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Morgarella [4.7K]
2 years ago
10

“Why can I sometimes see the moon during the day?“Why can I sometimes see the moon during the day?

Physics
1 answer:
asambeis [7]2 years ago
4 0

Answer:

its bc of the way earth spins

Explanation:

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A ventilating fan is operated by 0.5hp electric motor. How much work in joules can the fan do in 3 hours? (I need the answer asa
shtirl [24]

Answer:

Power= 0.5hp = 375W

T = 3hrs = 10800s

Power = Work done/ Time

Work = Power * Time = 375 * 10800 = 4050000J

3 0
3 years ago
Violet light (410 nm) and red light (685 nm) pass through a diffraction grating with d = 3.33 x 10-6 m. What is the angular sepa
jeyben [28]

Answer:10.03

Explanation:

right on acellus

7 0
3 years ago
Suppose the battery in a clock wears out after moving thousand coulombs of charge through the clock at a rate of 0.5 Ma how long
Ksivusya [100]

Answer:

Hello your question is poorly written below is the complete question

Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?

answer :

a) 231.48 days

b) n = 3.125 * 10^15

Explanation:

Battery moved 10,000 coulombs

current rate = 0.5 mA

<u>A) Determine how long the clock run on the battery. use the relation below</u>

q = i * t ----- ( 1 )

q = charge , i = current , t = time

10000 = 0.5 * 10^-3 * t

hence  t = 2 * 10^7 secs

hence the time = 231.48  days

<u>B) Determine how many electrons per second flowed </u>

q = n*e ------ ( 2 )

n = number of electrons

e = 1.6 * 10^-19

q = 0.5 * 10^-3 coulomb ( charge flowing per electron )

back to equation 2

n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )

hence : n = 3.125 * 10^15

8 0
2 years ago
Mo is on a baseball team and hears that a ball thrown at a 45 degree angle from the ground will travel the furthest distance. Ho
Galina-37 [17]

Answer:

Explanation:

Usually the angle between the y axis  and x axis is 90° and we know that for furthest travel the degree angle must be 45° with the horizontal, Mo must release the ball about halfway between straight ahead and straight up

3 0
3 years ago
Calculate the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 m, for a temperature of
max2010maxim [7]

Answer:

The rate of heat conduction through the layer of still air is 517.4 W

Explanation:

Given:

Thickness of the still air layer (L) = 1 mm

Area of the still air = 1 m

Temperature of the still air ( T) = 20°C

Thermal conductivity of still air (K) at 20°C = 25.87mW/mK

Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

Q =\frac{KA(\delta T)}{L}

Q =\frac{25.87*1*20}{1}

Q = 517.4 W

Therefore, the rate of heat conduction through the layer of still air is 517.4 W

6 0
3 years ago
Read 2 more answers
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