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Sindrei [870]
2 years ago
12

In the 25-ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of int

ensity 2500W/m2 at the floor of the facility. (This simulates the intensity of sunlight near the planet Venus.)A) Find the average radiation pressure (in pascals) on a totally absorbing section of the floor.B) Find the average radiation pressure (in atmospheres) on a totally absorbing section of the floor.C) Find the average radiation pressure (in pascals) on a totally reflecting section of the floor.D) Find the average radiation pressure (in atmospheres) on a totally reflecting section of the floor.
Physics
1 answer:
Ugo [173]2 years ago
3 0

Answer:

a. 8.33 x 10 ⁻⁶ Pa

b. 8.19 x 10 ⁻¹¹ atm

c. 1.65 x 10 ⁻¹⁰ atm

d. 2.778 x 10 ⁻¹⁴ kg / m²

Explanation:

Given:

a.

I = 2500 W / m² , us = 3.0 x 10 ⁸ m /s

P rad = I / us

P rad  = 2500 W / m² / 3.0 x 10 ⁸ m/s

P rad = 8.33 x 10 ⁻⁶ Pa

b.

P rad = 8.33 x 10 ⁻⁶ Pa *[  9.8 x 10 ⁻⁶ atm / 1 Pa ]

P rad = 8.19 x 10 ⁻¹¹ atm

c.

P rad = 2 * I / us = ( 2 * 2500 w / m²) / [ 3.0 x 10 ⁸ m /s ]

P rad = 1.67 x 10 ⁻⁵ Pa

P₁ = 1.013 x 10 ⁵ Pa /atm

P rad = 1.67 x 10 ⁻⁵ Pa / 1.013 x 10 ⁵ Pa /atm = 1.65 x 10 ⁻¹⁰ atm

d.

P rad  = I / us

ΔP / Δt = I / C² = [ 2500 w / m² ] / ( 3.0 x 10 ⁸ m/s)²

ΔP / Δt = 2.778 x 10 ⁻¹⁴ kg / m²

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allochka39001 [22]

Answer:

a.   q2 = 16.4μC, positive charge

b.   F = 0.900N

c.   downward

Explanation:

a. In order to calculate the charge of the unknown charge you use the following formula, for the electric force between two charges:

F_e=k\frac{q_1q_2}{r^2}            (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

r: distance between the charges = 0.300m

q1: charge 1 = -0.550 μC = 0.550*10^-6C

q2: charge 2 = ?

Fe: electric force = 0.900N

The force exerted in the second charge points upward, then, the sign of the second charge is positive because this charge is getting closer to the first one.

You solve the equation (1) for the second charge ans replace the values of the other parameters:

q_2=\frac{r^2F_e}{kq_1}=\frac{(0.300m)^2(0.900N)}{(8.98*10^9Nm^2/C^2)(0.550*10^{-6}C)}\\\\q_2=1.64*10^{-5}C\\\\q_2=16.4*10^{-6}C=16.4*10\mu C

The values of the second charge is 1.64 μC

b. By the third Newton Law, you have that the force exerted in the second charge is equal to the force exerted by the first charge on the second one.

The force exerted on the first charge is 0.900N

c. The charges are attracting between them, then, the force exerted on the first charge is pointing downward.

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Engineers who design battery-operated devices such as cell phones and MP3 players try to make them as efficient as possible. An
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efficiency= [useful energy transferred ÷ total energy supply]×100%

So, [5500÷10000]×100%=0.55×100

                                        =55%

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I could be wrong, but I'm pretty sure it's 144kg.
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An owl has a mass of 4.00 kg. It dives to catch a mouse, losing 800.00 J of its GPE. What was the starting height of the owl, in
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Answer:

height =20m

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Find the average power Pavg created by the force F in terms of the average speed vavg of the sled.
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Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The power created  is  P_{avg} =  F  *  v_{avg}

Explanation:

From the question we are told that

    The that the average power is  mathematically represented as

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Where W is  is the Workdone which is  mathematically represented as

         W =  F  *  s

      Where F is  the applies force and  s  is the displacement  due to the force  

        So  

                P_{avg} =  \frac{F *s  }{\Delta  t  }

Now this  displacement can be represented mathematically as  

            s =  v_{avg} *  \Delta  t

Where v_{avg } is the average  velocity and \Delta  t is the time  taken  

So  

            P_{avg} =  \frac{F *v_{avg} *  \Delta  t    }{\Delta  t  }

=>         P_{avg} =  F  *  v_{avg}

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