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Natasha2012 [34]
2 years ago
14

A current exists whenever electric charges move. If ΔQ is the net charge that passes through a surface during a time period Δt,

then the average current during this time interval is defined as average current = ΔQ Δt = Q2 − Q1 t2 − t1 . If we take the limit of this average current over smaller and smaller time intervals, we get what is called the current I at a given time t1: I = lim Δt→0 ΔQ Δt = dQ dt . Thus the current is the rate at which charge flows through a surface. The current in a wire is defined as the derivative of the charge: I(t) = Q'(t). What does b I(t) a dt represent?
Physics
1 answer:
jeka57 [31]2 years ago
5 0

Answer:

It represents the change in charge Q from time t = a to t = b

Explanation:

As given in the question the current is defined as the derivative of charge.

                                  I(t) = dQ(t)/dt ..... (i)

But if we take the inegral of the equation (i) for the time interval  from t=a to

t =b we get

                                   Q =∫_a^b▒〖I(t)  〗 dt

which shows the change in charge Q from time t = a to t = b. Form here we can say that, change in charge is defiend as the integral of current for specific interval of time.

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A shuttle on Earth has a mass of 4.5 E 5 kg. Compare its weight on Earth to its weight while in orbit at a height of 6.3 E 5 met
faltersainse [42]

Answer:

83%

Explanation:

On the surface, the weight is:

W = GMm / R²

where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.

In orbit, the weight is:

w = GMm / (R+h)²

where h is the height of the shuttle above the surface of the Earth.

The ratio is:

w/W = R² / (R+h)²

w/W = (R / (R+h))²

Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:

w/W = (6.4×10⁶ / 7.03×10⁶)²

w/W = 0.83

The shuttle in orbit retains 83% of its weight on Earth.

4 0
2 years ago
Recall the previous question and the scenario with Zamir and Talia finding their way through a maze. Why is their displacement t
Ad libitum [116K]

Sample Response: Zamir and Talia’s total distances are different because they walked different paths in the maze. Zamir took a longer path. However, they had the same displacement because they both ended at the same position.

4 0
2 years ago
Read 2 more answers
An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o
Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

T=2\pi\sqrt{\dfrac{m}{k}}

m is the mass of object

m=\dfrac{W}{g}

m=\dfrac{2.45}{9.8}

m = 0.25 kg

k=\dfrac{4\pi^2m}{T^2}

k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}

k = 24.09 N/m

or

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0
2 years ago
A baggage handler throws a 15 kg suitcase along the floor of an airplane luggage compartment with a speed of 1.2 m/s. The suitca
Hatshy [7]

Answer:

0.0367

Explanation:

The loss in kinetic energy results into work done by friction.

Since kinetic energy is given by

KE=0.5mv^{2}

Work done by friction is given as

W= umgd

Where m is the mass of suitacase, v is velocity of the suitcase, g is acceleration due to gravity, d is perpendicular distance where force is applied and u is coefficient of kinetic friction.

Making u the subject of the formula then we deduce that

u=\frac {v^{2}}{2gd}

Substituting v with 1.2 m/s, d with 2m and taking g as 9.81 m/s2 then

u=\frac {1.2^{2}}{2*9.81*2}=0.0366972477064\approx 0.0367

Therefore, the coefficient of kinetic friction is approximately 0.0367

7 0
2 years ago
A toroidal coil has a mean radius of 16 cm and a cross-sectional area of 0.25 cm2; it is wound uniformly with 1000 turns. A seco
Roman55 [17]

Answer:

Explanation:

Mutual inductance is equal to magnetic flux induced in the secondary coli due to unit current in the primary coil .

magnetic field in a torroid  B = μ₀ n I , n is number of turns per unit length and I is current .

B = 4π x 10⁻⁷ x (1000 / 2π x .16  )x 1 ( current = 1 A)

flux in the secondary coil

= B x area of face of coil x no of turns of secondary

= 4π x 10⁻⁷ x (1000 /2π x .16  ) .25 x 10⁻⁴ x 750

= 2 x 1000 x .25 x( 750 /.16) x 10⁻¹¹

2343.75 x 10⁻⁸

= 23.43 x 0⁻⁶ H.

.

6 0
2 years ago
Read 2 more answers
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