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alexdok [17]
3 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]3 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]3 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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Use the passage to answer the question.
Zielflug [23.3K]

The reasoning in the passage is : ( C ) The spots are the result of interference, which happens with waves but not particles

<h3>Light waves </h3>

Light passed through a slit and projected onto a screen will form a pattern which is very visible and the pattern might have dark spots if there is some form of interference. Interference is present in waves and not particles.

Hence we can conclude that the reasoning in the passage is The spots are the result of interference, which happens with waves but not particles

Learn more about light waves : brainly.com/question/25847009

8 0
2 years ago
Martha must carry a 45 N package up three flights of stairs. Each flight of stairs has a height of 2.3m, and the actual distance
Darina [25.2K]

Answer:

310.5 J

Explanation:

The total work done by Martha is equal to the increase in gravitational potential energy of the package, which is equal to

\Delta U = mg\Delta h

where

(mg) = 45 N is the weight of the package

\Delta h is the increase in height of the package

The package is carried up 3 flights of stairs, each one with a height of 2.3 m, so the total increase in heigth is

\Delta h = 3 \cdot 2.3 m=6.9 m

And so, the work done by Martha is

U=(45 N)(6.9 m)=310.5 J

3 0
3 years ago
. Imagine that you are standing at the center of a giant bowl of gelatin. What type of wave will you make across the top of the
vichka [17]
Transverse wave as the wave is going up and down no compressions
3 0
3 years ago
E<br> 3.6 What force is needed to give a mass of<br> 20 kg an acceleration of 5 m/s??
luda_lava [24]

Explanation:

  • Mass(m)= 20kg
  • Acceleration (a)= 5m/s²
  • Force(F)= ?

We know that,

  • F=ma
  • F=20×5
  • F=100N

Hence, the needed force is 100N.

6 0
2 years ago
Read 2 more answers
A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the
kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring  73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E  =  2000 N/C

Electric field due tor ring is guiven as

E = \frac{KQx}{[x^2+ R^2]^{3/2}}

E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}

E1 = 3714.672 N/C

electric field due to point charge q

E  =\frac[kq}{x^2}

E = \frac{9\times 10^9 \times q}{0.70^2}

E2 = 1.837\times 10^{10}\times q

now the eelctric charge at point P is

E = E1 + E22000 =  3714.672 + 1.837\times 10[10} \times q

solving for q

q = - 93.334 nC

7 0
3 years ago
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