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2 years ago

How does a message travel across the gap at a synapse?

Physics

1 answer:

IRINA_888 [86]2 years ago

4 0

It's sent through neurotransmitters.

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How come in free fall you feel weightless even though gravity is pulling down on you?

AysviL [449]

Feeling of Weight.

When walking, you feel the weight on your feet, therefore, your brain automatically refers to it as a source of weight.

In the air there is no platform to land on, therefore the brain does not have the conscience to register you getting pulled down.

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2 years ago

*How much energy is<br>transferred in lifting a 5 kg<br>Mass 3m

AlexFokin [52]

Answer:

147 J

Explanation:

The energy transferred to potential energy is :

U = m * g * h = (5 kg) * (9.8 m/s^2) * (3 m) = 147 J

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2 years ago

When energy is transformed or changed from one form into another, some of the energy will be lost to the environment as?

sasho [114]

Generally it is lost as heat.

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2 years ago

You slide a hockey puck across the ice in a hockey rink

REY [17]

Answer-

mk thats kool.

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2 years ago

Read 2 more answers

An isolated conducting sphere has a 17 cm radius. One wire carries a current of 1.0000020 A into it. Another wire carries a curr

notsponge [240]

14 ms is required to reach the potential of 1500 V.

<u>Explanation:</u>

The current is measured as the amount of charge traveling per unit time. So the charge of electrons required for each current is determined as the product of current with time.

$Charge = Current \times Time$

As two different current is passing at two different times, the net charge will be the different in current. So,

$\text { Charge }=(1.0000020-1.0000000) \times t=2 \times 10^{-6} \times t$

The electric voltage on the surface of cylinder can be obtained as the ratio of charge to the radius of the cylinder.

$V=\frac{k q}{R}$

Here $k = 9 * 10^9$ , q is the charge and R is the radius. As $q=2 \times 10^{-6} \times t$ and R =17 cm = 0.17 m, then the voltage will be

$V=\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}$

The time is required to find to reach the voltage of 1500 V, so

$1500 =\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}$

$\begin{aligned}&t=\frac{1500 \times 0.17}{\left(9 \times 10^{9} \times 2 \times 10^{-6}\right)}\\&t=14.1666 \times 10^{-3} s=14\ \mathrm{ms}\end{aligned}$

So, 14 ms is required to reach the potential of 1500 V.

3 0

2 years ago