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bulgar [2K]
4 years ago
14

What is the effective (rms) current value for an ac current with an amplitude of 10 a?

Physics
1 answer:
adelina 88 [10]4 years ago
5 0
The relationship between peak voltage and root mean square voltage for a symmetrical sine wave is:

rms value = peak value / √2

Using this formula

I(rms) = I(peak) / √2
I(rms) = 10 / √2

The effective or rms value of the current is 7.07 A
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The uncertainty in the position of an electron along an x axis is given as 68 pm. What is the least uncertainty in any simultane
umka21 [38]

Answer:

\Delta p_x=7.75\times 10^{-25}\ kg-m/s

Explanation:

Given that,

The uncertainty in the position of an electron along the x-axis is, \Delta x=68\ pm=68\times 10^{-12}\ m

We need to find the east uncertainty in any simultaneous measurement of the momentum component of this electron.

We know that the Heisenberg's uncertainty principle gives the relation between the uncertainty in position and the momentum of electron as :

\Delta p_x{\cdot}\Delta x\ge \dfrac{h}{4\pi }

Putting all the values, we get :

\Delta p_x{\cdot}\ge \dfrac{h}{4\pi \Delta x}\\\\\Delta p_x \ge \dfrac{6.63\times 10^{-34}}{4\pi \times 68\times 10^{-12}}\\\\\Delta p_x\ge 7.75\times 10^{-25}\ kg-m/s

So, the momentum component of this electrons is greater than 7.75\times 10^{-25}\ kg-m/s.

6 0
4 years ago
Which function is decreasing?
tangare [24]
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5 0
4 years ago
An object has a mass of 120 kg on the moon. What is the force of gravity acting on the object on the moon?
Art [367]
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5 0
3 years ago
Read 2 more answers
A 0.140-kg baseball is dropped from rest from a height of 1.8 m above the ground. It rebounds to a height of 1.4 m. What change
aliina [53]

Answer:

\Delta p=-1.56\ kg-m/s

Explanation:

It is given that,

Mass of the baseball, m = 0.14 kg

It is dropped form a height of 1.8 m above the ground. Let u is the velocity when it hits the ground. Using the conservation of energy as :

u=\sqrt{2gh}

h = 1.8 m  

u=\sqrt{2\times 9.8\times 1.8}

u = 5.93 m/s

Let v is the speed of the ball when it rebounds. Again using the conservation of energy to find it :

v=\sqrt{2gh'}

h' = 1.4 m  

v=-\sqrt{2\times 9.8\times 1.4}

v = -5.23 m/s

The change in the momentum of the ball is given by :

\Delta p=m(v-u)

\Delta p=0.14(-5.23-5.93)

\Delta p=-1.56\ kg-m/s

So, the change in the ball's momentum occurs when the ball hits the ground is 1.56 kg-m/s. Hence, this is the required solution.

4 0
4 years ago
Consider the following distinct forces:______________________________.
Kamila [148]

Answer:

1 and 2

Explanation:

It is given that, force exerted by air is negligible in any way.

Also, it is given that channel is in the shape of a segment of a circle with its center at O.

When it is within the frictionless channel at position 'Q',

A gravity exerts force on the ball in the downward direction. On the other hand, the channel pointing from Q to O also exerts a force on the ball.

However, there is no any force in the direction of motion. On the other hand, he channel pointing from O to Q does not exert a force on the ball.

5 0
4 years ago
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