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andrew11 [14]
3 years ago
11

If a person displaces a mass of60kg through 300 metres in 1Minute, what is the power??​

Physics
1 answer:
densk [106]3 years ago
7 0

Answer:

Power = 2940 Watts

Explanation:

Given the following data;

Mass = 60 kg

Height = 300 meters

Time = 1 minutes = 60 seconds

To find the power;

First of all, we would determine the potential energy possessed by the person due to height.

We know that acceleration due to gravity is equal to 9.8 m/s².

Potential energy can be defined as an energy possessed by an object or body due to its position.

Mathematically, potential energy is given by the formula;

P.E = mgh

Where,

P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

Substituting into the formula, we have;

P.E = 60 * 9.8 * 300

P.E = 176400

Next, we find the power using the following formula;

Power = energy/time

Power = 176400/60

Power = 2940 Watts

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The force of air resistance acts to oppose the motion of an object moving through the air. A ball is thrown upward and eventuall
ozzi

Answer:

For a (1) net force will be greater than the weight of the ball

For b (2) net force will be lesser than the weight of the ball

Explanation:

For (a):

For a linear motion of a system, one must have to understand, according to Newtons first law of motion, which is also known as law of inertia, a body which is at motion will continue to move or a body at rest will continue to rest until an external force is applied to it. In the given case, when ball goes upward, one thing is for sure, the net force is greater than the weight of the ball, because three forces are applied during upward motion:

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External force to throw ball upward

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Since ball is going upward, so net force is greater than both weight and air friction which are pulling ball downward.

For (b):

For a linear motion of a system, one must have to understand, according to Newtons first law of motion, which is also known as law of inertia, a body which is at motion will continue to move or a body at rest will continue to rest until an external force is applied to it. In the given case, when ball goes downward, one thing is for sure, the net force is lesser than the weight of the ball, because two forces are applied during downward motion:

gravity or weight which is pulling the ball downward,

air resistance, which is acting upward as it is creating friction between ball and air molecules, so creating hindrance in downward motion

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Net Force = Weight - Air friction

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4 0
3 years ago
A square-based shipping crate is being designed that must contain a volume of 16 ft3 . The material that is used for the base an
Vlada [557]

Answer:

Explanation:

Given

volume V=16 ft^3

Suppose base is square with side L

height of crate is h

Volume V=L^2\times h

16=L^2\times h

Cost of top and bottom area c_1=3L^2

Cost of Side area c_2=4Lh\times 2=8Lh=8L\times \frac{16}{L^2}=\frac{128}{L}

Total Cost C=c_1+c_2

Total Cost C=3L^2+\frac{128}{L}

Differentiate C w.r.t Length

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L^3=\frac{128}{6}

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6 0
3 years ago
A sinusoidal electromagnetic wave is propagating in vacuum. At a given point P and at a particular time, the electric field is i
olasank [31]

Answer:

a

The direction of the wave propagation is the negative  z -axis

b

The amplitude of  electric and magnetic field are  A_E= 3.35*10^5 V/m ,

A_M= 1.12 *10^{-3} T respectively

Explanation:

According to right hand rule, your finger (direction of electric field) would be pointing in the positive  x-axis  i.e towards your right let your palms be face toward the direction of the magnetic field i.e negative y-axis  (toward the ground ) Then anywhere your thumb stretched out is facing is the direction of propagation of the wave here in this case is the negative  z -axis

             The Intensity of the wave is mathematically represented as

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Given that I = 7.43 \frac{kW}{cm^2} =  7.43 \frac{*10^3}{*10^-{4} }= 7.43*10^7 \frac{W}{m^2}

Making E_{rms} the subject we have

                   E_{rms} = \sqrt{\frac{I}{0.5*c*\epsilon_o} }

Substituting values as given on the question

                E_{rms} = \sqrt{\frac{7.43 *10^7[\frac{W}{m^2} ]}{0.5 * 3.08*10^8 *8.85*10^{-12}} }

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                  A_E = \sqrt{2} * E_{rms}

                         = \sqrt{2} * 2.37*10^5

                        A_E= 3.35*10^5 V/m

The amplitude of the magnetic field is mathematically represented as

                       A_M = \frac{A_E}{c}

Substituting value

                      A_M = \frac{3.35 *10^5}{3.0*10^8}

                             A_M= 1.12 *10^{-3} T

7 0
3 years ago
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