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3 years ago

A constant friction force of 21 N acts on a 63-kg skier for 25 s on level snow. What is the skier’s change in velocity?

Physics

1 answer:

erica [24]3 years ago

4 0

The skier’s change in velocity is 8.33.../8.4 m/s.
21/63= 0.33...
0.33...x 25 s = 8.33.../8.4 m/s

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A 2-kg wheel rolls down the road with a linear speed of 15m/s. Find its trwansitional and rotational kinetic energies.

Elza [17]

Answer:

The translational kinetic energy is 225 J

The rotational kinetic energy is 225 J

Explanation:

Given;

mass of the wheel, m = 2-kg

linear speed of the wheel, v = 15 m/s

Transnational kinetic energy is calculated as;

E = ¹/₂MV²

where;

M is mass of the moving object

V is the velocity of the object

E = ¹/₂ x 2 x (15)²

E = 225 J

Rotational kinetic energy is calculated as;

E = ¹/₂Iω²

where;

I is moment of inertia

ω is angular velocity

$E = \frac{1}{2} I \omega^2\\\\E = \frac{1}{2} *mr^2*(\frac{v}{r})^2\\\\E = \frac{1}{2} *mr^2*\frac{v^2}{r^2} \\\\E = \frac{1}{2}mv^2$

E = ¹/₂ x 2 x (15)²

E = 225 J

Thus, the translational kinetic energy is equal to rotational kinetic energy

6 0

3 years ago

Option B.

Bezzdna [24]

Answer:

it would be...

Explanation:

7 0

2 years ago

USE K U E S

nata0808 [166]

Explanation:

Problem 1.

Initial speed of the runner, u = 0

Acceleration of the runner, $a=4.2\ m/s^2$

Time taken, t = 100 s

Let v is the speed of the runner now. Using the first equation of kinematics as :

$v=u+at$

$v=at$

$v=4.2\ m/s^2\times 100\ s$

v = 420 m/s

Problem 2.

Initial speed of the plane, u = 0

Distance covered, d = 300 m

Time taken, t = 25 s

Using the equation of kinematics as :

$d=ut+\dfrac{1}{2}at^2$

$d=\dfrac{1}{2}at^2$

$a=\dfrac{2d}{t^2}$

$a=\dfrac{2\times 300\ m}{(25\ s)^2}$

$a=0.96\ m/s^2$

Problem 3.

A ball free falls from the top of the roof for 5 seconds. Let it will fall at a distance of d. It is given by :

$d=ut+\dfrac{1}{2}gt^2$

$d=\dfrac{1}{2}\times 9.8\times (5)^2$

d = 122.5 meters

Let v is the final speed at the end of 5 seconds. Again using first equation of kinematics as :

$v=u+gt$

$v=9.8\ m/s^2\times 5\ s$

v = 49 m/s

Hence, this is the required solution.

3 0

3 years ago

A person walks at 2 metres per second for 8 seconds then at 1.5 metres per second for 8 seconds. What is the average speed

beks73 [17]

2 meters per second for 8 seconds

1.5 meters per second for 8 seconds

The average speed should be 1.8 or 1.7 but I think its 1.8

7 0

3 years ago

Simple question please help! A ball is thrown 19.0 m/s at an angle of 40.0º with the horizontal. Assume the ball is thrown at gr

d1i1m1o1n [39]

a. 1.51 s

In this part of the problem, we are only interested in the horizontal motion of the ball. Along the horizontal direction, the motion of the ball is a uniform motion with constant velocity, which is equal to the horizontal component of the initial velocity:

$v_x = v_0 cos \theta = (19.0 m/s)(cos 40^{\circ})=14.6 m/s$

So, the ball travels horizontally at a speed of 14.6 m/s; in order to cover the distance of d = 22.0 m that separates it from the wall, the time need is

$t=\frac{d}{v_x}=\frac{22.0 m}{14.6 m/s}=1.51 s$

b. 7.25 m

We now know that the ball takes 1.51 s to hit the wall, 22.0 away. Now we have to analyze the vertical motion of the ball, which is an accelerated motion with constant acceleration g =9.8 m/s^2 towards the ground (acceleration due to gravity).

The initial vertical velocity is

$v_{y0}=v_0 sin \theta=(19.0 m/s)(sin 40^{\circ})=12.2 m/s$

The vertical position of the ball at time t is given by the equation

$y(t)=v_{0y} t -\frac{1}{2}gt^2$

We know that the ball hits the wall at t=1.51 s, so if we substitute this value into the previous formula, we find at what height y the ball hits the wall:

$y(1.51 s)=(12.2 m/s)(1.51 s)-\frac{1}{2}(9.8 m/s^2)(1.51 s)^2=7.25 m$

7 0

3 years ago