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SCORPION-xisa [38]
3 years ago
7

The current theory of the structure of the

Physics
1 answer:
IRISSAK [1]3 years ago
3 0

1) The mass of the continent is 3.3\cdot 10^{21} kg

2) The kinetic energy of the continent is 624 J

3) The speed of the jogger must be 4 m/s

Explanation:

1)

We start by finding the volume of the continent. We have:

L = 5850 km = 5.85\cdot 10^6 m is the side

t = 35 km = 3.5\cdot 10^4 m is the depth

So the volume is

V=L^2 t = (5.85\cdot 10^6)^2 (3.5\cdot 10^4)=1.20\cdot 10^{18} m^3

We also know that its density is

d=2750 kg/m^3

Therefore, we can find the mass by multiplying volume by density:

m=dV=(2750)(1.20\cdot 10^{18})=3.3\cdot 10^{21} kg

2)

The kinetic energy of the continent is given by:

K=\frac{1}{2}mv^2

where

m=3.3\cdot 10^{21} kg is its mass

v = 3.2 cm/year is the speed

We have to convert the speed into m/s. We have:

3.2 cm = 0.032 m

1 year = 1(365)(24)(60)(60)=3.15\cdot 10^7 s

So, the speed is:

v=\frac{0.032 m}{3.15 \cdot 10^7 s}=1.02\cdot 10^{-9} m/s

So, we can now find the kinetic energy:

K=\frac{1}{2}(1.20\cdot 10^{21})(1.02\cdot 10^{-9})^2=624 J

3)

Here we have a jogger of mass

m = 78 kg

And the jogger has the same kinetic energy of the continent, so

K = 624 J

The kinetic energy of the jogger is given by

K=\frac{1}{2}mv^2

where v is the speed of the jogger.

Solving for v, we find the speed that the jogger must have:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(624)}{78}}=4 m/s

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

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Explanation:

According to the work-energy theorem, the work done on the hockey puck is equal to the change in kinetic energy of the puck.

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where

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Substituting numbers into the equation, we find the work done by the player on the puck:

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A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi
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To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

W = F * d

Where,

F= Force

d = Distance

The distance in this case is a composition between number of steps and the height. Then,

d=h*N, for h as the height of each step and N number of steps.

On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

V_f^2-V_i^2 = 2a\Delta X

Where,

V_f = Final velocity

V_i = Initial Velocity

a = Acceleration

\Delta X = Displacement

PART A) For the particular case of work we know then that,

W = F*d

W = m*g*(h*N)

W = 50*9.8*(0.3*30)

W = 4.41kJ

Therefore the Work to do that activity is 4.41kJ

PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

V_f^2-V_i^2 = 2a\Delta X

Here,

V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow3 steps in one second

v_f = 0

Replacing,

V_f^2-V_i^2 = 2a\Delta X

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Re-arrange for a,

a = -\frac{0.9^2}{2*30*0.3}

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At this point we can calculate the time, which is,

t = \frac{\Delta V}{a}

t = \frac{0-0.9}{-45*10^{-3}}

t = 20s

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

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