how does the frequency of a radio wave compare to the frequency of the vibrating electrons that produce it?

A race car accelerates at a constant rate from 20 m/s to 60 m/s in a time of 2 seconds. What is the car’s acceleration?

Vlada [557]

$\huge\fbox{Answer ☘}$

Given ,

initial velocity , u = 20 m/s

final velocity , v = 60 m/s

time taken = 2 seconds

Now ,

$\bold{acceleration = \frac{v - u}{t} } \\ \\ = > \bold{ \frac{60 - 20}{2} } \\ \\ = > \bold{\frac{40}{2} } \\ \\ \bold{ => 20 \: ms {}^{ - 2}}$

hope helpful~

5 0

2 years ago

Danny Diver weighs 500 N and steps off a diving board 10 m above the water. Danny hits the water with kinetic energy of

Morgarella [4.7K]

Answer:

Danny hits the water with kinetic energy of 5000 J.

Explanation:

Given that,

The Weight of Danny Diver,

F = 500 N

m*g= 500 N

He steps off a diving board 10 m above the water.

h=10 m

when Danny diver hits water he generates the kinetic energy.

We need to find the kinetic energy of the water.

Let kinetic energy is K.

K = m*g*h

Where g is acceleration due to gravity.

that g= 9.8 m/s^2

now substituting the values in above equation

K= (500) * 10

K= 5000 J

Hence,

he hits the water with kinetic energy of 5000 J.

Learn more about Kinetic energy here:

<u>brainly.com/question/15587458</u>

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7 0

2 years ago

8. What is the mass of an object if a force of 34 N produces an acceleration of 4 m/s/s?

coldgirl [10]

Answer:

<h3>The answer is 8.5 kg</h3>

Explanation:

The mass of the object can be found by using the formula

$m = \frac{f}{a} \\$

where

f is the force

a is the acceleration

So we have

$m = \frac{34}{4} = \frac{17}{2} \\$

We have the final answer as

Hope this helps you

7 0

3 years ago

Read 2 more answers

An object is thrown directly downward from the top of a very tall building. The speed of the object just as it is released is 17

Softa [21]

Answer:

distance cover is = 102.53 m

Explanation:

Given data:

speed of object is 17.1 m/s

$t_1 = 3.32 sec$

$t_2 = 5.08 sec$

from equation of motion we know that

$d_1 = vt_1 + \frac{1}{2} gt_1^2$

where d_1 is distance covered in time t1

so $d_1 = 17.1 \times 3.32 + \frac{1}{2} 9.8 \times 3.32^2$ =

$d_1 = 110.78 m$

$d_2 = vt_2 + \frac{1}{2} gt_2^2$

where d_2 is distance covered in time t2

$d_2 = 17.1 \times 5.08 + \frac{1}{2}\times9.8 \times 5.08^2$

$d_2 = 213.31 m$

distance cover is = 213.31 - 110.78 = 102.53 m

3 0

3 years ago

A Carnot engine operates with an efficiency of 26.0% when the temperature of its cold reservoir is 281 K. Assuming that the temp

VLD [36.1K]

Answer:

The temperature of cold reservoir should be 246.818 K for efficiency of 35%

Explanation:

In first case we have given efficiency of Carnot engine = 26 % = 0.26

Temperature of cold reservoir $T_L=281K$

We know that efficiency of Carnot engine is given by

$\eta =1-\frac{T_L}{T_H}$

$0.26 =1-\frac{281}{T_H}$

$T_H=379.72K$

For second Carnot engine efficiency is given as 35% = 0.35

And temperature of hot reservoir is same so $T_H=379.72K$

So $0.35=1-\frac{T_L}{379.72}$

$T_L=246.818K$

So the temperature of cold reservoir should be 246.818 K for efficiency of 35%

4 0

3 years ago