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2 years ago

Rahman drops two stones P and Q each of mass 1 kg and 2 kg simultaneously from

Physics

1 answer:

kolbaska11 [484]2 years ago

5 0

This question involves the use of the equations of motion for vertical motion.

The time taken by the stones P and Q to reach the ground is the same, that is "2 s".

The velocity with which Q hits the ground is "20 m/s".

The time taken by the stones to reach the ground can be calculated by using the second equation of motion for the vertical motion:

$h = v_it+\frac{1}{2}gt^2$

For both the stones P and Q:

h = height = 20 m

$v_i$ = initial velocity = 0 m/s

t = time = ?

g = acceleration due to gravity = 10 m/s²

Therefore,

$20\ m = (0\ m/s)t+\frac{1}{2}(10\ m/s^2)(t)^2\\\\t = \sqrt{\frac{20\ m}{5\ m/s^2}}$

t = 2 s

Hence, the time taken by both the stones to reach the ground is the same.

To find the final velocity of stone Q we will use the first equation of motion for the vertical motion:

$v_f=v_i+gt\\v_f = 0\ m/s+(10\ m/s^2)(2\ s)\\v_f = 20\ m/s$

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

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A ball thrown in the air will never go as far as physics ideally would predict because?

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Air resistance fefrgtdnjrcbrdg

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3 years ago

A swimming pool has the shape of a box with a base that measures 30 m by 10 m and a uniform depth of 2.5 m. How much work is req

pav-90 [236]

Answer:

18,375 kJ

Explanation:

Parameters given:

Density of water: 1000 kg/m³

Acceleration due to gravity = 9.8 m/s²

Dimensions of pool = 30m × 10m × 2.5m

Depth of pool = 2.5 m

First we need to find the mass of the water in the pool:

Density = mass/volume

Mass = density * volume

Since the pool is full of water, the volume of the water is equal to the volume of the pool:

Volume = 30 * 10 * 2.5 = 750 m³

Mass = 1000 * 750

Mass = 750,000 kg

We can now find the force required to pump the water out of the pool:

F = m * g

Where m = mass of water

g = acceleration due to gravity

F = 750,000 * 9.8 = 7,350,000 N

The work needed to be done is the product of the force required by the distance the water would need to move to be pumped out (depth of pool):

Work = F * d

Work = 7,350,000 * 2.5

Work = 18,375,000 J = 18,375 kJ

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3 years ago

Read 2 more answers

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at In a few mi

Nataliya [291]

Answer:

2274 J/kg ∙ K

Explanation:

The complete statement of the question is :

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

$m_{m}$ = mass of metal = 400 g

$c_{m}$ = specific heat of metal = ?

$T_{mi}$ = initial temperature of metal = 100 °C

$m_{a}$ = mass of aluminum cup = 100 g

$c_{a}$ = specific heat of aluminum cup = 900.0 J/kg ∙ K

$T_{ai}$ = initial temperature of aluminum cup = 15 °C

$m_{w}$ = mass of water = 500 g

$c_{w}$ = specific heat of water = 4186 J/kg ∙ K

$T_{wi}$ = initial temperature of water = 15 °C

$T$ = Final equilibrium temperature = 40 °C

Using conservation of energy

heat lost by metal = heat gained by aluminum cup + heat gained by water

$m_{m} c_{m} (T_{mi} - T) = m_{a} c_{a} (T - T_{ai}) + m_{w} c_{w} (T - T_{wi} ) \\(400) (100 - 40) c_{m} = (100) (900) (40- 15) + (500) (4186) (40 - 15)\\ c_{m} = 2274 Jkg^{-1}K^{-1}$