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3 years ago

12

Which formula uses mass in the equation?

1 answer:

Naya [18.7K]3 years ago

5 0

The mass can be calculated by dividing the net force acting on an object by the acceleration of the object. When talking about net force, we use the units kilogram meter per second squared. This is also known as a Newton. The units for acceleration is meters per seconds squared, and the units for mass are kilograms.

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Why does it take the water in a swimming pool some time to heat up during a hot day?

BartSMP [9]

Because the top layer of a pool will be warmer than the bottom layer, that why filtration is important to cycle the water evenly.

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4 years ago

What is the force acting on an object of mass 10 kg moving with a uinform velocity of 10 m/s ?

Rus_ich [418]

If its uniform there is no force.

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8 0

3 years ago

Muscular Endurance is the ability of a muscle to continue to perform without fatigue.

Art [367]

Answer:

True.

Explanation:

Defenintion of Muscular Endurance:

The ability of a muscle (or set of muscles) to perform a repeated action without tiring.

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3 years ago

Lead hs a density of 11.3 g/cm3 and a mass of 282.5 g . what is its volume?

JulsSmile [24]

D = m / V

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4 0

4 years ago

26.0 g of copper pellets are removed from a 300∘C oven and immediately dropped into 120 mL of water at 21.0 ∘C in an insulated c

Scilla [17]

Answer:

The new water temperature is 26.4 °C

Explanation:

Given;

mass of copper, $M_{cu}$ = 26 g = 0.026 kg

temperature of copper, t = 300 °C

volume of water, V = 120 mL = 0.12 L

temperature of water, t = 21 °C

density of water, ρ = 1 kg/L

mass of water = density x volume

mass of water = (1 kg/L) x 0.12 L = 0.12 kg

heat lost by copper = heat gained by water

Both copper and water reach final temperature, T

Heat gained by water, $Q_w$ = $m_w$ cΔθ = $m_w C(T - t)$

$Q_w = m_w C(T - t)\\\\Q_w = 0.12*4200(T-21)\\\\Q_w = 504(T-21)$

Heat lost by copper is given by;

$Q_{cu} = m_{cu}C(300-T)\\\\Q_{cu} = 0.026*385(300-T)\\\\Q_{cu} = 10.01(300 - T)$

$Q_{cu} = Q_w$

504(T- 21) = 10.01(300 - T)

504 T - 10584 = 3003 - 10.01 T

504 T + 10.01 T= 3003 + 10584

514.01 T = 13587

T = (13587) / 514.01

T = 26.4 °C

Therefore, the new water temperature is 26.4 °C

7 0

3 years ago