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PIT_PIT [208]
3 years ago
7

How many minutes would it take a light wave to travel from the planet Venus to Earth? (Average distance from Venus to Earth = 28

million miles).
Physics
1 answer:
Gnesinka [82]3 years ago
6 0

Answer:

t=2.51min

Explanation:

The time taken by the light to travel a given distance is defined as:

t=\frac{d}{c}

Here c is obviously the speed of light. Now we convert the average distance form Venus to Earth to meters:

28*10^{6}mi*\frac{1609.34}{1mi}=4.51*10^{10}m

Finally, we calculate the minutes taken by the light to travel from Venus to Earth:

t=\frac{4.51*10^{10}m}{3*10^8\frac{m}{s}}\\t=150.33s*\frac{1min}{60s}=2.51min

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I believe the answer is a test tube.
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2 years ago
How much work is required to lift a 2 kilogram mass to a height of 10 meters
Semmy [17]

Given mass = 2kg, height = 10m,g = 9.8.


We know that Work done W = FD


= > W = (mg)(D)


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Hope this helps!

4 0
3 years ago
What is the average speed of a bowling ball that travels a distance of 5.0 m in 2.58 seconds?
Gre4nikov [31]

Answer:

<em><u>1</u></em><em><u>.</u></em><em><u>9</u></em><em><u>4</u></em><em><u> </u></em><em><u>m</u></em><em><u>/</u></em><em><u>s</u></em><em><u> </u></em><em><u>(</u></em><em><u>2</u></em><em><u>d</u></em><em><u>p</u></em><em><u>)</u></em>

Explanation:

Speed= Distance/Time

5.0/2.58 = 1.937984496

= <em><u>1.94</u></em> m/s (<em><u>2dp</u></em>)

7 0
3 years ago
A 4.00-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 27.0 N is require
NikAS [45]

Answer:

a)

135Nm⁻¹

b)

0.925 Hz

c)

1.2ms⁻¹

d)

0 m

e)

6.7ms⁻²

f)

\pm 0.2 m

Explanation:

a)

F = force required to hold the object at rest connected with stretched spring = 27 N

x = stretch in the spring from equilibrium position = 0.2 m

k = force constant of the spring

force required to hold the object at rest is same as the spring force , hence

F = k x

k = \frac{F}{x}

inserting the values

k = \frac{27}{0.2} = 135 Nm⁻¹

b)

frequency of the oscillations is given as

f =\frac{1}{2\pi }\sqrt{\frac{k}{m}}

inserting the values

f =\frac{1}{2(3.14) }\sqrt{\frac{135}{4}}\\f = 0.925 Hz

c)

A = Amplitude of oscillations = 0.2 m

w = angular frequency

Angular frequency is given as

w = 2\pi f = 2 (3.14) (0.925) = 5.8 rads⁻¹

Maximum speed of oscillation is given as

v_{max} = Aw

v_{max} = (0.2)(5.8)\\v_{max} = 1.2 ms⁻¹

d)

maximum speed of the object occurs at the equilibrium position, hence

x = 0 m

e)

Maximum acceleration of oscillation is given as

a_{max} = Aw^{2}

a_{max} = (0.2)(5.8)^{2}\\a_{max} = 6.7ms⁻²

f)

maximum acceleration occurs when the object is at extreme positions, hence

x = \pm 0.2 m

4 0
3 years ago
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