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3 years ago

What provides most of the force to allow the runners to move?

Physics

1 answer:

kobusy [5.1K]3 years ago

4 0

Answer:

When you are running the most important force that you should understand is friction. Friction is a force that opposes movement between two objects, but for runners friction makes you faster. Friction gives you a better and more efficient way to use your energy into speed.

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If the same net force is applied to a 7kg

sergij07 [2.7K]

(2nd variant, 7kg is 3 times faster than 21kg)

3 0

3 years ago

You are given two infinite, parallel wires each carrying current I. The wires are separated by a distance d, and the current in

hoa [83]

Answer:F/L = μ0*I²/2πd

Explanation:

Both wires carries current of the value (I) and are separated by the distance (d).

The length of both wires is (L).

The first wire creates a magnetic field of the magnitude according to Bio-savart law

B=(Uo * I)/2πd

This magnetic field has the ability to exert a force F on the second wire.

This force (F) is given as

F=BIL* sin θ

θ here is 90 because the magnetic field is perpendicular to the length of the second wire.

Thus the value of sin 90 = 1

Hence F= BIL

But B=(Uo * I)/2πd.

We have that F= B=[(Uo * I)/2πd] x IL

Thus we have that F= [Uo* I²/2πd] x L

By rearranging, and bringing L to the left hand side of the equation, we have that

F/L = [Uo* I²/2πd]

6 0

3 years ago

Newton’s third law says that every time there is an ___________force, there is also a__________ force that is __________ in size

lesantik [10]

Newton’s third law of motion says that every time there is an ____action_______ force, there is also a ______reaction________ force that is ___equal________ in size and acts in the ___opposite____________ direction. This means that forces ALWAYS occur in __pairs________.

Explanation:

3 0

3 years ago

What is the magnitude (in N/C) and direction of an electric field that exerts a 4.00 x 10^−5 N upward force on a −2.00 µC charge

emmasim [6.3K]

Answer:

The magnitude of electric field is $1.25\times10^{14}\ N/C$ in downward.

Explanation:

Given that,

Force $F= 4.00\times10^{-5}\ N$

Charge q= -2.00 μC

We know that,

Charge is negative, then the electric field in the opposite direction of the exerted force.

We need to calculate the magnitude of electric field

Using formula of electric force

$F = qE$

$E = \dfrac{F}{q}$

$E=\dfrac{4.00\times10^{-5}}{-2.00\times1.6\times10^{-19}}$

$E=-1.25\times10^{14}\ N/C$

Negative sign shows the opposite direction of electric force.

Hence, The magnitude of electric field is $1.25\times10^{14}\ N/C$ in downward.

8 0

4 years ago

Calculate the magnitude of the normal force on a 25.2 kg block in the following circumstances. (Enter your answers in N.) HINT (

irga5000 [103]

Answer:

when the body is resting N = 246.96 N

when the body is resting on a tilted surface N = 212.12 N.

when the body is in a elevator N = 317.036 N

Explanation:

when the block is resting on a stationary surface the normal force is balanced by the weight of the body.

weight of the body = mg = 25.2×9.8 = 246.96 N

therefore normal force = 246.98 N.

when the block is resting on a tilted surface the normal force will be balanced by the $\cos \theta$ component of the weight where Ф is the angle of inclination.

therefore N = mg $\cos \Phi$

N= 212.12 N.

when the block is resting on a elevator that is accelerated upward the normal force will be the sum of weight and force due to acceleration ma

therefore N = 246.98 + 25.2×2.78

N = 317.036 N

5 0

3 years ago