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3 years ago

9

What must be applied to move a load?

2 answers:

jarptica [38.1K]3 years ago

5 0

A larger force than the object in question would be required to move an object. If the force is smaller, the object will not move and the force will be repelled.

Misha Larkins [42]3 years ago

3 0

An applied force must be applied to move a load. This applied force must be large enough to overcome any opposing forces in order to move the load.

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Which formula can be used to find the X component of the resultant vector?

julia-pushkina [17]

Answer:

If you are simply looking for the X component then the most applicable formula from the choices given is Tx + Ux+ Vx. This means that you will add all x-components. For example: If a man walking along the x-axis walks 10 meters to the right, 5 back and 2 meters forward, what is the resultant vector?

4 0

2 years ago

Which of the following statements describes Newton's first law?

jeka94

Answer:

F = MA

Explanation:

OP you didn't give us any examples, but force equals mass times acceleration is Newton's First Law.

Dropping a ball (mass) from the top of a building can show gravity, a form of acceleration.

5 0

3 years ago

what will be the speed of a solid sphere of mass 2.0 kilograms and radius 15.0 centimeters when it reaches the bottom of a incli

Yuki888 [10]

Explanation:

Below is an attachment containing the solution.

4 0

3 years ago

A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and ou

zepelin [54]

Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}$

$E=6.88\times10^{7}\ N/C$

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d). For, r = 7.00 cm

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}$

$E=5.43\times10^{6}\ N/C$

The electric field outside the solid conducting sphere and direction is away of solid sphere.

Hence, This is the required solution.

6 0

3 years ago

1.a. identifying what are the two major motions of earth as it travels through space

Lina20 [59]

Ehh, Earth rotates around its axis and Earth revolves around the sun as it travels through space.

6 0

3 years ago

Read 2 more answers