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zhannawk [14.2K]

3 years ago

14

A body of mass 5 kg is at rest what would be the magnitude of force which will make the body move with the velocity of 10 metre

per second to the power -1
AND end of 0.5 second

1 answer:

kondor19780726 [428]3 years ago

5 0

Answer:

===========100N is required force

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Three students are having a conversation. It goes like this. Allison says: "This problem says, can you add a resistor to the cir

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Answer:

Nima and Natasha are absolutely correct.

Explanation:

When connecting two resistors in series, their resistances add:

$R_{eq}=R_{1}+R_{2}$

which means that whenever we add a resistance in series, their magnitudes will add, giving us a resistance that is greater than the original resistance, which will demand less current from the battery because of ohm's law:

$I=\frac{V}{R}$

So, the greater the resistance, the smaller the current.

When connecting two resistors in parallel, the reciprocal of ther resistances add:

$\frac{1}{R_{eq}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$

or

$R_{eq}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}$

The equivalent resistance will always be less than the smallest resistor in the circuit, so the equivalent resistance will always decrease as more resistors are added. A decrease in the resistance means that the current will increase.

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Protons have ___charge; they have equal amounts of positive and negative ____charges?

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The moon Phobos orbits Mars

shepuryov [24]

$27.9816 \times 10^{3} s$ is the period of orbit.

<u>Explanation: </u>

The equation that is useful in describing satellites motion is Newton form after Kepler's Third Law. The period of the satellite (T) and the average distance to the central body (R) are related as the following equation:

$\frac{T^{2}}{R^{3}}=\frac{4 \times \pi^{2}}{G \times M_{c e n t r a l}}$

Where,

T is the period of the orbit

R is the average radius of orbit

G is gravitational constant $6.673 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}$

Here, given data

$M=6.23 \times 10^{23} \mathrm{kg}$

$R=9.38 \times 10^{6} \mathrm{m}$

Substitute the given values, we get T as

$\frac{T^{2}}{\left(9.38 \times 10^{6}\right)^{3}}=\frac{4 \times(3.14)^{2}}{\left(6.673 \times 10^{-11}\right) \times 6.23 \times 10^{23}}$

$T^{2}=\frac{4 \times 9.8596 \times 825.29 \times 10^{18}}{41.57 \times 10^{12}}$

$T^{2}=\frac{32548.12 \times 10^{18-12}}{41.57}=782.97 \times 10^{6}$

Taking square root, we get

$T=27.9816 \times 10^{3} s$

4 0

3 years ago