All categories

3 years ago

Newton would agree that all objects on Earth exert a

Physics

1 answer:

Dafna11 [192]3 years ago

5 0

The answer is true .....

You might be interested in

What is a key phrase for remembering the order of the planets?

horsena [70]

My very eager mother just served us nine pizzas

7 0

2 years ago

How long have advances in astronomy been occurring? Only for the past few years

solniwko [45]

I'd have to say that the list of choices doesn't go far enough.

Advances in Astronomy have been occurring for at least the past two millennia (2000 years). Maybe longer.

3 0

3 years ago

IF YOU ANSWER THESE 4 QUESTIONS! I WILL GIVE YOU BRAINEST!!! 17 POINTS!!!

melisa1 [442]

Answer:

1.d

2.d

3.c

4.c

Okay, d & d, c & c. Iknow what you're thinking, I guessed. Nope! I just took a good look and chosed what I thought was the best answer

Explanation:

Have a great day! Byeeee

4 0

2 years ago

WATS THE BEST ANIME <br>VERY IMPORTANT PLS ANSWER!!!!!!!

WARRIOR [948]

There isn’t a best anime , but in my opinion any anime on Netflix is good other than DBZ too many episodes and storyline gets messed up.

6 0

3 years ago

You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t

Sladkaya [172]

Answer:

0.767

Explanation:

The work done on the truck by the frictional drag force is given by

$W=-Fd$

where

F is the magnitude of the frictional force

d = 38.0 m is the maximum displacement allowed for the truck

The negative sign is due to the fact that the force of friction is opposite to the motion of the truck

The force of friction can also be written as:

$F=\mu mg$

where

$\mu$ is the coefficient of kinetic friction between the truck and the lane

m is the mass of the truck

g is the acceleration of gravity

So we can rewrite the work done as

$W=-\mu mg d$ (1)

According to the work-energy theorem, the work done by friction is equal to the change in kinetic energy of the truck:

$W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (2)

where

v = 0 is the final velocity of the truck

u = 23.9 m/s is the initial velocity of the truck

By combining (1) and (2) we get

$-\frac{1}{2}mu^2 = -\mu mg d$

And solving for $\mu$ , we find the minimum coefficient of kinetic friction able to stop the truck in a distance d:

$\mu = \frac{u^2}{2gd}=\frac{23.9^2}{2(9.8)(38.0)}=0.767$

7 0

3 years ago