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Mumz [18]
3 years ago
6

What are the answers to the question

Physics
1 answer:
Vitek1552 [10]3 years ago
7 0

Answer: the link isnt loading

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What parts are found in an electric generator? Check all that apply.
Blizzard [7]

Answer:

an armature a permanent magnet brushes slip rings

Explanation:

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3 years ago
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If the net external force acting on a system is zero , then the total momentum of the system is zero
Neporo4naja [7]
If net external force acting on the system is zero, momentum is conserved. That means, initial and final momentum are same → total momentum of the system is zero.
8 0
4 years ago
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How would a spinning disk's kinetic energy change if its moment of inertia was five times larger but its angular speed was five
Keith_Richards [23]

Answer:

<em>The kinetic energy of a spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.</em>

<em></em>

Explanation:

Let us first consider the initial characteristics of the angular motion of the disk

moment of inertia = I

angular speed = ω

For the second case, we consider the characteristics to now be

moment of inertia = 5I  (five times larger)

angular speed = ω/5  (five times smaller)

Recall that the kinetic energy of a spinning body is given as

KE = \frac{1}{2}Iw^{2}

therefore,

for the first case, the K.E. is given as

KE = \frac{1}{2}Iw^{2}

and for the second case, the K.E. is given as

KE = \frac{1}{2}(5I)(\frac{w}{5} )^{2}   = \frac{5}{50}Iw^{2}

KE = \frac{1}{10}Iw^{2}

<em>this is one-tenth the kinetic energy before its spinning characteristics were changed.</em>

<em>This implies that the kinetic energy of the spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.</em>

6 0
3 years ago
- In Einstein's famous equation E = me?, describing the
mel-nik [20]

Answer:

option C is correct

Explanation:

8 0
3 years ago
A load of 800 N is lifted using a block and tackle having 5 pulleys. If the applied effort is 200 N, calculate
ser-zykov [4K]

Explanation:

Load=800N

Effort=200N

1. Mechanical Advantage = LOAD/EFFORT

= 800N/200N

= 4

2 Velocity Ratio = no. Of pulleys =5

3. Efficiency = Mechanical advantage / velocity ratio × 100%

= (4/5)×100%

=80%

4. output work= load×load distance

= 800N × 5m

= 4 × 1000J

5. Efficiency = (output work/input work) ×100%

Or, 80% = (4000J/input work) ×100%

Or, 80%/100% = 4000J/inputwork

Or, 4/5 = 4000J/inputwork

Or, input work =4000J × 5/4

Input work = 5×1000J

I hope it helped! ;-)

8 0
3 years ago
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