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seropon [69]
3 years ago
5

17. For how long should a force of 130 N be applied to an object of mass 50 kg to change its speed from 20 m/s to 60 m/s?

Physics
1 answer:
ohaa [14]3 years ago
5 0

Answer:

c. 15.4 s

Explanation:

Given the following data;

Mass, m = 50kg

Force, F = 130N

Initial velocity, u = 20m/s

Final velocity, v = 60m/s

To find the time;

First of all, we would solve for acceleration using the formula below;

Force = mass * acceleration

130 = 50*acceleration

Acceleration = 130/50

Acceleration = 2.6m/s²

Now, we would use the first equation of motion to find the time.

V = U + at

60 = 20 + 2.6t

2.6t = 60 - 20

2.6t = 40

t = 40/2.6

Time, t = 15.39 ≈ 15.4 seconds.

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Answer: a) 7.1 * 10^3 N; b) -880 N directed out of the curve.

Explanation: In order to solve this problem we have to use the Newton laws, then we have the following:

Pcos 15°-N=0

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from the first we obtain N, the normal force

N=750Kg*9.8* cos (15°)= 7.1 *10^3 N

Then to calculate the frictional force (f) we can use the second equation

f=P sin (15°) -m*ac where ac is the centripetal acceletarion which is equal to v^2/r

f= 750 *9.8 sin(15°)-750*(85*1000/3600)^2/150= -880 N

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2 years ago
In Niels Bohr's 1913 model of the hydrogen atom, the single electron is in a circular orbit of radius 5.29×10⁻¹¹m and its speed
Svet_ta [14]

The magnitude of the magnetic moment due to the electron's motion is 87.87 * 10^{-37}.

<h3>What is magnetic moment?</h3>

The magnetic pull and direction of a magnet or other object that produces a magnetic field are referred to as the magnetic moment in electromagnetism. Things that have magnetic moments include electromagnets, permanent magnets, various compounds, elementary particles like electrons, and a number of celestial objects (such as many planets, some moons, stars, etc).

The term "magnetic moment" really refers to the magnetic dipole moment of a system, which is the portion of the magnetic moment that can be represented by an equivalent magnetic dipole or a pair of magnetic north and south poles that are only very slightly apart. The magnetic dipole component is adequate for sufficiently small magnets or over sufficiently large distances.

Calculations:

radius= 5.29 * 10^{-11} m\\

velocity=2.9* 10^{6} m/s

Working formula, M=N/A

I=\frac{charge flow }{time taken} =\frac{e}{time taken\\}

T= \frac{2xr}{v} =\frac{2xx * 5.29 * 10^{-11} }{2.9* 10^{6} }

   =15.16 * 10^{-5} s

I= \frac{1.6 * 10^{-19} }{15.16 * 10^{-5} }= 0.10 * 10^{-14}

                     =1 * 10^{-15} C

M=1x (1* 10^{-15} * (5.29 * 10^{-11} )^{2}

  =87.87 * 10^{-37}

To learn more about magnetic moment ,visit:

brainly.com/question/14298729

#SPJ4

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Answer:

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Answer to your query is provided below

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Explanation:

Explanation for the same is attached in image

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