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Annette [7]
4 years ago
14

Work & Power Problems

Physics
1 answer:
Artyom0805 [142]4 years ago
6 0

Answer:

1. Unit for work is measured in Joules.

Formula for work is power * time taken

2 The unit for force is Newton

A work done in lifting a jar of water

3. The unit for measuring power is Joules.

The formula for measuring power is work done/time taken

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Two light beams can interfere in a way that results in a less-intense beam of light. This destructive interference supports whic
riadik2000 [5.3K]

Answer: the wave model

Explanation: I just took the test

3 0
3 years ago
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A blue supergiant star has surface temperature 27 kK and has total power output 100000 times that of the Sun.
Free_Kalibri [48]

Answer:

Wien peak ( λmax ) is 107.40 nm

radius of super giant is 1.086 ×10^{10} m

Explanation:

given data

temperature 27 kK

power = 100000 times of Sun

Sun radius =  6.96 × 10^8 m

to find out

Wien peak ( λmax ) and radius of supergiant (r)

solution

we will apply here first wien law to find Wien peak  that is

λmax = b / t

λmax = 2.9 × 10^{-3} / 27000 = 1.0740 × 10^{-7}

so Wien peak ( λmax ) is 107.40 nm

and

now we apply steafay law that is

P = σ × A × T^{4}   .........................1

and we know total power output 100000 time of Sun

so we say  

4πr²sT^{4}  = 100000 × 4πR²sTs^{4}

r² = 100000  × R²Ts^{4} / T^{4}

put here value

r² = 100000  × (6.96×6000^{8} )² × 6000^{4} / 27000^{4}

r² = 1.18132 ×10^{20}

r = 1.086 ×10^{10} m

so radius of super giant is 1.086 ×10^{10} m

3 0
4 years ago
A boy flies a kite with the string at a 30∘ angle to the horizontal. The tension in the string is 4.5 N. Part A Part complete Ho
zimovet [89]

Answer:

Work done is zero

Explanation:

given data

Angle of kite with horizontal =  30 degree

tension in the string =  4.5 N

WE KNOW THAT

Work =  force * distance

horizontal force =  Tcos\theta = 4.5*cos30 = 3.89 N

DISTANCE = 0 as boy stands still. therefore

work done = 3.89 *0 = 0

3 0
3 years ago
If 1.34 ✕ 1020 electrons move through a pocket calculator during a full day's operation, how many coulombs of charge moved throu
ioda

Given :

Number of operations move through a pocket calculator during a full day's operation , n=1.34 \times 10^{20} .

To Find :

How many coulombs of charge moved through it .

Solution :

We know , charge in  one electron is :

e^-=-1.6\times 10^{-19}\ coulombs

So , charge on n electron is :

C=e^-\times n\\C=-1.6\times 10^{-19}\times 1.34\times 10^{20} \ C\\C=-21.44\ C

Therefore , -21.44 coulombs of charge is moved through it .

Hence , this is the required solution .

3 0
3 years ago
Need help to get this question right!!
Charra [1.4K]

Answer:

16 times as strong

Explanation:

From the question given above, the following assumptions were made:

Initial Force (F₁) = F

Initial distance apart (r₁) = r

Final distance apart (r₂) = ¼r

Final force (F₂) =?

Next, we shall obtain a relationship between the force and the distance apart. This can be obtained as follow:

F = GM₁M₂ / r²

Cross multiply

Fr² = GM₁M₂

If G, M₁ and M₂ are kept constant, then,

F₁r₁² = F₂r₂²

Finally, we determine the new force as follow:

Initial Force (F₁) = F

Initial distance apart (r₁) = r

Final distance apart (r₂) = ¼r

Final force (F₂) =?

Fr² = F₂ × (¼r)²

Fr² = F₂ × r²/16

Fr² = F₂r² / 16

Cross multiply

16Fr² = F₂r²

Divide both side by r²

F₂ = 16Fr² / r²

F₂ = 16F

From the calculations made above, we can see that the new force is 16 times the original force.

Thus, the new force is 16 times stronger.

8 0
3 years ago
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