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disa [49]
2 years ago
15

Formula: KE = 1/2 my?

Physics
1 answer:
9966 [12]2 years ago
3 0

Answer:

22400 Joules

Explanation:

Apply the formula:

KE = 1/2 . 40 . 1120

KE = 20 . 1120

KE = 22400 Joules

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What is the function of enter key​
garri49 [273]

Answer:

what key

Explanation:

4 0
3 years ago
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What is another way of saying “getting the smallest possible force”?
Dmitriy789 [7]

Answer:

MOMENTUM

Explanation:

another way of saying getting the smallest force possible is the word " MOMENTUM".

momentum is the ability to keep maintaining,incresing or itself developing to move at constant speed or to increase the speed.

7 0
3 years ago
How much current must be applied across a 60 Ω light bulb filament in order for it to consume 55 W of power? Unserious answers w
sergij07 [2.7K]

Answer: current I = 0.96 Ampere

Explanation:

Given that the

Resistance R = 60 Ω 

Power = 55 W

Power is the product of current and voltage. That is

P = IV ...... (1)

But voltage V = IR. From ohms law.

Substitutes V in equation (1) power is now

P = I^2R

Substitute the above parameters into the formula to get current I

55 = 60 × I^2

Make I^2 the subject of formula

I^2 = 55/60

I^2 = 0.92

I = sqr(0.92)

I = 0.957 A

Therefore, 0.96 A current must be applied.

4 0
3 years ago
while spinning in a centrifuge a 70.0 kg astronaut experiences an acceleration of 5.00 g, or five times the acceleration due to
ollegr [7]

Answer:

Explanation:

70.0(5.00)(9.81) = 3,433.5 = 3430 N

4 0
2 years ago
Only 35 % of the intensity of a polarized light wave passes through a polarizing filter. What is the angle between the electric
Nana76 [90]

Answer:

The angle between the electric field and the axis of the filter is 54⁰

Explanation:

Apply the equation for intensity of light through a polarizer.

I = I_oCos^2 \theta

where;

I is the intensity of the transmitted light

I₀ is the intensity of the incident light

θ is the incident angle

If only 35 % of the intensity of a polarized light wave passes through a polarizing filter, then the ratio of the intensity of the transmitted light to that of the intensity of the incident light is given by;

\frac{I}{I_o}  = Cos^2 \theta\\\\\frac{35}{100} =  Cos^2 \theta\\\\Cos^2 \theta = 0.35\\\\Cos\theta = \sqrt{0.35} \\\\Cos\theta = 0.5916\\\\\theta = Cos^{-1}(0.5916)\\\\\theta  = 54 ^0

Therefore, the angle between the electric field and the axis of the filter is 54⁰

3 0
2 years ago
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