Besides gravity, what factor keeps the moon and Earth in orbit?

The wheel of a stationary exercise bicycle at your gym makes one rotation in 0.670 s. Consider two points on this wheel: Point P

lesya [120]

Answer:

a) For P: $v=0.938\frac{m}{s}$

For Q: $v = 1.876\frac{m}{s}$

b) For P:

$a_{rad}=8.80\frac{m}{s^{2}}$

for Q:

$a_{rad}=17.60\frac{m}{s^{2}}$

c) As the distance from the axis increases then speed increases too.

Explanation:

a) Assuming constant angular acceleration we can find the angular speed of the wheel dividing the angular displacement θ between time of rotation:

$\omega =\frac{\theta}{t}$

One rotation is 360 degrees or 2π radians, so θ=2π

$\omega =\frac{2\pi}{0.670} =9.38\frac{rad}{s}$

Angular acceleration is at every point on the wheel, but speed (tangential speed) is different and depends on the position (R) respect the rotation axis, the equation that relates angular speed and speed is:

$v = \omega R$

for P:

$v = 9.38\frac{rad}{s}*0.1m=0.938\frac{m}{s}$

for Q:

$v = 9.38\frac{rad}{s}*0.2m=1.876\frac{m}{s}$

b) Centripetal acceleration is:

$a_{rad}= \frac{v^2}{R}$

for P:

$a_{rad}= \frac{(0.938)^2}{0.1}=8.80\frac{m}{s^{2}}$

for Q:

$a_{rad}= \frac{(1.876)^2}{0.2}=17.60\frac{m}{s^{2}}$

c) As seen on a) speed and distance from axis is $v = \omega R$ because ω is constant the if R increases then v increases too.

3 0

3 years ago

A car, starting from the origin, travels

sweet [91]

Answer:

The net displacement of the car is 3 km West

Explanation:

Please see the attached drawing to understand the car's trajectory: First in the East direction for 4 km (indicated by the green arrow that starts at the origin (zero), and stops at position 4 on the right (East).

Then from that position, it moves back towards the West going over its initial path, it goes through the origin and continues for 3 more km completing a moving to the West a total of 7 km. This is indicated in the drawing with an orange trace that end in position 3 to the left (West) of zero.

So, its NET displacement considered from the point of departure (origin at zero) to the final point where the trip ended, is 3 km to the west.

8 0

3 years ago

Read 2 more answers

A detrital rock is named

levacccp [35]

Sedimentary rocks also known as clastic sedimentary rock

3 0

3 years ago

Read 2 more answers

A force Fx=cx-3.00x2 acts on a virus as the virus moves along an x axis, with F measured in Newtons, x in meters, and c a consta

andrey2020 [161]

Answer:

4

Explanation:

We are given that

$F_x=cx-3x^2$

K.E at x=0 m=20 J

K.E at x=3 m=11 J

We have to find the value of c.

By work energy theorem

Work done=Change in kinetic energy

W= $\int_{0}^{3} Fdx=\int_{0}^{3}(cx-3x^2)dx$

$W=[\frac{cx^2}{2}-x^3]^{3}_{0}$

$W=\frac{9c}{2}-27$

$\Delta K=11-20=-9 J$

$-9=\frac{9c}{2}-27$

$-9+27=\frac{9c}{2}$

$18=\frac{9c}{2}$

$c=\frac{2}{9}\times 18=4$

4 0

3 years ago

A ball of mass 0.160 kg is dropped from a height of 2.25 m. When it hits the ground it compresses 0.087 m.

Studentka2010 [4]

A) 6.64 m/s downward

B) 0.026 s

C) -40.9 N

Explanation:

A)

We can solve this problem by using the law of conservation of energy.

In fact, since the total mechanical energy of the ball must be conserved, this means that the initial gravitational potential energy of the ball before the fall is entirely converted into kinetic energy just before it reaches the floor.

So we can write:

$PE=KE\\mgh = \frac{1}{2}mv^2$

where

m = 0.160 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 2.25 m is the initial height of the ball

v is the final velocity of the ball before hitting the ground

Solving for v, we find:

$v=\sqrt{2gh}=\sqrt{2(9.8)(2.25)}=6.64 m/s$

And the direction of the velocity is downward.

B)

The motion of the ballduring the collision is a uniformly accelerated motion (= with constant acceleration), so the time of impact can be found by using a suvat equation:

$s=(\frac{u+v}{2})t$