The

A 65 N boy sits on a sled weighing 52 N on a horizontal surface. The coefficient of friction between the sled and the snow is 0.

pogonyaev

Answer:

1.40 N

Explanation:

The magnitude of the frictional force is given by:

$F=\mu N$

where

$\mu$ is the coefficient of friction

N is the magnitude of the normal reaction

The coefficient of friction for this problem is $\mu=0.012$ . The magnitude of the normal reaction is equal to the combined weight of the boy and the sled, because the surface is horizontal, so

$N=65 N+52 N=117 N$

Therefore, the frictional force is

$F=\mu N=(0.012)(117 N)=1.40 N$

3 0

3 years ago

A small bouncy ball with a momentum of 8 kg∙m/s to the left approaches head-on a large door at rest. The ball bounces straight b

vlada-n [284]

Answer:

-14 kg m/s

Explanation:

Taking the direction "to the left" as positive direction, the initial momentum of the ball is

p1 = +8 kg m/s

while the final momentum is

p2 = -6 kg m/s

so the change in momentum is

$\Delta p = p_2 - p_1 = -6kg m/s - (8 kg m/s) = -14 kg m/s$

According to the impulse theorem, the impulse exerted on the ball is equal to the change in momentum of the ball, so:

$I_1 = -14 kg m/s$ (which means 14 kg m/s to the right)

While the impulse that the ball exerted on the ball is equal and opposite in direction, so:

$I_2 = + 14 kg m/s$ (which means towards the left)

4 0

4 years ago

How many Medals in the Luge does the U.S. have?

Dafna11 [192]

Answer:

American athletes have won a total of 2,523 medals (1,022 of them gold) at the Summer Olympic Games and another 305 (105 of them gold) at the Winter Olympic Games, making the United States the most prolific medal-winning nation in the history of the Olympics.

Explanation:

Hope this helps! ^^

7 0

3 years ago

A centrifuge accelerates uniformly from rest to 15000 rpm in 330 s . Through how many revolutions did it turn in this time?

eduard

Answer:

The number of revolutions turned by the centrifuge is 8250 revolutions.

Explanation:

Given;

number of revolution per minutes, ω = 15000 rpm

time of motion, t = 330 s = 5.5 minutes

The number of revolutions turned by the centrifuge is given by;

$N = \frac{1500 \ Rev}{minutes} *5.5 \ minutes\\\\N = 8250 \ revolutions$

Therefore, the number of revolutions turned by the centrifuge is 8250 revolutions.

4 0

3 years ago

Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g

tekilochka [14]

Answer: a) It will take more time to return to the point from which it was released

Explanation: To determine how long it takes for the ball to return to the point of release and considering it is a free fall system, we can use the given formula:

$d=v_{0}.t + \frac{1}{2} .a.t^{2}$ , where:

d is the distance the ball go through;

v₀ is the initial velocity, which is this case is 0 because he releases the ball;

a is acceleration due to gravity;

t is the time necessary for the fall;

Suppose <em>h</em> is the height from where the ball was dropped.

On Earth:

h=0.t + $\frac{1}{2}.10.t^{2}$

h = 5t²

$t_{T}$ = $\sqrt{\frac{h}{5} }$

On the other planet:

h = 0.t + $\frac{1}{2}.30.t^{2}$

h = 15.t²

$t_{P}$ = $\sqrt{\frac{h}{15} }$

Comparing the 2 planets:

$\frac{t_{T} }{t_{P} } = \frac{\sqrt{\frac{h}{5} } }\sqrt{{\frac{h}{15} } }$

$\frac{t_{T} }{t_{P} }$ = $\sqrt{3}$ or $t_{T} = \sqrt{3}.t_{P}$

Comparing the two planets, on the massive planet, it will take more time to fall the height than on Earth. In consequence, it will take more time to return to the initial point, when it was released.

5 0

3 years ago