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4 years ago

How is power calculated? multiply force and displacement divide force by displacement multiply work and time divide work by time

Physics

2 answers:

erastovalidia [21]4 years ago

8 0

Answer:

B on edge

Explanation:

Savatey [412]4 years ago

6 0

Answer:

divide work by time

Explanation:

Power=work/time.

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What happens to particles in a surface wave (such as a water wave)?

AlekseyPX

Your question: What happens to particles in a surface wave (such as a water wave)?

My answer: B, They move perpendicular and parallel to the direction of wave motion.

Explanation: This is practically the answer the question below (or above me) tried to explain.

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3 years ago

Read 2 more answers

Suddenly a worker picks up the bag of gravel. Use energy conservation to find the speed of the bucket after it has descended 2.3

fiasKO [112]

Explanation:

A worker picks up the bag of gravel. We need to find the speed of the bucket after it has descended 2.30 m from rest. It is case of conservation of energy. So,

$\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh}$

h = 2.3 m

$v=\sqrt{2\times 9.8\times 2.3} \\\\v=6.71\ m/s$

So, the speed of the bucket after it has descended 2.30 m from rest is 6.71 m/s.

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3 years ago

One way to cool a gas is to let it expand. When a certain gas under a pressure of 5.00 x 10^6 Pa at 25.°C is allowed to expand t

hichkok12 [17]

Answer:

The final temperature is T2= 5.35°C

Explanation:

Apply the Gay-lussacs's law we have

$\frac{P1}{T1} = \frac{P2}{T2}$

P1, initial pressure= 5.00 x 10^6 Pa

T1, initiation temperature= 25.°C

P2, final pressure= 1.07 x 10^6 Pa

T2, final temperature= ?

$\frac{5.00 * 10^6}{25} =\frac{1.07 *10^6}{T2} \\$

Cross multiplying and making T2 subject of formula we have

$T2 =\frac{1.07 *10^6*25}{5.00 * 10^6} \\\\T2= \frac{26.75}{5} \\T2= 5.35$

T2= 5.35°C

6 0

3 years ago

What minimum heat is needed to bring 250 g of water at 20 ∘C to the boiling point and completely boil it away? The specific heat

Amiraneli [1.4K]

Answer:633.8 KJ

Explanation:

Given

mass of water $\left ( m\right )=250gm$

Initial temperature $\left ( T_i\right )=20^{\circ}C$

Final temperature $\left ( T_f\right )=100^{\circ}C$

Specific heat of water $\left ( c \right )$ =4190 J/kg-k

heat of vaporization $\left ( L\right )=22.6\times 10^5 J/kg$

Heat required for process $\left ( Q\right )$ =heat to raise water temperature from 20 to 100 +Heat to vapourize water completely

Q=mc $\left ( T_f-T_i\right )+mL$

Q= $0.25\times 4190\times \left ( 100-20\right )+0.25\times 22\times 10^5$

Q= $\left ( 0.838+5.5\right )\times 10^5$

Q= $6.338\times 10^5J=633.8 KJ$

4 0

4 years ago

Which statement describes between a cup if cold water and a cup of boiling water

marishachu [46]

The thermal energy keeps these two changing their temperature over time.

The surrounding temperature (environment) clearly effects the cold water AND the boiling water.

Eventually, both would reach an equilibrium in temperature. <span />

4 0

3 years ago