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3 years ago

15

Lucifer is sitting in a tree that is 15 m above the ground. She jumps straight off the tree in the horizontal direction and land

s 7 m away from the base of the tree. How fast did she jump off the tree?

1 answer:

-BARSIC- [3]3 years ago

4 0

Answer:

huppp huppp huppp huppp huppp huppp

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Alpha particles Ichargeq q = + 2e mass m=6.8*10^ -27 kg) at 17*10^ 4 m/s What magnetic field strength would be required to bend

Rasek [7]

Answer:

Alpha particles Ichargeq q = + 2e mass m=6.8*10^ -27 kg) at 17*10^ 4 m/s What magnetic field strength would be required to bend them into a circular path of radiuse c = 0.25m

Explanation:

Alpha particles Ichargeq q = + 2e mass m=6.8*10^ -27 kg) at 17*10^ 4 m/s What magnetic field strength would be required to bend them into a circular path of radiuse c = 0.25m ok

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4 years ago

Outside the Solar System is there any gravitational pull from the Sun?

babunello [35]

Answer:

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3 years ago

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Which of the following are elements in both civil and criminal trials. (select all that apply)

Ganezh [65]

Answer:

a an b

Explanation:

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3 years ago

A balloon that has been rubbed on wool has a negative charge. When the balloon is held over—but not touching—some bits of paper,

alexandr402 [8]

Answer:

This process is known as static electricity

Explanation:

When balloon is rubbed on the wool , a negative charge is usually formed. When the balloon is then held over—but not touching—some bits of paper, some of the bits jump up to the balloon. The bits which jump up to the balloon are positive charged and we know that unlike charges attract.

This process explains static electricity which is when the charges on a material are not in a balanced state.

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3 years ago

You have a battery marked " 6.00 V 6.00 V ." When you draw a current of 0.383 A 0.383 A from it, the potential difference betwee

Archy [21]

Answer:

V = 4.81 V

Explanation:

As the potential difference between the battery terminals, is less than the rated value of the battery, this means that there is some loss in the internal resistance of the battery.
We can calculate this loss, applying Ohm's law to the internal resistance, as follows:

$V_{rint} = I* r_{int}$

The value of the potential difference between the terminals of the battery, is just the voltage of the battery, minus the loss in the internal resistance, as follows:

$V = V_{b} - V_{rint} = 5.03 V = 6.0 V - 0.383 A* r_{int}$

We can solve for rint, as follows:

$r_{int} = \frac{V_{b}-V}{I} =\frac{6.0V-5.03V}{0.383A} = 2.53 \Omega$

When the circuit draws from battery a current I of 0.469A, we can find the potential difference between the terminals of the battery, as follows:

$V = V_{b} - V_{rint} = 6.0 V - 0.469 A* 2.53 \Omega= 6.0 V - 1.19 V = 4.81 V$

As the current draw is larger, the loss in the internal resistance will be larger too, so the potential difference between the terminals of the battery will be lower.

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3 years ago