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alexdok [17]
4 years ago
9

when a seagull picks an oyster up into the sky and then lets it drop on the rocks below to open the shell; where is the oysters

potential energy greatest? where is it’s kinetic energy greatest?
Physics
2 answers:
Sati [7]4 years ago
5 0

Answer:

When a seagull picks an oyster up into the sky and then lets it drop on the rocks below to open the shell; where is the oyster's potential energy greatest? Where is its kinetic energy greatest?

Potential energy is greatest at maximum height; kinetic energy is greatest just before the oyster strikes the ground.

Explanation:

vazorg [7]4 years ago
3 0

Answer:

Potential energy is greatest right before the bird drops it. Kinetic energy is greatest the instant before the oyster hits the ground

Explanation:

Potential energy is the energy stored in an object you store energy in an object by raising it higher in the air.

Kinetic energy is moving energy as the oyster falls it loses the stored energy and it goes into the moving energy which is kinetic.  

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kaheart [24]
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<h3>B ) The force applied on the smaller gear is transmitted without any loss to the larger gear .</h3><h3 /><h3>C ) the direction of motion can be changed without changing the direction of the applied force .</h3>

D ) the system would continue to move without any further, after and initial force has set in motion.

6 0
3 years ago
What is the speed of a wave on a string with a wavelength of 1.75 m and a frequency of 2.0 Hz
deff fn [24]

Answer:

V=3.5 m/s

Explanation:

V=(F)(W)

V=(2)(1.75)

V= 3.5 m/s

7 0
3 years ago
A volumen constante un gas ejerce una presión de 880 mmHg a 20o Celsius dentro de una olla a presión ¿Qué temperatura habrá si e
jasenka [17]

Answer: Hence, the final temperature is 350 K

Explanation :

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1\text{ and }T_1 are the initial pressure and temperature of the gas.

P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:

P_1=880mmHg\\T_1=20^0C=(20+273)K=293K\\P_2=1050mmHg\\T_2=?

Putting values in above equation, we get:

\frac{880mmHg}{293K}=\frac{1050mmHg}{T_2}\\\\T_2=350K

Hence, the final temperature is 350 K

8 0
3 years ago
A U-shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully poured into the left arm of
Gekata [30.6K]

Answer:

a) P=2450\ Pa

b) \delta h=23.162\ cm

Explanation:

Given:

height of water in one arm of the u-tube, h_w=25\ cm=0.25\ m

a)

Gauge pressure at the water-mercury interface,:

P=\rho_w.g.h_w

we've the density of the water =1000\ kg.m^{-3}

P=1000\times 9.8\times 0.25

P=2450\ Pa

b)

Now the same pressure is balanced by the mercury column in the other arm of the tube:

\rho_w.g.h_w=\rho_m.g.h_m

1000\times 9.8\times 0.25=13600\times 9.8\times h_m

h_m=0.01838\ m=1.838\ cm

<u>Now the difference in the column is :</u>

\delta h=h_w-h_m

\delta h=25-1.838

\delta h=23.162\ cm

7 0
3 years ago
What is the acceleration of a 59 kg object pushed with a force of 500 Newton's ?
Igoryamba

Answer:

a=500/50=10 m/s^2

Explanation:

f=m•a     a=f/m

a=500/50= 10 m/s^2

4 0
3 years ago
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