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Fynjy0 [20]
3 years ago
9

​In Figure 4.24, a current of 0.3 A flows through the conductor CD, and a charge of 4C passes through a cross-section AB of the

conductor. Find the time taken by the charge to pass through AB​
Physics
2 answers:
olga2289 [7]3 years ago
8 0

Answer:

13.33 seconds

Explanation:

I = Q/t

t = Q/I = 4/0.3 = 13.33 seconds

strojnjashka [21]3 years ago
7 0

This is your answer . I hope it will help you .

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A pump uses a piston of 15 cm diameter that moves at 2.0 cm/s as it pushes a fluid through a pipe. what is the speed of the flui
Nitella [24]
The important point here is that volumetric flow rate in the pump and the pipe is the same.

Q = AV, where Q = Volumetric flow  rate, A = Cross sectional area, V = velocity

Q (pump) = (π*15^2)/4*2 = 353.43 cm^3/s
Q (pipe) = (π*(3/10)^2)/4*V = 0.071V

Q (pump) = Q (pipe)
0.071V = 353.43 => V = 5000 cm/s

Therefore, the flow  of water in the pipe is 5000 cm/s.
4 0
3 years ago
By quadrupling the mass of the body hanging on the thread what will become the period of oscillation?​
nikklg [1K]

Well, the tension in the thread will probably quadruple, but the hanging body will continue to just hang there.  

The question gives us no evidence that it is doing any oscillating, and there's no reason for it to start just because it suddenly got heavier.

6 0
3 years ago
Suppose that you have been chosen for a space mission to a distant planet. Due to the length of time you'll be away from Earth y
Contact [7]

Answer:

I should be active for 15 hours to meet the physical activity requirement.

Explanation:

Since time dilates in moving objects, we use the formula t = t₀/√(1 - β²) where t = time in space vehicle, t₀ = time on earth = 9 hours and β = v/c where v = speed of space vehicle = 0.8c.

So, t = t₀/√(1 - β²)

t = 9/√(1 - (v/c)²)

= 9/√(1 - (0.8c/c)²)

= 9/√(1 - (0.8)²)

= 9/√(1 - (0.64)

= 9/√0.36

= 9/0.6

= 15 hr

So, according to a timer on the space vehicle, I should be active for 15 hours to meet the physical activity requirement.

8 0
3 years ago
Ans of this question A test charge of 1 couloumb moved from 30cm against the field of intensity 50N/c find the energy store in i
UkoKoshka [18]

Answer:

A. Zero

Explanation:

Given data,

The charge of the test charge, q = 1 C

The distance the charge moved against the filed of intensity, x = 30 cm

                                                                                                        = 0.3 m

The electric field intensity, E = 50 N/C

The energy stored in the charge at 0.3 m is given by the formula,

                                V = k q/r

Where,                        

                                     = 9 x 10⁹ Nm²C⁻²

The charge is moved from the potential V₁ to V₂ at 30 cm

Substituting the given values in the above equation

                            V₁ = 9 x 10⁹ x 30 / 0.3

                                =  1.5 x 10¹² J

And,

                            V₂ = 1.5 x 10¹² J

The energy stored in it is,

                             W = V₂ - V₁

                                  = 0

Hence, the energy stored in the charge is, W = 0        

6 0
3 years ago
Which scenarios are examples of physical changes? Select three options
pashok25 [27]

Answer:

A kid becoming an adult

A leg becoming bruised

A person's blood pressure raising because they are running

need a picture to answer specific questions.

Explanation:

All of these are physical changes. Hope that this helps you and have a great day :)

4 0
3 years ago
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