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2 years ago

Hello people~

Physics

2 answers:

Wewaii [24]2 years ago

5 0

Answer:

A or B

(a) is always at +ve potential.

(b) is always at zero potential.

Explanation:

reason 

a conductor with positive charge is always positive and zero potential

hope it's helpful

galina1969 [7]2 years ago

4 0

Answer:

Explanation:

Let's see the formula

V=kq/r

If q is positive it yeilds V or potential as positive hence it is at zero potential

Option B is correct

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Ou are given a 25.3 µf capacitor that is connected to a 13.0 v dc power supply. what will be the charge that is stored on this c

Black_prince [1.1K]

charge stored in the capacitor=3.29 x 10⁻⁴ C

Explanation:

we use the formula

Q= C V

Q= charge

C= capacitor=25.3 μF= 25.3 x 10⁻⁶ F

V= voltage= 13 V

Q=(25.3 x 10⁻⁶ ) (13)

Q= 3.29 x 10⁻⁴ C

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3 years ago

Two children, A and B, fire identical 10.0g all bearings from a catapult. the elastic band of each catapult extended by 0.10m an

sergiy2304 [10]

Answer:

The speed decreases and the frequency remains the same.

Explanation:

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2 years ago

Why do we need optics now a days?

AveGali [126]

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3 years ago

An unknown number of identical light bulbs are connected to a 15 V battery in parallel. The current through the battery is 2 A.

zaharov [31]

Answer:

An unknown number of identical light bulbs are connected to a 15 V battery in parallel. The current through the battery is 2 A. If the light bulbs are connected to the battery in series, the current through the battery is 5 mA. How many bulbs are there?

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Explanation:

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3 years ago

A pelican flying along a horizontal path drops a fish from a height of 4.7 m. The fish travels 9.3 m horizontally before it hits

slavikrds [6]

Answer:

(A) 9.5 m/s

(B) 5.225 m

Explanation:

vertical height (h) = 4.7 m

horizontal distance (d) = 9.3 m

acceleration due to gravity (g) = 9.8 m/s^{2}

initial speed of the fish (u) = 0 m/s

(A) what is the pelicans initial speed ?

lets first calculate the time it took the fish to fall

s = ut + $(\frac{1}{2}) at^{2}$

since u = 0

s = $(\frac{1}{2}) at^{2}$

t = $\sqrt{\frac{2s}{a} }$

where a = acceleration due to gravity and s = vertical height

t = $\sqrt{\frac{2 x 4.7 }{9.8} }$ = 0.98 s

pelicans initial speed = speed of the fish

speed of the fish = distance / time = 9.3 / 0.98 = 9.5 m/s

initial speed of the pelican = 9.5 m/s

(B) If the pelican was traveling at the same speed but was only 1.5 m above the water, how far would the fish travel horizontally before hitting the water below?

vertical height = 1.5 m

pelican's speed = 9.5 m/s

lets also calculate the time it will take the fish to fall

s = ut + $(\frac{1}{2}) at^{2}$

since u = 0

s = $(\frac{1}{2}) at^{2}$

t = $\sqrt{\frac{2s}{a} }$

where a = acceleration due to gravity and s = vertical height

t = $\sqrt{\frac{2 x 1.5 }{9.8} }$ = 0.55 s

distance traveled by the fish = speed x time = 9.5 x 0.55 = 5.225 m

8 0

3 years ago