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Alex787 [66]
3 years ago
14

When NASA was communicating with astronauts on the Moon, the time from sending on the Earth to receiving on the moon was 1.33 s.

Find the distance from Earth to the Moon. (The speed of radio waves is 3.00 ´ 108 m/s.)
Physics
1 answer:
frez [133]3 years ago
5 0

To solve this problem it is necessary to apply the concepts related to the kinematic equations of movement description, which determine the velocity, such as the displacement of a particle as a function of time, that is to say

v = \frac{x}{t}\rightarrow x = v*t

Where,

x = Displacement

v = Velocity

t = Time

Our values are given as,

v=3*10^8m/s

t = 1.33 s

Replacing we have that,

x=v*t

x=(3*10^8)(1.33)

x = 399'000.000m

Therefore the distance from Earth to the Moon is 399.000 km

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What is the density of a substance that has a mass of 453g and a volume of 224mL?
Mrrafil [7]
Density is mass per unit volume. In this case, the unit is g/ml
Mass=453g                              ρ=453g/224ml
Volume=224ml                         ρ=2.022g/ml(rounded to nearest hundredth)
Density=?
ρ=mass/volume
Therefore the density of the substance is 2.022g/ml

6 0
3 years ago
Hola me pueden ayudar con este problema de física....por favor mirar la imagen.. gracias
makkiz [27]
Si la velocidad es 3 m/s, y ellos quieren saber la distancia despues 2 segundos, necesita que multiplicar 2 y 3.

La respeusta debiera ser 6m.
8 0
3 years ago
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1
notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is  F_{net}=  3.22*10^{-5} \ J

Explanation:

From the question we are told that

    The third charge is  q_3 =  55 nC =  55 *10^{-9} C

    The position of the third charge is  x = -1.220 \ m

     The first charge is q_1 =  -16 nC  =  -16 *10^{-9} \ C

     The position of the first charge is x_1 =  -1.650m

      The second charge is  q_2 =  32 nC  =  32 *10^{-9} C

      The position of the second charge is  x_2 =   0  \ m  

The distance between the first and the third charge is

      d_{1-3} =  -1.650 -(-1.220)

     d_{1-3} = -0.43 \ m

The force exerted on the third charge by the first is  

     F_{1-3} =  \frac{k  q_1 q_3}{d_{1-3}^2}

Where k is the coulomb's constant with a value  9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

      F_{1-3} =  \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}

       F_{1-3} = 4.28 *10^{-5} \ N

 The distance between the second and the third charge is      

  d_{2-3} =  0- (-1.22)

   d_{2-3} =1.220 \ m

The force exerted on the third charge by the first is mathematically evaluated as

       F_{2-3} =  \frac{k  q_2 q_3}{d_{2-3}^2}

substituting values

       F_{2-3} =  \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}

       F_{2-3} =  1.06*10^{-5} N

The net force is

      F_{net} =  F_{1-3} -F_{2-3}

substituting values

    F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}

    F_{net}=  3.22*10^{-5} \ J

6 0
3 years ago
The mass of a basketball is three times greater than the mass of a softball . Compare the momentum’s of a softball and a basketb
Lemur [1.5K]
They are moving the softball of basketball
4 0
3 years ago
Christina drives his moped 7 kilometers North. She stops for lunch and then drives 5 kilometers East. What distance did she cove
Mashutka [201]

Answer:

She covers the distance is 12 km.

The magnitude of displacement is 8.6 km.

The direction of her displacement is north east.

Explanation:

Given that,

Christina drives his moped 7 kilometers North and stop for lunch and then drive 5 km east.

We need to calculate the total distance

Using formula of distance

d=d_{1}+d_{2}

Put the value into the formula

d=7+5

d=12\ km

We need to calculate the magnitude of displacement

Using formula of displacement

D=xi+yj

D=5i+7j

D=\sqrt{5^2+7^2}

D= 8.6\ km

The direction of her displacement is north east.

Hence, She covers the distance is 12 km.

The magnitude of displacement is 8.6 km.

The direction of her displacement is north east.

6 0
3 years ago
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