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3 years ago

A total charge of 7.5 mC passes through a cross-sectional area of a wire in 0.9 s. What is the current in the wire

Physics

1 answer:

Elan Coil [88]3 years ago

4 0

Answer:

the current in the wire is 0.008333333 A

Explanation:

The computation of the current in the wire is as follows;

Current in the wire is

= Total charge ÷ cross sectional area of the wire

= 7.5 × 10^-3C ÷ 0.9s

= 0.008333333 A

Hence, the current in the wire is 0.008333333 A

We simply applied the above formula so that the correct current of the wire could come

And, the same is considerd and relevant too

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Martha wants to calculate an object's velocity. What will she need to do?

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Answer:

you divide the distance by the time it takes to travel that same distance, then you add your direction to it.

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2 years ago

Tsunamis are fast-moving waves often generated by underwater earthquakes. In the deep ocean their amplitude is barely noticable,

Andrei [34K]

Answer:

a) V = 195.70 m/s

b) f=3.02 × 10⁻⁴ Hz

c) T = 3311.25 seconds

Explanation:

Given:

Wavelength, λ = 646 Km = 646000 m

Distance traveled = 3410 Km = 3410000 m

Time = 4.84 h = 4.84 × 3600 s = 17424 seconds

a) The speed (V) of the wave is given as

V = distance / time

V = 3410000 m/ 17424 seconds

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b) The frequency (f) of the wave is given as:

f = V / λ

f= 195.70 / 646000

f=3.02 × 10⁻⁴ Hz

c) The time period (T) is given as:

T = 1/ f

T = 1/ (3.02 × 10⁻⁴) Hz

T = 3311.25 seconds

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3 years ago

Would a ship in Lake Ontario (fresh water) float higher or lower in the water than in the Atlantic Ocean (salt water)?

ohaa [14]

Atlantic Ocean (salt water)

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2 years ago

À A car moves with an initial velocity of 18 m/s due north. Find the velocity of the car after 7 Os if

Nonamiya [84]

Answer:

(a) $v_f=28.5m/s$

(b) $v_f=7.5m/s$

Explanation:

Hello.

(a) In this case since the car is moving at an initial velocity of 18 m/s due north, the final velocity is computed considering the acceleration as positive since it is due north as well:

$v_f=v_0+at=18m/s+1.5m/s^2*7s\\\\v_f=28.5m/s$

(b) In this case, since the car is moving due north by the acceleration is due south it is undergoing a slowing down process, thereby the acceleration is negative therefore the final velocity turns out:

$v_f=v_0+at=18m/s-1.5m/s^2*7s\\\\v_f=7.5m/s$