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nordsb [41]
3 years ago
15

A certain x-ray tube requires a current of 7 mA at a voltage of 80 kV. The rate of energy dissipation is:

Physics
1 answer:
Artyom0805 [142]3 years ago
3 0

Answer:

560 watts

The rate of energy dissipation is 560 W

Explanation:

Rate of energy dissipation is the rate of energy consumption in the x-ray tube.

P = VI

Given;

Voltage V = 80 kV = 80,000

Current I = 7mA = 0.007 A

Substituting the given values;

P = 80,000 V × 0.007A

P = 560 Joules per second

P = 560 watts

The rate of energy dissipation is 560 W

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hammer [34]

Answer:

a) = 3.94 m

b) = 3.15 m

Explanation:

Given

Mass of the block, m = 242 g

Force constant, k = 1.62 kN/m

Compression of the spring, x = 10 cm

Angle of inclination = 60°

a) if we equate the energy at the bottom of the ramp to the energy at a distance d up the ramp, we have

1/2kx² = mgh where, h = dsinΦ

1/2kx² = mgdsinΦ

1/2 * 1.62*10^3 * 0.1² = 0.242 * 9.8 * dsin 60

1/2 * 16.2 = 2.3716 * d sin 60

d sin 60 = 8.1 / 2.3716

0.866 d = 3.415

d = 3.415 / 0.866

d = 3.94 m

b) net force on the block = mgd sin 60 + µ mgd cos 60

8.1 = d[mg sin 60 + µ mg cos 60]

8.1 = d [0.242 * 9.8 * 0.866 + 0.44 * 0.242 * 9.8 * 0.5]

8.1 = d (2.05 + 0.52)

8.1 = 2.57 d

d = 8.1 / 2.57

d = 3.15 m

3 0
3 years ago
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A current of 0.92 a flows in a wire. how many electrons are flowing past any point in the wire per second? the charge on one ele
Fantom [35]
The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval, \Delta t:
I= \frac{Q}{\Delta t}
First we need to find the net charge flowing at a certain point of the wire in one second, \Delta t=1.0 s. Using I=0.92 A and re-arranging the previous equation, we find
Q=I \Delta t= (0.92 A)(1.0 s)=0.92 C

Now we know that each electron carries a charge of e=1.6 \cdot 10^{-19} C, so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
N= \frac{Q}{q} = \frac{0.92 C}{1.6 \cdot 10^{-19} C}=5.75 \cdot 10^{18}
3 0
3 years ago
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Ray Of Light [21]

Answer:

I believe C

Explanation:

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5 0
3 years ago
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Trey is conducting an experiment to determine which fish food causes the most goldfish activity over a period of two weeks. He i
mr_godi [17]

Answer:

This question is incomplete as it lacks options, the options are:

Which change would be the most effective way to improve the validity of this experiment?

A) Increase the number of goldfish in each tank.

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One of the ways to improve an experiment, which is lacking in this case is to have a CONTROL GROUP. Based on the fish foods used for each group, it is obvious that all groups in this experiment are experimental. Hence, creating another group that is fed regular fish food will serve as the CONTROL GROUP in order to compare with the experimental groups. This will improve the validity of the experiment.

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Answer:

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Explanation:

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3 years ago
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