3 years ago

He wrote a book of aerobics

Physics

2 answers:

vampirchik [111]3 years ago

4 0

Answer:

George Washington

Explanation:

ifk

GaryK [48]3 years ago

3 0

Answer:

kenneth H. cooper sorry i need 20 mores letters

You might be interested in

A wire is stretched just to its breaking point by a force F . A longer wire made of the same material has the same diameter. The

ICE Princess25 [194]

Answer: No

Explanation:

The force F required is equal to the Force exerted in stretching the first material since conditions are the same

8 0

3 years ago

Read 2 more answers

The sensitivity of a measuring instrument is the value of the smallest quantity that can be read or estimated with it. What is t

nirvana33 [79]

Answer:

The smallest part of a millimeter that can be read with a digital caliper with a four digit display is 0.02mm. Thus, it has to be converted to centimetre. So, divide by 10, we then have 0.02/10= *0.002cm* not mm.

6 0

4 years ago

What does the REVOLUTION of Earth around the Sun bring us and how long does it take?

olasank [31]

Answer:

takes 365 days and it bring us seasons such as, spring,winter and fall.

7 0

3 years ago

(???) Is iz allergic to peanut butter jigglys, I JIGGLE IN THE WIND!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Vadim26 [7]

Uh are u okay? G tell ur parents i-

7 0

3 years ago

Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri

Ainat [17]

Answer:

The taken is $t_A = 19.0 \ s$

Explanation:

Frm the question we are told that

The speed of car A is $v_A = 22 \ m/s$

The speed of car B is $v_B = 29.0 \ m/s$

The distance of car B from A is $d = 300 \ m$

The acceleration of car A is $a_A = 2.40 \ m/s^2$

For A to overtake B

The distance traveled by car B = The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is

$d = v_B * t_A$

Where $t_B$ is the time taken by car B

Now this can also be represented as using equation of motion as

$d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300$

Now substituting values

$d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

Equating the both d

$v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

substituting values

$29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A =1.2 t_A^2 - 300$

$1.2 t_A^2 - 7 t_A - 300 = 0$

Solving this using quadratic formula we have that

$t_A = 19.0 \ s$

7 0

3 years ago