Which electromagnetic wave has the lowest frequencies (less than 3..*.10 © hertz)?

As the distance between two objects increases, the gravitational force of attraction between them will

jeka94

That force DEcreases. This is a big part of the reason why the Earth attracts you with more force than Jupiter does.

4 0

3 years ago

A 7.00 kg bowling ball is held 2.00 m above the ground. Using g- 9.8 m/s^2, how much energy does the bowling ball have due to it

Alexeev081 [22]

It should be B.137 just took the test.

3 0

3 years ago

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

Match the measurement with the prober SI unit.

padilas [110]

Answer:

Your question is incomplete

8 0

3 years ago

A boy slides a book across the floor using a force of five and over a distance of 2M what is the kinetic energy of the book afte

Alinara [238K]

Answer: 10Nm or 10J

Explanation:

Given the following :

Force (f) = 5

Distance (d) = 2m

Calculate the kinetic energy assuming no friction

Work done = force × distance

Work done = 5N × 2m = 10Nm

Recall :

Work done = ΔK.E ( change in kinetic energy)

Therefore, kinetic energy of the book after sliding = ΔK. E, which is equal to work done.

Hence, K. E of book after sliding is 10Nm

5 0

3 years ago