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dolphi86 [110]
3 years ago
15

Lightning is an example of what phenomenon?

Physics
1 answer:
eduard3 years ago
8 0
A release of a large amount of energy
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In calisthenics, the resistance is ultimately provided by __________. A. free weights B. springs and elastic bands C. gravity D.
tangare [24]

Answer:

I think the answer is b am sorry if it is wrong

Explanation:

5 0
2 years ago
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Some neodymium glass lasers can provide 100TW of power in 1.0 ns pulses at a wavelength of 0.26 micrometers. how much energy is
Aleksandr-060686 [28]

Answer:

 E = 10⁵ J

Explanation:

given,

Power, P = 100 TW

           = 100 x 10¹² W

time, t = 1 ns

           = 1 x 10⁻⁹ s

The energy of a single pulse is:-

Energy = Power x time

 E = P t

 E = 100 x 10¹² x 1 x 10⁻⁹

 E = 10⁵ J

The energy contained in a single pulse is equal to 10⁵ J

7 0
3 years ago
Why is a virus NOT classified as a living thing?
Shkiper50 [21]

The correct answer would be B. A virus is living BUT   however it does NOT have all of the 7 characteristics of life.

Hope that helped ^^

4 0
3 years ago
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Consult interactive solution 2.22 before beginning this problem. a car is traveling along a straight road at a velocity of +30.0
Inessa05 [86]

Let a_1 be the average acceleration over the first 2.46 seconds, and a_2 the average acceleration over the next 6.79 seconds.

At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let v be the velocity of the car after the first 2.46 seconds.

By definition of average acceleration, we have

a_1=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

a_2=\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}

and we're also told that

\dfrac{a_1}{a_2}=1.66

(or possibly the other way around; I'll consider that case later). We can solve for a_1 in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:

1.66a_2=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

\implies1.66\left(\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}\right)=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

Now we can solve for v. We find that

v=20.8\,\dfrac{\mathrm m}{\mathrm s}

In the case that the ratio of accelerations is actually

\dfrac{a_2}{a_1}=1.66

we would instead have

\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}=1.66\left(\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}\right)

in which case we would get a velocity of

v=24.4\,\dfrac{\mathrm m}{\mathrm s}

6 0
3 years ago
An alternating voltage of 25.0 V is connected across the primary coil of a transformer.
uranmaximum [27]

Answer:

A

Explanation:

Wattage = E * I

E = 25

I = 5

Wattage = 25 * 5

Wattage = 125

The secondary, in an ideal transformer has the same wattage.

125 = 50 * I           Divide by 50

125/50 = I

I = 2.5

7 0
3 years ago
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