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alisha [4.7K]
3 years ago
13

A calculator has a resistance of 22 Ω. What is the power rating for this calculator when connected to a 1.5 V battery?

Physics
1 answer:
GREYUIT [131]3 years ago
3 0

Answer:

33

Explanation:

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Just after a motorcycle rides off the end of a ramp and launches into the air, its engine is turning counterclockwise at 8325 re
alexgriva [62]

Answer:

\frac{Ie}{lm} = 1.10*10^{-3}

Explanation:

GIVEN DATA:

Engine operating speed nf = 8325 rev/min

engine angular speed ni= 12125 rev/min

motorcycle angular speed N_m= - 4.2 rev/min

ratio of moment of inertia of engine to motorcycle is given as

\frac{Ie}{lm} = \frac{-N}{(nf-ni)}

\frac{Ie}{lm} = \frac{-(-4.2)}{(12125 - (8325))}

\frac{Ie}{lm} = 1.10*10^{-3}

5 0
3 years ago
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What is the atmospheric pressure 1.00 km above the surface of Venus? Express your answer in Earth-atmospheres.
EleoNora [17]

Below is an attachment containing the solution.

4 0
2 years ago
Substances which naturally attract each other called what
grandymaker [24]

Answer:

Ferromagnetism, physical phenomenon in which certain electrically uncharged materials strongly attract others. Two materials found in nature, lodestone (or magnetite, an oxide of iron, Fe3O4) and iron, have the ability to acquire such attractive powers, and they are often called natural ferromagnets.

Hope this help :)

5 0
2 years ago
. A chemistry student’s height is measured at 68.5 inches. How tall is the student in cm?
tatiyna

1 inch = 2.54 centimeters

All we need to do is multiply.

68.5 * 2.54 = 173.99cm

Best of Luck!

3 0
3 years ago
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If the mass of the block is 5 kg and the speed 7 m/s, what is the work done <br> on the block?
Kamila [148]

Answer:

A block of mass M = 5 kg is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force F = 40N is applied, the acceleration of the block will be then (g=10ms

2 ).

Mass of the block=5kg

Coeffecient of friction=0.2

external applied force, F=40N

The angle at which the force is applied=30degree

So the horizontal component of force=Fcos30=40×

23 =20 3 N

While the uertical component of the force acting in upward direction=Fsin30=40× 21

​ =20N

The normal reaction from the surface (N)=mg−Fsin30=50−20=30N

So the ualue of limiting friction=μN=0.2×30=6N

Hence the net horizontal force on the block=Fcos30=μN=20

3

​

N−6N=28.64N

The horizontal acceleration of the block=

m

Fcos30−μN = 528.64

​  =5.73m/s 2

8 0
3 years ago
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