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Elis [28]
2 years ago
9

A garage door opener has a power rating of 350 watts. If the door is in operation for 30 seconds, how many joules of energy are

used? (Hint: Remember that 1 watt = 1 joule/sec)
Physics
1 answer:
Ksju [112]2 years ago
7 0
Because of the hint we can conclude what equation we need to solve this problem. We have power and duration that means that we need to express energy:

1 joule = 1watt * 1 second
or
E (energy) = P (power) * t (time duration)
E = 350 * 30 = 10500 joules.
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nika2105 [10]

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What is true about X-rays and microwaves?
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radio waves,microwaves, terahertz radiation, infrared radiation, visible light, ultraviolet radiation,X-rays<span> and gamma </span>rays<span>.</span>
7 0
3 years ago
In what way are gravitational and electrical forces similar?
Nadusha1986 [10]

Answer:

D. Both occur between objects independently whether they are in contact  or not.

Explanation:

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F=G\frac{m_1 m_2}{r^2}

where G is the gravitational constant, m1 and m2 are the two masses, and r is the distance between the two masses.

- The electrical force is a force that is exerted between two (or more) objects having electrical charge. It can be either attractive or repulsive, depending on the sign of the two charges, and its magnitude is given by

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6 0
3 years ago
A simple and common technique for accelerating electrons is shown in the figure, where there is a uniform electric field between
Sunny_sXe [5.5K]

Answer : 4.483\times 10^{15}\ m/s^2.

Explanation:

It is given that,

Electric field strength, E=2.55\times 10^{4}\ N/C

We know that,

Charge of electron, q=1.6\times 10^{-19}\ C

Mass of electron, m=9.1\times 10^{-31}\ kg

From the definition of electric field, F=qE...............(1)

According to Newton's second law, F = ma..........(2)

From equation (1) and (2)

ma=qE

a=\dfrac{qE}{m}

a=\dfrac{1.6\times 10^{-19}\ C\times 2.55\times 10^{4}\ N/C }{9.1\times 10^{-31}\ kg}

a=0.4483\times 10^{16}\ m/s^2

or

a=4.483\times 10^{15}\ m/s^2

So, the horizontal component of acceleration of an electron is 4.483\times 10^{15}\ m/s^2.

Hence, it is the required solution.

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