A student in the Biomechanics class has decided that she would like to make her arms

What happens to the force between two charges if one of the charges are doubled?

Julli [10]

Answer:

The new force between the charges becomes double of the initial force.

Explanation:

The force acting between charge particles is given by :

$F=k\dfrac{q_1q_2}{r^2}$

k is electrostatic constant

r is distance between charges

If one of the charges are doubled, then, q₁ = 2q₁

The new force becomes,

$F'=\dfrac{2kq_1q_2}{r^2}\\\\F'=2F$

So, the new force between the charges becomes double of the initial force.

7 0

3 years ago

100 °C is a greater temperature than which of the following?

ohaa [14]

When somebody hands you a Celsius°, it's easy to find the equivalent Fahrenheit°.

Fahrenheit° = (1.8 · Celsius°) + 32° .

So 100°C works out to 212°F.

It's also easy to find the equivalent Kelvin. Just add 273.15 to the Celsius.

So now you can see that 100°C is equal to A and D,
and it's less than B .

The only one it's greater than is C .

6 0

2 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

Which rule(s) is used for combining velocities?

DIA [1.3K]

Answer: B. Vector rules

Explanation:

4 0

3 years ago

A man pulls on his dog's leash to keep him from running after a bicycle. Which term best describes this example? Select one: A.

madreJ [45]

C. Negative force. The dog isn't going to learn that way.

6 0

3 years ago