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natita [175]
3 years ago
12

A net charge of 5.0 coulombs passes a point on a

Physics
2 answers:
liraira [26]3 years ago
7 0

Answer: 1.0 x 10^2

Explanation:

cupoosta [38]3 years ago
6 0

Answer : The average value of current is 1.0\times 10^2\ A.

Explanation :

It is given that,

Net charge, q = 5 C

Time, t = 0.05 s

The electric current is defined as the rate of change of electric charge.

I=\dfrac{q}{t}

I=\dfrac{5\ C}{0.05\ s}

I = 100\ A

or

I=1.0\times 10^2\ A

The average current is 1.0\times 10^2\ A.

Hence, this is the required solution.

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77julia77 [94]

You calculated the density correctly.

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-- If the object is MORE dense than the fluid, it sinks.

-- If the object is LESS dense than the fluid, it floats.

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4 years ago
ANSWER FAST FOR BRAINLIEST!
Ne4ueva [31]

Answer:

the last one

Explanation:

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3 years ago
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A 6000 kg roller coaster goes around a loop of radius 30 m at 6 m/s. What is the centripetal acceleration?
Margarita [4]

Answer:

The answer to the question is 7200

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3 years ago
In the first stage of a two-stage Carnot engine, energy is absorbed as heat Q1 at temperature T1 = 500 K, work W1 is done, and e
Nataly [62]

Answer:

Efficiency = 52%

Explanation:

Given:

First stage

heat absorbed, Q₁ at temperature T₁ = 500 K

Heat released, Q₂ at temperature T₂ = 430 K

and the work done is W₁

Second stage

Heat released, Q₂ at temperature T₂ = 430 K

Heat released, Q₃ at temperature T₃ = 240 K

and the work done is W₂

Total work done, W = W₁ + W₂

Now,

The efficiency is given as:

\eta=\frac{\textup{Total\ work\ done}}{\textup{Energy\ provided}}

or

Work done = change in heat

thus,

W₁ = Q₁ - Q₂

W₂ = Q₂ - Q₃

Thus,

\eta=\frac{(Q_1-Q_2)\ +\ (Q_2-Q_3)}{Q_1}}

or

\eta=1-\frac{(Q_1-Q_3)}{Q_1}}

or

\eta=1-\frac{(Q_3)}{Q_1}}

also,

\frac{Q_1}{T_1}=\frac{Q_2}{T_2}=\frac{Q_3}{T_3}

or

\frac{T_3}{T_1}=\frac{Q_3}{Q_1}

thus,

\eta=1-\frac{(T_3)}{T_1}}

thus,

\eta=1-\frac{(240\ K)}{500\ K}}

or

\eta=0.52

or

Efficiency = 52%

8 0
4 years ago
A powerful missile reaches a speed of 5 kilometers per second in 10 seconds after its launch. What is the average acceleration o
bixtya [17]

Answer:

D

Explanation:

5 Km=5000m

so Δv=5000 m/sec

a=Δv/Δt

=5000/10

a=500 m/sec²                 as 500÷1000=0.5 Km

a=0.5 km/sec²

so D is the right answer.

3 0
3 years ago
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