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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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4 0
3 years ago
The carbon isotope 14C is used for carbon dating of archeological artifacts. 14C(mass 2.34×10−26kg) decays by the process known
Nookie1986 [14]

Answer:

2240.92365 m/s

Explanation:

m_1 = Mass of electron = 9.11\times 10^{−31}\ kg

v_1 = Speed of electron = 5.7\times 10^7\ m/s

p_2 = Neutrino has a momentum = 7.3\times 10^{-24}\ kg m/s

M = total mass = 2.34\times 10^{-26}\ kg

In the x axis as the momentum is conserved

Mv_x=m_1v_1\\\Rightarrow v_x=\dfrac{m_1v_1}{M}\\\Rightarrow v_x=\dfrac{9.11\times 10^{−31}\times 5.7\times 10^7}{2.34\times 10^{-26}}\\\Rightarrow v_x=2219.10256\ m/s

In the y axis

Mv_y=p_2\\\Rightarrow v_y=\dfrac{p_2}{M}\\\Rightarrow v_y=\dfrac{7.3\times 10^{-24}}{2.34\times 10^{-26}}\\\Rightarrow v_y=311.96581\ m/s

The resultant velocity is

R=\sqrt{v_x^2+v_y^2}\\\Rightarrow R=\sqrt{2219.10256^2+311.96581^2}\\\Rightarrow R=2240.92365\ m/s

The recoil speed of the nucleus is 2240.92365 m/s

3 0
3 years ago
It is observed that the time for the ball to strike the ground at b is 2.5 s. determine the speed at which the ball was thrown.
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6 0
2 years ago
The transverse standing wave on a string fixed at both ends is vibrating at its fundamental frequency of 250 Hz. What would be t
hodyreva [135]

Answer:

Explanation:

fundamental frequency, f = 250 Hz

Let T be the tension in the string and length of the string is l ans m be the mass of the string initially.

the formula for the frequency is given by

f=\frac{1}{2l}\sqrt{\frac{Tl}{m}}    .... (1)

Now the length is doubled ans the tension is four times but the mass remains same.

let the frequency is f'

f'=\frac{1}{2\times 2l}\sqrt{\frac{4T\times 2l}{m}}    .... (2)

Divide equation (2) by equation (1)

f' = √2 x f

f' = 1.414 x 250

f' = 353.5 Hz

7 0
2 years ago
A pebble is released from rest at a certain height and falls freely, reaching an impact speed of 6 m/s at the floor. Next, the p
Anna71 [15]

Answer:

Explanation:

Let h be the height .

initial velocity in first case u = 0

final velocity v = 6 m /s

acceleration due to gravity g = 9.8 m /s²

v² = u² + 2 g h

6² = 0 + 2 x 9.8 x h

h = 1.837 m .

For second case u = 3 m /s

v² = u² + 2 gh

= 3² + 2 x 1.837 x 9.8

= 9 + 36

= 45 m

v = 6.7 m /s

8 0
3 years ago
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