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Y_Kistochka [10]
2 years ago
11

What is the correct answer?

Physics
2 answers:
Anastaziya [24]2 years ago
4 0

Answer:

i think 2 but not sure

Explanation:

timama [110]2 years ago
3 0

Answer:

P(1)=0

Explanation:

Sorry if its wrong

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Round your answers to one decimal place.this parallel circuit has two resistors at 15 and 40 ohms. what is the total resistance?
lubasha [3.4K]
1) The equivalent resistance of two resistors in parallel is given by:
\frac{1}{R_{eq}}= \frac{1}{R_1}+ \frac{1}{R_2}
so in our problem we have
\frac{1}{R_{eq}} =  \frac{1}{15 \Omega}+ \frac{1}{40 \Omega}=0.092 \Omega^{-1}
and the equivalent resistance is
R_{eq} =  \frac{1}{0.092 \Omega^{-1}}=10.9 \Omega

2) If we have a battery of 12 V connected to the circuit, the current in the circuit will be given by Ohm's law, therefore:
I= \frac{V}{R_{eq}}= \frac{12 V}{10.9 \Omega}=1.1 A
4 0
2 years ago
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Which simple machine is a doorknob?
Anestetic [448]

The answer is wheel and axle

3 0
3 years ago
A ball is rolling across the floor at a constant speed. What will happen to the ball if it is exposed
ELEN [110]
I think it’s speed will increase, if I understood it correctly
7 0
2 years ago
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A spacecraft of mass 1500 kg orbits the earth at an altitude of approximately 450 km above the surface of the earth. Assuming a
PSYCHO15rus [73]

Answer:

1.28 x 10^4 N

Explanation:

m = 1500 kg, h = 450 km, radius of earth, R = 6400 km

Let the acceleration due to gravity at this height is g'

g' / g = {R / (R + h)}^2

g' / g = {6400 /  (6850)}^2

g' = 8.55 m/s^2

The force between the spacecraft and teh earth is teh weight of teh spacecraft

W = m x g' = 1500 x 8.55 = 1.28 x 10^4 N

8 0
3 years ago
When two point charges are a distance d part, the electric force that each one feels from the other has magnitude F. In order to
Serjik [45]

Answer:

E) d/sqrt2

Explanation:

The initial electric force between the two charge is given by:

F=k\frac{q_1 q_2}{d^2}

where

k is the Coulomb's constant

q1, q2 are the two charges

d is the separation between the two charges

We can also rewrite it as

d=\sqrt{k\frac{q_1 q_2}{F}}

So if we want to make the force F twice as strong,

F' = 2F

the new distance between the charges would be

d'=\sqrt{k\frac{q_1 q_2}{(2F)}}=\frac{1}{\sqrt{2}}\sqrt{k\frac{q_1 q_2}{(2F)}}=\frac{d}{\sqrt{2}}

so the correct option is E.

8 0
2 years ago
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