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3 years ago

What is the force on a body of mass 20 kg moving with an acceleration of 3 m/s2 ?

Physics

2 answers:

Ahat [919]3 years ago

7 0

Answer:

$\boxed{Force \: on \: body = 60 \: N}$

Given:

Mass of body = 20 kg

Acceleration = 3 m/s²

Explanation:

Force = mass × acceleration

= 20 × 3

= 60 kg m/s² or 60 N

zubka84 [21]3 years ago

5 0

Hey!!

Solution,

Mass(m)=20kg

Acceleration due to gravity(g)=3 m/s^2

Force=?

Now,

Force=m*g

 = 20*3

 = 60 Newton

The force is 60 Newton 

Hope it helps

Good luck on your assignment..

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During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound. The time between seeing

hodyreva [135]

Answer:

d = 1700 meters

Explanation:

During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound. The time between seeing the lighting and hearing the sound was 5 second, t = 5 seconds

Speed of sound, v = 340 m/s (say)

Let d is the distance of the colliding cloud from the observer. The distance covered by the object. It is given by :

$d=v\times t$

$d=340\ m/s\times 5\ s$

d = 1700 meters

So, the distance of the colliding cloud from the observer is 1700 meters. Hence, this is the required solution.

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If you quadruple the temperature of a black body, by what factor will the total energy radiated per second per square meter incr

arlik [135]

In modern physics, as it was called "Stefan-Boltzmann law", the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of the black body's temperature T

as:

$P\alpha T^4$

where: P is the power (total energy radiated per second per square meter) and T is the temperature of a black body.

then we can make a ratio between the state of before quadruple (with subscript 1) and after (with subscript 2) as:

$\frac{P_{1} }{P_{2} } =\frac{T_{1}^4 }{T_{2}^4}$

$T_{2}=4T_{1}$

Then

$\frac{P_{1} }{P_{2} } =\frac{T_{1}^4 }{(4T_{1})^4}$

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2 years ago

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2 years ago

Suppose the battery in a clock wears out after moving thousand coulombs of charge through the clock at a rate of 0.5 Ma how long

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Answer:

Hello your question is poorly written below is the complete question

Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?

answer :

a) 231.48 days

b) n = 3.125 * 10^15

Explanation:

Battery moved 10,000 coulombs

current rate = 0.5 mA

A) Determine how long the clock run on the battery. use the relation below

q = i * t ----- ( 1 )

q = charge , i = current , t = time

10000 = 0.5 * 10^-3 * t

hence t = 2 * 10^7 secs

hence the time = 231.48 days

B) Determine how many electrons per second flowed

q = n*e ------ ( 2 )

n = number of electrons

e = 1.6 * 10^-19

q = 0.5 * 10^-3 coulomb ( charge flowing per electron )

back to equation 2

n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )

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