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guajiro [1.7K]
2 years ago
15

Two rocks, a and b, are thrown horizontally from the top of a cliff. rock a has an initial speed of 10 meters per second and roc

k b has an initial speed of 20 meters per second. how would you describe the time it takes for the rocks to reach the ground and the horizontal distance they travel? both rocks hit the ground at the same time and at the same distance from the base of the cliff. both rocks hit the ground at the same time, but rock b lands twice as far as rock a from the base of the cliff. both rocks hit the ground at the same time, but rock a lands twice as far as rock b from the base of the cliff. rock b hits the ground before rock a, and rock b lands twice as far as rock a from the base of the cliff.
Physics
1 answer:
Arte-miy333 [17]2 years ago
5 0

Both rocks hit the ground at the same time and at the same distance from the base of the cliff. Option A is correct.

<h3>What is speed ?</h3>

Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. it is denoted by u for the initial speed while u for the final speed. its SI unit is m/sec.

If the  resistance is ignored, things falling near the Earth's surface have the same estimated acceleration due to gravity of 9.8 meters per second.

As a result, the objects' acceleration is the same, and their velocity is growing at a consistent pace.

So that both rocks hit the ground at the same time and at the same distance from the base of the cliff.

Hence, option A is correct.

To learn more about the sped, refer to the link;

brainly.com/question/7359669

#SPJ1

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The current is reduced to half of its original value.

Explanation:

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        I_{1} = \frac{V}{R_{int} +r_{L} }

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       I_{1} = \frac{V}{R_{int} +r_{L} } = \frac{V}{2*r} (1)

  • When the battery ages, if the internal resistance triples, the new current can be found using Ohm's Law again:

       I_{2} = \frac{V}{R_{int} +r_{L} } = \frac{V}{(3*r) +r} = \frac{V}{4*r}  (2)

  • We can find the relationship between I₂, and I₁, dividing both sides, as follows:

        \frac{I_{2} }{I_{1} } = \frac{V}{4*r} *\frac{2*r}{V} = \frac{1}{2}

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Change in electric potential is given by

E=q(V_2-V_1)\\\Rightarrow E=(19e-18e)(0-(-85.5\times 10^{-3})\\\Rightarrow E=1.6\times 10^{-19}\times 85.5\times 10^{-3}\\\Rightarrow E=1.368\times 10^{-20}\ J

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