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Alex787 [66]
4 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]4 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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which of the following statements is a reasonable conclusion about the Arctic ecosystem? a)The number of offspring produced is a
Vitek1552 [10]

Answer:

A is the answer

Explanation:

6 0
3 years ago
Suppose you have two meter sticks, one made of steel and one made of invar (an alloy of iron and nickel), which are the same len
Mekhanik [1.2K]

Answer:

  • The difference in length for steel is 2.46 x 10⁻⁴ m
  • The difference in length for invar is 1.845 x 10⁻⁵ m

Explanation:

Given;

original length of steel, L₁ = 1.00 m

original length of invar, L₁ = 1.00 m

coefficients of volume expansion for steel, \gamma_{st.} =  3.6 × 10⁻⁵ /°C

coefficients of volume expansion for invar, \gamma_{in.} =  2.7 × 10⁻⁶ /°C

temperature rise in both meter stick, θ = 20.5°C

Difference in length, can be calculated as:

L₂ = L₁ (1 + αθ)

L₂  = L₁ + L₁αθ

L₂  - L₁ = L₁αθ

ΔL = L₁αθ

Where;

ΔL is difference in length

α is linear expansivity = \frac{\gamma}{3}

Difference in length, for steel at 20.5°C:

ΔL =  L₁αθ

Given;

L₁ = 1.00 m

θ = 20.5°C

\alpha = \frac{\gamma}{3} = \frac{3.6*10^{-5}}{3} = 1.2*10^{-5} /^oC

ΔL  = 1 x 1.2 x 10⁻⁵ x 20.5 = 2.46 x 10⁻⁴ m

Difference in length, for invar at 20.5°C:

ΔL =  L₁αθ

Given;

L₁ = 1.00 m

θ = 20.5°C

\alpha = \frac{\gamma}{3} = \frac{2.7*10^{-6}}{3} = 0.9*10^{-6}/^oC

ΔL  = 1 x 0.9 x 10⁻⁶ x 20.5 = 1.845 x 10⁻⁵ m

8 0
3 years ago
If sin=cos <br>tan=?<br>cot=?<br>​
Nezavi [6.7K]

Answer:

sin=cos

tan=sin/cos

cos=sin

Explanation:

I've just answer what ive known

I hope its hepls to you

6 0
3 years ago
A double-slit experiment is set up using red light (λ = 706 nm). A first order bright fringe is seen at a given location on a sc
Elanso [62]

Answer:

λ = 470.66 nm

Explanation:

for bright fringey_m = \frac{m\lambda D}{d}

D= distance between slit and screen

d= distance between the slits

for first order bright fringe m = 1,

        y_1 = \frac{1\lambda D}{d}

         y_1 = {706*D}{d}

for dark fringe,we have

y_m = {(m + 1/2)\lambda D}{d}    

Now to get the dark fringes at the same location we should have;

(706)D/d = (m + 1/2)λD/d    

put m = 1

(1 + 1/2)λ = (706)

λ = 470.66 nm

6 0
4 years ago
PLEASE SOLVE QUICKLY!!!<br> Solve for A, B, and C from graph
hram777 [196]

A = 59.35cm

B = 196.56g

C = 74.65g

<u>Explanation:</u>

We know,

x = \frac{L}{\frac{W}{F} +1}

and L = x+y

1.

Total length, L = 100cm

Weight of Beam, W = 71.8g

Center of mass, x = 49.2cm

Added weight, F = 240g

Position weight placed from fulcrum, y = ?

L-y = \frac{L}{\frac{W}{F}+1 } \\100 - y = \frac{100}{\frac{71.8}{49.2}+1 } \\100 - y = \frac{100}{1.46+1}\\\\100 - y = \frac{100}{2.46} \\100-y = 40.65\\\\y = 59.35cm

Therefore, position weight placed from fulcrum is 59.35cm

2.

Total length, L = 100cm

Center of mass, x = 47.8 cm

Added weight, F = 180g

Position weight placed from fulcrum, y = 12.4cm

Weight of Beam, W = ?

47.8 = \frac{100}{\frac{W}{180}+1 }\\\47.8  = \frac{100}{\frac{W+180}{180} } \\\\47.8 = \frac{100 X 180}{W+180}\\ \\47.8W + 47.8 X 180 = 18000\\47.8W  = 18000 - 8604\\W = \frac{9396}{47.8}\\ W = 196.56g

Therefore, weight of the beam is 196.56g

3.

Total length, L = 100cm

Center of mass, x = 50.8 cm

Position weight placed from fulcrum, y = 9.8cm

Weight of Beam, W = 72.3g

Added weight, F = ?

50.8 = \frac{100}{\frac{72.3}{F}+1 }\\\ 50.8  = \frac{100}{\frac{72.3+F}{F} } \\\\50.8 = \frac{100 X F}{72.3+F}\\ \\50.8 X 72.3 + 50.8 X F = 100F\\\\3672.84 = 100F-50.8F\\3672.84 = 49.2F\\F = 74.65g

Therefore, Added weight F is 74.65g

A = 59.35cm

B = 196.56g

C = 74.65g

4 0
3 years ago
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