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Svetlanka [38]
3 years ago
15

The current in a wire varies with time according to the relationship 1=55A−(0.65A/s2)t2. (a) How many coulombs of charge pass a

cross section of the wire in the time interval between t = 0 and t = 7.5 s? (b) What constant current would transport the same charge in the same time interval?
Physics
2 answers:
bulgar [2K]3 years ago
6 0

Answer:

A) Q = 321 C

B) I = 42.8 A

Explanation:

We are given the relationship between current and time to be;

I = (55 - 0.65A/s²)t²

A) We are to find the charge transferred(Q) from t=0 s to t= 7.5 s.

Q represents the current which is the net charge flowing through the area in this period of time and it's given by the formula ;

ΔQ = I•dt

So, ΔQ = (55 - 0.65A/s²)t²•dt

To obtain Q, let's integrate ΔQ;

∫ΔQ = ∫(55 - 0.65A/s²)t²•dt at boundary of t = 0 and 7.5s

Q = 55At - (0.65/3)(A/s²)t³ at boundary of t = 0 and 7.5s

Thus,

Q = 55A(7.5s) - (0.65/3)(A/s²)(7.5s)³

Q = 412.5 A.s - 91.4 A.s ≈ 321 A.s

Now,A.s is also known as Columbs(C)

Thus, Q = 321 C

B) Current flow is given by the equation;

I = Q/t

Now, t will be 7.5s because it is the certain time that represents the current.

Thus, plugging in the relevant values ;

I = 321/7.5 = 42.8 A

mario62 [17]3 years ago
3 0

Answer:

(A) Q = 321.1C (B) I = 42.8A

Explanation:

(a)Given I = 55A−(0.65A/s2)t²

I = dQ/dt

dQ = I×dt

To get an expression for Q we integrate with respect to t.

So Q = ∫I×dt =∫[55−(0.65)t²]dt

Q = [55t – 0.65/3×t³]

Q between t=0 and t= 7.5s

Q = [55×(7.5 – 0) – 0.65/3(7.5³– 0³)]

Q = 321.1C

(b) For a constant current I in the same time interval

I = Q/t = 321.1/7.5 = 42.8A.

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Peter’s body supplies a force of 500 N to run up a 10-m hill in 10 s. How much power is involved in Peter’s run up the hill? Exp
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Explanation:

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Power is mathematically expressed as:

P=\frac{W}{t} (1)

Where t is the time during which work W  is performed.

On the other hand, the Work W done by a Force F refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.  It is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy). Therefore, 1 Joule is the work done by a force of 1 Newton when moving an object, in the direction of the force, along 1 meter (1J=(1N)(1m)=Nm  ).

When the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:  

W=(F)(d) (2)

In this case, we have the following data:

F=500 N

d=10 m

t=10 s

So, let's calculate the work done by Peter and then find how much power is involved:

From (2):

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Substituting (4) in (1):

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A car accelerates along a straight road from rest to 90 km/h in 5.0 s. A) Find the magnitude of its average acceleration. B) Fin
egoroff_w [7]

Answer:

a)   A=5m/s^2

b)   V=25m/s

Explanation:

From the question we are told that:

Velocity v=90km/h=25m/s

Time t=5.0s

a)

Generally From Newton's motion equation for Average acceleration is   mathematically given by

  A=\frac{V-U}{t}

Since its from Rest Therefore

  U=0

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b)

Generally the equation for Average Velocity is mathematically given by

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 AT t=1.0s\\\\V=0+5(1)

 V=5m/s

 AT t=2.0s\\\\V=0+5(2)

 V=10m/s

 AT t=5.0s\\\\V=0+5(5)

 V=25m/s

 

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