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Charra [1.4K]
3 years ago
11

How do I calculate the speed of light and the distance between the earth and the moon?

Physics
1 answer:
MAVERICK [17]3 years ago
7 0

To calculate the distance to a star, astronomers observe it from different places along Earth's orbit around the Sun. If they measure the object's position several months apart, their "two eyes" will have a separation of well over 100 million miles!

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3.1
Ilia_Sergeevich [38]

Answer:

The value is  C = 30729\  c

Explanation:

From the question we are told that

  The power rating of the stove is  P = 4.4 KW  = 4.4 *10^{3} \  W

   The duration of its use everyday is t_1  = 70 \ minutes  = 1.167 \ hours    

    The rating of the light bulbs is P_2'  = 150 W

   The number is n = 7

   The power rating of the total  bulb is  P_2 = 7 * 150 = 1050 \  W

    The duration of its use everyday is  t_2  = 7 hours

    The power rating of miscellaneous appliance P_3 = 1.8 \ KW = 1.8 *10^{3} \  W

    The duration of its use everyday is t_3 = 1 hour

     The power rating of hot water P = 4 KW  = 4 *10^{3} \  W

      The duration of its use everyday is t_4 = 120 \ minutes  = 2 hours

Generally the total electrical energy used in 1 month is mathematically represented as

    E = P_1 * t_1 * 30 + P_2 * t_2 * 30 + P_3 * t_3 * 30+ P_4 * t_4 * 30

=>  E = 4.4*10^{3} * 1.167  * 30 + 1050 * 7 * 30 + 1.8 *1 * 30+ 4*10^{3} * 2 * 30

=>  E = 614598  \ W \cdot h

=>   E = 614.6  \  K W \cdot h

Generally the monthly electricity  bill is mathematically represented as

    C = 614.6 * 50

=> C = 30729\  c

8 0
3 years ago
(04.03 MC)
serious [3.7K]
I belive what your looking for is oxygen

5 0
3 years ago
Read 2 more answers
A cord of mass 0.65 kg is stretched between two supports 28 m apart. if the tension in the cord is 150 n, 2 how long will it tak
MaRussiya [10]
Ans: Time <span>taken by a pulse to travel from one support to the other = 0.348s
</span>
Explanation:
First you need to find out the speed of the wave.

Since
Speed = v = \sqrt{ \frac{T}{\mu} }

Where
T = Tension in the cord = 150N
μ = Mass per unit length = mass/Length = 0.65/28 = 0.0232 kg/m

So
v = \sqrt{ \frac{150}{0.0232} } = 80.41 m/s

Now the time-taken by the wave = t = Length/speed = 28/80.41=0.348s
5 0
3 years ago
You have two square metal plates with side lengths of (6.50 C) cm. You want to make a parallel-plate capacitor that will hold a
gtnhenbr [62]

Answer:

The necessary separation between  the two parallel plates is 0.104 mm

Explanation:

Given;

length of each side of the square plate, L = 6.5 cm = 0.065 m

charge on each plate, Q = 12.5 nC

potential difference across the plates, V = 34.8 V

Potential difference across parallel plates is given as;

V = \frac{Qd}{L^2 \epsilon_o} \\\\d = \frac{V L^2 \epsilon_o}{Q}

Where;

d is the separation or distance between the two parallel plates;

d = \frac{VL^2 \epsilon_o}{Q} \\\\d =  \frac{34.8*(0.065)^2 *8.854*10^{-12}}{12.5*10^{-9}} \\\\d = 0.000104 \ m\\\\d = 0.104 \ mm

Therefore, the necessary separation between  the two parallel plates is 0.104 mm

6 0
3 years ago
17. (a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air ( 3.0×106
sleet_krkn [62]

Answer:

Explanation:

Distance between plates d = 2 x 10⁻³m

Potential diff applied = 5 x 10³ V

Electric field = Potential diff applied /  d

= 5 x 10³  / 2 x 10⁻³

= 2.5 x 10⁶ V/m

This is less than  breakdown strength for air  3.0×10⁶ V/m

b ) Let the plates be at a separation of d .so

5 x 10³ / d = 3.0×10⁶ ( break down voltage )

d = 5 x 10³  / 3.0×10⁶

= 1.67 x 10⁻³ m

= 1.67 mm.

5 0
3 years ago
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