PLEASE HELP!

Determine the mass of the earth from the known period and distance of the moon.

Free_Kalibri [48]

High School Physics 5+3 pts Previous question Next question Naomi wants to know the effect of different colors of light on the growth of plants. She believes that plants can survive best in white light. She buys 25

6 0

3 years ago

A 3.0 X 10^3 grams rock swings in a circle with a diameter of 1000 cm. Given the constant speed around the circle is 17.90 mph (

marissa [1.9K]

Answer:

a = 12.8 m/s^2

Explanation:

To find the centripetal acceleration you use the following formula:

$a_c=\frac{v^2}{r}$ (1)

v: tangential speed of the rock = 17.90mph

r: radius of the orbit = 1000cm/2 = 500cm = 0.5m

You first change the units of the tangential velocity in order to replace the values of v and r in the equation (1):

$17.90mph*\frac{1609.34m}{1\ m}*\frac{1\ h}{3600\ s}=8m/s$

Then, you use the equation (1):

$a_c=\frac{(8m/s)^2}{5m}=12.8\frac{m}{s^2}$

hence, the centripetal acceleration is 12.8 m/s^2

3 0

3 years ago

Under what circumstances does distance traveled equal magnitude of displacement? What is the only case in which magnitude of dis

Sedaia [141]

Answer: Please find the answer in the explanation

Explanation:

Under what circumstances does distance traveled equal magnitude of displacement?

When a body's motion is linear in one direction. Or a body moving in a straight line without turning back.

What is the only case in which magnitude of displacement and distance are exactly the same?

When the body is moving in a straight line with without changing direction or without turning back.

6 0

3 years ago

Calculate the acceleration of a mobile that at 4s is 32m from the origin, knowing that its initial speed is 10m / s.

ehidna [41]

Answer:

5.5 m/s^2

Explanation:

I believe this is the answer > using the formula a= v-v0/t

Hope this helps!

7 0

2 years ago

Read 2 more answers

A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The

Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}$

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

$F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg$

Here, ${\mu _{\rm{s}}}$ is the coefficient of static friction, mm is mass of the object, and $g$ is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass ${m_1}$

Substitute for in the expression of maximum static friction ${F_{\rm{f}}} = {\mu _{\rm{s}}}mg$

${F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g$

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute for , [/tex]{m_1}[/tex] for in the equation .

${F_{\rm{f}}} = {m_1}a$

Substitute ${\mu _{\rm{s}}}{m_1}g$ for ${F_{\rm{f}}}$ in the equation ${F_{\rm{f}}} = {m_1}a$

${\mu _{\rm{s}}}{m_1}g = {m_1}a$

Rearrange for a.

$a = {\mu _{\rm{s}}}g$

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right)a$

Substitute for in the above equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0

3 years ago