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allochka39001 [22]
3 years ago
14

A body is moving along a circular path with variable speed, it has both radial and tangential acceleration.

Physics
1 answer:
belka [17]3 years ago
5 0

Answer:

True;   ar = v^2 / R      Radial acceleration because it moves in a circular path

    at = α R = tangential acceleration because its speed changes

a = at + ar    total acceleration equals sum of radial and tangential

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What creates the changes in step 1,2,3 of the nitrogen cycle
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3 years ago
If a star is hotter is the wavelength shorter and higher energy?
insens350 [35]

Answer:

That is true, the hotter a star is the shorter of wavelength it emits and the higher. It also depends on the mass of the star because the bigger it is, the faster it fuses its hydrogen fuel. Faster burning of fuel means more energy is released and so results in higher temperatures.

Hope this helps! Good luck :)

Explanation:

3 0
3 years ago
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A basketball player makes a jump shot. The 0.599 kg ball is released at a height of 2.18 m above the floor with a speed of 7.05
7nadin3 [17]

Answer:

W_{drag} = 4.223\,J

Explanation:

The situation can be described by the Principle of Energy Conservation and the Work-Energy Theorem:

U_{g,A}+K_{A} = U_{g,B} + K_{B} + W_{drag}

The work done on the ball due to drag is:

W_{drag} = (U_{g,A}-U_{g,B})+(K_{A}-K_{B})

W_{drag} = m\cdot g\cdot (h_{A}-h_{B})+ \frac{1}{2}\cdot m \cdot (v_{A}^{2}-v_{B}^{2})

W_{drag} = (0.599\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.18\,m-3.10\,m)+\frac{1}{2}\cdot (0.599\,kg)\cdot [(7.05\,\frac{m}{s} )^{2}-(4.19\,\frac{m}{s} )^{2}]

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7 0
4 years ago
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In the future, people will only enjoy one sport: Electrodes. In this sport, you gain points when you cause metallic discs hoveri
pochemuha

Answer:

  • Disk C and Disk D

Explanation:

The total charge in the disks

q_1 + q_2 = q_{total}

must be conserved before and after bringing them together.

Lets equate the sum of the initial charge with the sum of the final for the disk:

q_{1_i} + q_{2_i} = q_{1_f} + q_{2_f} = 2 * (+8.5) \mu C

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So, the initial charges must sum +17 μC.

Now, as there are no charges over +17 μC, this means that both charges must be positive.

As the only positive charges are q_C and q_D, this disk must be the ones we are looking for. And, as we can see, they sum 17 μC:

q_{C} + q_{D} = 5 \mu C + 12 \mu C = 17 \mu C

3 0
3 years ago
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Allushta [10]

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Mark me as Brainliest

Explanation:

7 0
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