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gtnhenbr [62]
4 years ago
8

What is the weight of a 10-kilogram object on Earth?

Physics
1 answer:
Rasek [7]4 years ago
8 0
98 N weight on object
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A clam dropped by a seagull takes 3.0 seconds to hit the ground. What is the seagull's approximate height above the ground at th
ankoles [38]
<h2>The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 9.81 m/s²  

        Time, t = 3 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 0 x 3 + 0.5 x 9.81 x 3²

                      s = 44.145 m

The seagull's approximate height above the ground at the time the clam was dropped is 4 m

4 0
4 years ago
Which statement is true about the force of gravity?
kumpel [21]

Answer:

The moon's gravity pulls the Earth to make tides.

Explanation:

The Moons Gravity Pulls On The Earth With Different  Strenght Making High Tide And Low Tide.

Hope This Helps!

4 0
2 years ago
How did Archimedes measure the mass of the block of gold?
MA_775_DIABLO [31]
Archimedes found a piece of gold and a piece of silver with exactly the same mass. He dropped the gold  into a bowl filled to the brim with water and measured the volume of water that spilled out. Then he did the same thing with the piece of solver. Although both metals had the same mass, the silver gad a larger volume; therefore, it displaced more water than the gold did. That's because the silver was less dense than gold. Afterwards he applied the same method to the crown for the king he served who had got a new crown from a jeweler who gave it to him. Archimedes found a piece of pure gold that had the same mass as the crown. He placed the pure gold chuck and the crown in water, one at a time. The crown displaced more water the piece of gold. Therefore, its density was less than pure gold.  
5 0
3 years ago
An air bubble released by a remotely operated underwater vehicle, 120 m below the surface of a lake, has a volume of 1.40 cm3. T
Usimov [2.4K]

Answer:

17.7 cm^3

Explanation:

depth, h = 120 m

density of water, d = 1000 kg/m^3

V1 = 1.4 cm^3

P1 = P0 + h x d x g

P2 = P0

where, P0 be the atmospheric pressure

Let V2 be the volume of the bubble at the surface of water.

P0 = 1.01 x 10^5 Pa

P1 = 1.01 x 10^5 + 120 x 1000 x 9.8 = 12.77 x 10^5 Pa

Use

P1 x V1 = P2 x V2

12.77 x 10^5 x 1.4 = 1.01 x 10^5 x V2

V2 = 17.7 cm^3

Thus, the volume of bubble at the surface of water is 17.7 cm^3.

7 0
3 years ago
Tulongan nyo ko guys​
Whitepunk [10]
Ahhhh this is a USA server not philipino server so yeah sorry
6 0
3 years ago
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