / 2

8 0

3 years ago

a person throws a ball in such a way that its speed is zero at one particular point in its path. How did the person throw the ba

Keith_Richards [23]

vertical! or in other words, up!

6 0

4 years ago

An old clock has a spring that must be wound to make the clock hands move. Which statement describes the energy of the spring an

kondaur [170]

<span>D) The spring has potential energy and the hands have kinetic energy. </span>

4 0

3 years ago

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A solid conducting sphere is given a positive charge q. How is the charge q distributed in or on the sphere?

leva [86]

Answer:

Explanation:

the sphere is solid and conducting, so the charge is uniformly distributed over its volume.

7 0

3 years ago

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A car traveling at speed v takes distance d to stop after the brakes are applied. What is the stopping distance if the car is in

Vikki [24]

49d

<h3>Further explanation</h3>

This case is about uniformly accelerated motion.

<u>Given:</u>

The initial speed was v takes distance d to stop after the brakes are applied.

<u>Question:</u>

What is the stopping distance if the car is initially traveling at speed 7.0v?

Assume that the acceleration due to the braking is the same in both cases. Express your answer using two significant figures.

<u>The Process:</u>

The list of variables to be considered is as follows.

$\boxed{u \ or \ v_i = initial \ velocity}$
$\boxed{u \ or \ v_t \ or \ v_i = terminal \ or \ final \ velocity}$
$\boxed{a = acceleration \ (constant)}$
$\boxed{d = distance \ travelled}$

The formula we follow for this problem are as follows:

$\boxed{ \ v^2 = u^2 + 2ad \ }$

a = acceleration (in m/s²)
u = initial velocity
v = final velocity
d = distance travelled

Step-1

We substitute v as the initial speed, distance of d, and zero for final speed into the formula.

$\boxed{ \ 0 = v^2 + 2ad \ }$

$\boxed{ \ v^2 = -2ad \ }$

Both sides are divided by -2d, we get $\boxed{ \ a = \Big( -\frac{v^2}{2d} \Big) \ . . . \ (Equation-1) \ }$

Step-2

We substitute 7.0v as the initial speed, zero for final speed, and Equation-1 into the formula.

$\boxed{ \ 0 = (7.0v)^2 + 2 \Big( -\frac{v^2}{2d} \Big)d' \ }$

Here d' is the stopping distance that we want to look for.

$\boxed{ \ 2 \Big( \frac{v^2}{2d} \Big)d' = (7.0v)^2 \ }$

We crossed out 2 in above and below.

$\boxed{ \ \Big( \frac{v^2}{d} \Big)d' = 49.0v^2 \ }$

We multiply both sides by d.

$\boxed{ \ v^2 d' = 49.0v^2 d \ }$

We crossed out v^2 on both sides.

$\boxed{\boxed{ \ d' = 49.0d \ }}$

Hence, by using two significant figures, the stopping distance if the car is initially traveling at speed 7.0v is 49d.

<h3>Learn more</h3>

Determine the acceleration of the stuffed bear brainly.com/question/6268248
Particle's speed and direction of motion brainly.com/question/2814900
About the projectile motion brainly.com/question/2746519

Keywords: a car traveling at speed v, takes distance d to stop after the brakes are applied, the stopping distance, if the car is initially traveling at speed 7.0v, the acceleration due to the braking is the same, two significant figures.

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3 years ago

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