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BARSIC [14]
3 years ago
13

Is it possible to calculate the displacement based on an elapsed time from a position time graph?

Physics
1 answer:
Nataly [62]3 years ago
5 0
No I don’t think so. But it worth a try tho. Try it out.
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Which of the following would be a valid method to increase the buoyant force acting on an object?
shepuryov [24]
I think it’s B hope this helps.
4 0
2 years ago
Read 2 more answers
After a model rocket reached its maximum height, it then took 5.0 seconds to return to the launch site. what is the approximate
kifflom [539]
Air resistance is ignored.
g = 9.8 m/s².
At maximum height, the vertical velocity is zero.

Let h =  the maximum height reached.
Let u =  the vertical launch velocity.

Because ot takes 5.0 seconds to reach maximum height, therefore
(u m/s) - (9.8 m/s²)*(5 s) = 0 
u = 49 m/s

The maximum height reached is
h = (49 m/s)*(5 s) - (1/2)*(9.8 m/s²)*(5 s)²
   = 122.5 m

Answer: 122.5 m
3 0
3 years ago
What is one way to take advantage of creating even more kinetic energy on this particular technologically advanced field
ch4aika [34]

Incomplete question. However, I provided a brief about Kinetic energy generation.

<u>Explanation:</u>

Interestingly, Kinetic energy in simple terms refers to the energy possessed by a body in motion.

It is often calculated using the formula E =  \frac{1}{2}MV^{2}

A good example of creating even more kinetic energy is a hand crank toy car that moves after you wind it a little, when the car moves it is generating another measure of K.E.

7 0
2 years ago
The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increa
Nostrana [21]

Answer: The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increased by 2.0 kg, the period is found to be 3.0 s.  Then the mass m is 0.625kg.

Explanation: To find the answer, we need to know more about the simple harmonic motion.

<h3>What is simple harmonic motion?</h3>
  • A particle is said to execute SHM, if it moves to and fro about the mean position under the action of restoring force.
  • We have the equation of time period of a SHM as,

                                          T=2\pi \sqrt{\frac{m}{k} }

  • Where, m is the mass of the body and k is the spring constant.
<h3>How to solve the problem?</h3>
  • Given that,

               T_1=2s\\m_1=m\\m_2=m+2kg\\T_2=3s

  • We have to find the value of m,

              T_1=2\pi \sqrt{\frac{m}{k} } \\T_2=2\pi \sqrt{\frac{m+2}{k} } \\\frac{T_1}{T_2} =\sqrt{\frac{m}{m+2} }\\\frac{2}{3} =\sqrt{\frac{m}{m+2} }\\\\

               m=\frac{5}{8} =0.625kg

Thus, we can conclude that, the mass m will be 0.625kg.

Learn more about simple harmonic motion here:

brainly.com/question/28045110

#SPJ4

3 0
2 years ago
An electron is released from rest in a weak electric field given by E =-2.30 x 10-10 N/Cj. After the electron has traveled a ver
Goshia [24]

Explanation:

It is given that,

An electron is released from rest in a weak electric field of, E=2.3\times 10^{-10}\ N/C

Vertical distance covered, s=1\ \mu m=10^{-6}\ m

We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :

v^2-u^2=2as

v^2=2as.............(1)

Electric force is F_e and force of gravity is F_g. As both forces are acting in downward direction. So, total force is:

F=mg+qE

F=9.1\times 10^{-31}\times 9.8+1.6\times 10^{-19}\times 2.3\times 10^{-10}

F=4.57\times 10^{-29}\ N

Acceleration of the electron, a=\dfrac{F}{m}

a=\dfrac{4.57\times 10^{-29}\ N}{9.1\times 10^{-31}\ kg}

a=50.21\ m/s^2

Put the value of a in equation (1) as :

v=\sqrt{2as}

v=\sqrt{2\times 50.21\times 10^{-6}}

v = 0.010 m/s

So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.

6 0
3 years ago
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