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zheka24 [161]
3 years ago
9

As the diver jumps, the board bends down. What kind of the energy does the board now have?

Physics
1 answer:
Vlada [557]3 years ago
8 0
Hi there!

The energy that is about to or can be released from an object after energy has been transferred to it is called potential energy. In this case, as the diver jumps, potential energy is stored in the board as it bends. The potential energy is then released as kinetic energy when the board springs back up and the diver actually jumps. 

Hope this helps!
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2 years ago
Newton’s second law states that the acceleration of an object is dependent on 1 variable which is the net force
Darya [45]
I think it’s false. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and then mass of the object.
7 0
2 years ago
B. On a separate sheet of paper, describe the different ways of generating electric power. ​
Afina-wow [57]

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4 0
2 years ago
PLEASE PLEASE HELP
pav-90 [236]

if in series one lightbulb burns out the rest are unable to turn on.

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6 0
3 years ago
A circuit consists of a series combination of 6.50 −kΩ and 4.50 −kΩ resistors connected across a 50.0-V battery having negligibl
Vlada [557]

Answer:

Part A: 16.1 V

Part B: 20.5 V

Part C: 21.5%

Explanation:

The voltmeter is in parallel with the 4.5-kΩ resistor and the combination is in series with the 6.5-kΩ resistor. The equivalent resistance of the parallel combination is given as

\dfrac{1}{R_E}=\dfrac{1}{4.50}+\dfrac{1}{10.0}

R_E=\dfrac{4.50\times10.0}{4.50+10.0} = 3.10

Part A

The voltmeter reading is the potential difference across the parallel combination. This is found by using the voltage-divider rule.

V_1 = \dfrac{3.10}{3.10+6.50}\times50.0 = \dfrac{3.10}{9.60}\times50.0 = 16.1 \text{ V}

Part B

Without the voltmeter, the potential difference across the 4.5-kΩ resistor is found using the same rule as above:

V_2 = \dfrac{4.50}{4.50+6.50}\times50.0 = \dfrac{4.50}{11.0}\times50.0 = 20.5 \text{ V}

Part C

The error in % is given by

\dfrac{20.5-16.1}{20.5}\times100\% = \dfrac{4.4}{20.5}\times100\% = 21.5\%

4 0
3 years ago
Read 2 more answers
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