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trasher [3.6K]
3 years ago
6

Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of

light to get from the Sun to the Earth (in s) is written in scientific notation. Note: the speed of light is approximately 3.0×108 m/s. Enter m and n, separated by commas.
Physics
1 answer:
emmainna [20.7K]3 years ago
3 0

Answer:

5,2

Explanation:

From the question we are told that:

Speed of light C=3.0×10^8 m/s.

Generally the equation for Average Speed is mathematically given by

V_{avg}=\frac{d}{t}

Where

d=Distance between the Earth and the sun

d=1.5*10^11m

Therefore

t=\frac{d}{V_{avg}}

t=\frac{1.5*10^11m}{3.0×10^8 m/s.}

t=5*10^2s

Since m and n is given in the form of

m*10^n

Therefore

m=5 & n=2

5,2

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Answer:5.17 m/s

Explanation:

Given

let u be the speed at cliff initial point

range over cliff is 1.45 m

and range of projectile is given by

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u=3.77 m/s

Conserving Energy

E_{bottom}=E_{initial\ point\ at\ cliff}

Kinetic energy=Kinetic energy +Potential energy gained

Let v be the initial velocity

\frac{mv^2}{2}=mgh+\frac{mu^2}{2}

v^2=u^2+2gh

v=\sqrt{u^2+2gh}

v=\sqrt{3.77^2+2\time 9.8\times 0.64}

v=\sqrt{26.75}=5.17 m/s

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Maksim231197 [3]

You can download answer here

tinyurl.com/wpazsebu

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3 years ago
If the radius of a coinis 20cm, find its<br>surface area -​
meriva

Answer:

800\picm²

Explanation:

Given parameters:

radius of the coin  = 20cm

Unknown:

Surface area of the coin  = ?

Solution:

A coin is made up of two faces;

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Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 24 kg and the larger bottom crate has a m
marusya05 [52]

Answer:

The sum of all forces for the two objects with force of friction F and tension T are:

(i) m₁a₁ = F

(ii) m₂a₂ = T - F

1) no sliding infers: a₁ = a₂= a

The two equations become:

m₂a = T - m₁a

Solving for a:

a = T / (m₁+m₂) = 2.1 m/s²

2) Using equation(i):

F = m₁a = 51.1 N

3) The maximum friction is given by:

F = μsm₁g

Using equation(i) to find a₁ = a₂ = a:

a₁ = μs*g

Using equation(ii)

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4) The kinetic friction is given by: F = μkm₁g

Using equation (i) and the kinetic friction:

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4 0
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A uniform electric field of 2 kNC-1 is in the x-direction. A point charge of 3 μC initially at rest at the origin is released. W
ANEK [815]
The electric force acting on the charge is given by the charge multiplied by the electric field intensity:
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F=(3 \cdot 10^{-6} C)(2000 N/C)=0.006 N

The initial kinetic energy of the particle is zero (because it is at rest), so its final kinetic energy corresponds to the work done by the electric force for a distance of x=4 m:
K(4 m)=W=Fd=(0.006 N)(4 m)=0.024 J
8 0
3 years ago
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