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kiruha [24]
3 years ago
9

Question 6 (15 points)

Physics
1 answer:
aleksandrvk [35]3 years ago
3 0

Answer:

that would be newtons 3rd law

Explanation:

because its how it is

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masya89 [10]

the answer would be 4

8 0
3 years ago
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The next four questions refer to the situation below.
Anna11 [10]

Answer:

 t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in},      t_{out} = \frac{D}{v_s +v_r}

Explanation:

This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

its speed is

         v_{sg 1} = v_{sr} + v_{rg}

The subscripts are s for the swimmer, r for the river and g for the Earth

with the velocity constant we can use the relations of uniform motion

           v_{sg1} = D / t_{out}

           D = v_{sg1}  t_{out}

now let's analyze when the swimmer turns around and returns to the starting point

        v_{sg 2} =  v_{sr}  - v_{rg}

         v_{sg 2} = D / t_{in}

         D = v_{sg 2}  t_{in}

with the distance is the same we can equalize

           v_{sg1} t_{out} = v_{sg2} t_{in}

          t_{out} =  t_{in}

           t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in}

This must be the answer since the return time is known. If you want to delete this time

            t_{in}= D / v_{sg2}

we substitute

            t_{out} = \frac{v_s - v_r}{v_s+v_r} ()

            t_{out} = \frac{D}{v_s +v_r}

7 0
2 years ago
A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten
babymother [125]

Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

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A proton is observed to have an instantaneous acceleration 11*10^11. what is the magnitude of e of the electric field at the pro
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The magnitude of the electric field at the proton's location is 10,437.5 N/C.

<h3>What the magnitude of the electric field?</h3>

The size of the electric field is basically characterized as the power per charge on the test charge. On the off chance that the electric field strength is meant by the image E. Very much like gravity, electric fields work the same way. In any case, while gravity generally draws in, an electric field, then again, can either rebuff or draw in. By and large, the Electric Field submits to the super-position guideline. the all out Electric Field from various charges is equivalent to the amount of the electric fields from each charge separately. An electric field is the actual field that encompasses electrically charged particles and applies force on any remaining charged particles in the field, either drawing in or repulsing them.

Learn more about the magnitude of the electric field, visit

brainly.com/question/26898699

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The farmer driving this tractor is spreading pesticides on the crops. What type of pollution is this an example of?
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Nonenvironmental source
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