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Radda [10]
3 years ago
12

If an object is in freefall it will experience air blank which generates a drag force

Physics
2 answers:
kogti [31]3 years ago
5 0

Answer:

Resistance

Explanation:

I believe it is air resistance but I could be wrong.

tia_tia [17]3 years ago
4 0

The question is fishing for "air resistance", but that's a mistake by whoever wrote the question.

"Free fall" means that gravity is the ONLY force on the body. If there's any air resistance, then it's not free fall.

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Answer:

d = 96 meters

Explanation:

a = acceleration

t = time

d = distance

d =  \frac{1}{2}  \times a \times  {t}^{2}

d =  \frac{1}{2}  \times 3 \times  {8}^{2}

d = 96

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A thin plastic rod of length 2.5 m is rubbed all over with wool, and acquires a charge of 75 nC, distributed uniformly over its
Colt1911 [192]

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3 years ago
Imagine a 10kg block moving with a velocity of 20m/s to the left.
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Kinetic energy , KE= [1/2]m*v^2

m = 10 kg
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3 0
3 years ago
A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o
gulaghasi [49]

A. 0.77 A

Using the relationship:

P=\frac{V^2}{R}

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega

The two light bulbs are connected in series, so their equivalent resistance is

R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega

The two light bulbs are connected to a voltage of

V  = 240 V

So we can find the current through the two bulbs by using Ohm's law:

I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A

B. 142.3 W

The power dissipated in the first bulb is given by:

P_1=I^2 R_1

where

I = 0.77 A is the current

R_1 = 240 \Omega is the resistance of the bulb

Substituting numbers, we get

P_1 = (0.77 A)^2 (240 \Omega)=142.3 W

C. 42.7 W

The power dissipated in the second bulb is given by:

P_2=I^2 R_2

where

I = 0.77 A is the current

R_2 = 72 \Omega is the resistance of the bulb

Substituting numbers, we get

P_2 = (0.77 A)^2 (72 \Omega)=42.7 W

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

P=\frac{E}{t}

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

7 0
3 years ago
How and why planets orbit the sun and how and why a planets speed changes orbit
anygoal [31]

the gravitonal pull

Explanation:

7 0
3 years ago
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