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eduard
3 years ago
14

A student connects four AA batteries (1.5 V each) in series to light up a light bulb. The circuit has a resistance of 50 Ω. How

much current flows through the circuit?
Physics
2 answers:
VLD [36.1K]3 years ago
5 0

Answer:

0.3

Explanation:

V = I R

1.5= I (50)

I = 1.5/50

I = 0.3 J

Therefore, 0.3joule current flows through the circuit.

Hope this helped

Phantasy [73]3 years ago
5 0

Answer:

0.3

Explanation:

V = I R

1.5= I (50)

I = 1.5/50

I = 0.3 J

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2/25/20 or 2/28/20 Dispatch #53
mixer [17]

Answer:

Power = 21[W]

Explanation:

Initial data:

F = 35[N]

d = 18[m]

In order to solve this problem we must remember the definition of work, which tells us that it is equal to the product of a force for a distance.

Therefore:

Work = W = F*d = 35*18 = 630 [J]

And power is defined as the amount of work performed in a time interval.

Power = Work / time

Time = t = 30[s]

Power = 630/30

Power = 21 [W]

3 0
3 years ago
How ocean currents effect temperature and the amount of moisture of the air mass above coastlines
GarryVolchara [31]
Hotter ocean tempatures mean more moisture in the dense air mass
8 0
3 years ago
How much power does it take to lift a 30.0 n box 10.0 m high in 5.00 s, if you must apply a 62n force to lift the box?
Illusion [34]
Power is defined as the rate at which the body is doing work:
P=\frac{W}{t}
Work is defined as displacement done by the force times that displacement:
W=F\cdot h
We know that we need 62N to move the box, so when we apply this force along the path of 10m we have done:
W=62N\cdot10m=620J
of work.
Now we just divide that by 5s to get how much power is required:
P=\frac{620J}{5s}=124W
5 0
3 years ago
Which of the following is a unit of rotational speed?
gladu [14]
C. Rotations per second
Or normally we'd use Radians Per second  

_Brainliest if helped!!
3 0
3 years ago
A car of mass m, traveling at constant speed, rides over the top of a round hill. How do the normal force of the road on the car
dybincka [34]

Answer:

The normal force will be lower than the gravitational force acting on the car. Therefore the answer is N < mg, which is <em>option B</em>.

Explanation:

Over a round hill, the centripetal force acting toward the the radius of the hill supports the gravitational force (mg) of the car. This notion can be expressed mathematically as follows:

At the top of a round hill

Normal force = Gravitational force - centripetal force

At the foot of a round hill

Normal Force = centripetal force + Gravitational force

4 0
3 years ago
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