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3 years ago

What is number 30,31 in this picture?

Physics

1 answer:

Maslowich3 years ago

6 0

On question 30, that is a displacement- time graph (DT). On this type of graph the gradient is equal to the velocity. B has the steepest gradient, then A and finally C

Now velocity is a vector quantity so it has a direction and speed ( speed doesn't have a fixed direction.)

on the DT graph im going to assume that movement B is a positive velocity with A and C being negative.
So by ranking these: A is the most negative, C is the least negative and B has to be the greatest as it is the only positive velocity.

Q31, The same type of graph is present, by looking at the gradients we can rank the largest and smallest velocities- speeds in the case of the question.
i'll skip my working out as its the same as before:

C, B, A and then D

the same idea as on Q30 applies to Q31 part b,

D,C,B then A

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$\displaystyle G\frac{M}{r^2} =a$

We are solving for the radius of the new planet, so we can rearrange the equation:

$\displaystyle r=\sqrt{\frac{GM}{a} }$

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2 years ago

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3 years ago

PLZ HELP

Nadya [2.5K]

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

2) The kinetic energy of the continent is 1683 J

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Explanation:

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$d=5850 km = 5.85\cdot 10^6 m$

and depth

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So its volume is

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We also know that the density of the continent is

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Therefore, we can calculate its mass as:

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

The kinetic energy of the continent is given by

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

where

m is its mass

v is its speed

We have already calculate its mass, while the speed is

v = 3.2 cm/year

We have to convert into SI units first, as follows:

$v=3.2 \frac{cm}{year} \cdot \frac{1}{100 cm/m} \cdot \frac{1}{(365 d/y)(24h/d)(60min/h)(60 s/min)}=1.01\cdot 10^{-9} m/s$

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So, the kinetic energy of the continent is

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Here we have a jogger having the same kinetic energy of the continent, so

$K=1683 J$

And the kinetic energy of the jogger can be expressed as

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where

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$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1683)}{78}}=6.57 m/s$

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