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maksim [4K]
3 years ago
5

a physics student throws a stone horizontally off a cliff. one second later, he throws a second identical stone in exactly the s

ame way. which of the following statements is true?
Physics
1 answer:
Virty [35]3 years ago
8 0

The second stone hits the ground exactly one second after the first.

The distance traveled by each stone down the cliff is calculated using second kinematic equation;

h = v_0_yt + \frac{1}{2} gt^2

where;

  • <em>t is the time of motion </em>
  • <em />v_0_y<em> is the initial vertical velocity of the stone = 0</em>

h = \frac{1}{2} gt^2

The time taken by the first stone to hit the ground is calculated as;

t_1 = \sqrt{\frac{2h}{g} }

When compared to the first stone, the time taken by the second stone to hit the ground after 1 second it was released is calculated as

t_2 = \sqrt{\frac{2h}{g} } + 1

t_2 = t_1 + 1

Thus, we can conclude that the second stone hits the ground exactly one second after the first.

"<em>Your question is not complete, it seems be missing the following information;"</em>

A. The second stone hits the ground exactly one second after the first.

B. The second stone hits the ground less than one second after the first

C. The second stone hits the ground more than one second after the first.

D. The second stone hits the ground at the same time as the first.

Learn more here:brainly.com/question/16793944

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  • Work=9J

\\ \sf\longmapsto W=Fd

\\ \sf\longmapsto d=\dfrac{W}{F}

\\ \sf\longmapsto d=\dfrac{9}{12}=0.72m

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2 years ago
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muminat

Power P is the rate at which energy is generated or consumed and hence is measured in units that represent energy E per unit time t. This is:

P = E/t

Solving for t:

t = E/P

t = 6007 J / 500 W

t = 12.014 s

<h2>t ≅ 12 s</h2>

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3 years ago
A well lagged copper calorimeter of mas 120g contains 70g of water and 10g ice both at 0°C . Dry steam at 100°C is passed in unt
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Answer:

7.6 g

Explanation:

"Well lagged" means insulated, so there's no heat transfer between the calorimeter and the surroundings.

The heat gained by the copper, water, and ice = the heat lost by the steam

Heat gained by the copper:

q = mCΔT

q = (120 g) (0.40 J/g/K) (40°C − 0°C)

q = 1920 J

Heat gained by the water:

q = mCΔT

q = (70 g) (4.2 J/g/K) (40°C − 0°C)

q = 11760 J

Heat gained by the ice:

q = mL + mCΔT

q = (10 g) (320 J/g) + (10 g) (4.2 J/g/K) (40°C − 0°C)

q = 4880 J

Heat lost by the steam:

q = mL + mCΔT

q = m (2200 J/g) + m (4.2 J/g/K) (100°C − 40°C)

q = 2452 J/g m

Plugging the values into the equation:

1920 J + 11760 J + 4880 J = 2452 J/g m

18560 J = 2452 J/g m

m = 7.6 g

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The answer would be 10ms 2

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