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Alexxandr [17]
3 years ago
10

A hiker walks 2.00 km north and then 3.00 km east, all in 2.50 hours. Calculate the magnitude and direction of the hiker’s (a) d

isplacement (in km) and (b) average velocity (in km/h) during those 2.50 hours. (c) What was her average

Physics
1 answer:
tino4ka555 [31]3 years ago
5 0

Answer:

Incomplete third question

I think it should be

C. What was her average speed

Explanation:

Check attachment for solution

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Write any two different between work and power?​
attashe74 [19]

Answer:

1. Work Is a type of Physical Activity  

2. Power is Basically Having control of Society

Explanation:

3 0
3 years ago
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The temperature of a solution will be estimated by taking n independent readings and averaging them. Each reading is unbiased, w
Viktor [21]

Answer:

68 readings.

Explanation:

We need to take this problem as a statistic problem where the normal distribution table help us.

We can start considerating that X is the temperature of the solution, then

0.9 = P(|\bar{x}-\mu|

0.9 = P(\frac{|\bar{x}-\mu|}{\frac{\sigma}{\sqrt{n}}}

0.9 = P(|Z|

For a confidence level of 90% our Z_{critic} is 1.645

Therefore,

\frac{0.1}{\frac{\sigma}{\sqrt{n}}} = 1.645

Substituting for \sigma = 5 and re-arrange for n, we have that n is equal to

n=(\frac{1.645\sigma}{0.1})^2

n=\frac{(1.645)^2(0.5)^2}{0.1^2}

n=67.65

n=68

We need to make 68 readings for have a probability of 90% and our average is within 0.1\°\frac

3 0
3 years ago
After crossing the finish line, a race car slows down from 47 m/s to 32m/s in 3seconds. What is the car’s acceleration?
Lisa [10]

Answer: -5 m/s^2

Explanation: a = v - u/t

                         = 32 - 47/3

                         = -15/3

                         = -5 m/s^2

7 0
3 years ago
Problem 8: Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric fi
yanalaym [24]

Answer:

10581.59 V

Explanation:

We are given that

Magnetic field=B=0.65 T

Speed of electron=v=6.1\times 10^7m/s

Charge on electron, q=e=1.6\times 10^{-19} C

Mass of electron,m_e=9.1\times 10^{-31} kg

We have to find the potential difference in volts required in the first part of the experiment to accelerate electrons.

V=\frac{v^2m_e}{2e}

Where V=Potential difference

m_e=Mass of electron

v=Velocity of electron

Using the formula

V=\frac{(6.1\times 10^7)^2\times 9.1\times 10^{-31}}{2\times 1.6\times 10^{-19}}

V=10581.59 V

Hence, the potential difference=10581.59 V

8 0
3 years ago
Someone pls help fast
Kruka [31]
<h3>You forgot to add question...Add questions before asking so we can help</h3>
4 0
3 years ago
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