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Rina8888 [55]
3 years ago
7

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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Ocean breezes are due to which method of heat transfer?
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An electron is accelerated within a particle accelerator using a 100 MV electric potential. The 100 MeV electron moves along an
Delicious77 [7]

Answer:

The length of the tube is 3.92 m.

Explanation:

Given that,

Electric potential = 100 MV

Length = 4 m

Energy = 100 MeV

We need to calculate the value of \gamma

Using formula of relativistic energy

E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)

Put the value into the formula

1.6\times10^{-15}= 9.1\times`10^{-31}\times9\times10^{16}(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)

(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)=\dfrac{1.6\times10^{-15}}{9.1\times10^{-31}\times9\times10^{16}}

Here, \gamma-1=(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)

\gamma-1=0.01953

\gamma=0.01953+1

\gamma=1.01953

We need to calculate the length

Using formula of length

L'=\dfrac{L}{\gamma}

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L'=\dfrac{4}{1.01953}

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8 0
3 years ago
Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g
Ne4ueva [31]

Answer:

C

Explanation:

- Let acceleration due to gravity @ massive planet be a = 30 m/s^2

- Let acceleration due to gravity @ earth be g = 30 m/s^2

Solution:

- The average time taken for the ball to cover a distance h from chin to ground with acceleration a on massive planet is:

                                 t = v / a

                                 t = v / 30

- The average time taken for the ball to cover a distance h from chin to ground with acceleration g on earth is:

                                 t = v / g

                                 t = v / 9.81

- Hence, we can see the average time taken by the ball on massive planet is less than that on earth to reach back to its initial position. Hence, option C

7 0
2 years ago
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Katena32 [7]

"Accuracy" would be the best option from the list regarding the property of a measurement that is best estimated from the percent error, since the higher the error is the lower the accuracy.


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