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kaheart [24]
3 years ago
7

Will give brainliest!!

Physics
1 answer:
givi [52]3 years ago
4 0
A. Move the candle to the right, or the focal point to the left.

For a convex lens, the closer an object is to the focal point, the larger it’s image is (and therefore the greater the magnification is). The two ways you could make the candle be closer to the focal point are to move the candle to the right, or the focal point to the left.
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My sis needs help with this and I don't wanna help her!!! Can you guys help her????
Katena32 [7]

Answer and Explanation:

Cup 1. The ice cubes are in a solid state.

Cup 2: The cup has water that is liquified.

Cup 3: The cup is empty and only contains free floating air particles.

***If you found my answer helpful, please give me the brainliest. :) ***

7 0
3 years ago
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What is meaning of convection​
crimeas [40]

Answer:

Convection is heat transfer through the movement of liquids and gases.

3 0
3 years ago
What force is needed to move a 5 kg mass with an acceleration of 5 m/s²?
ElenaW [278]

Answer:

b)25N

Explanation:

F=ma

F=(5kg)(5m/s^2)

F=25N

4 0
2 years ago
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Why do you think it is important that the U.S. Constitution defines citizenship?
velikii [3]

Answer:

It's important that the U.S. Constitution defines citizenship because it guarantees it to people who are born within the United States.

Explanation:

It"subjects to the jurisdiction thereof".

5 0
3 years ago
A sealed tank containing seawater to a height of 12.8 m also contains air above the water at a gauge pressure of 2.90 atm . Wate
exis [7]

Answer:

The velocity of water at the bottom, v_{b} = 28.63 m/s

Given:

Height of water in the tank, h = 12.8 m

Gauge pressure of water, P_{gauge} = 2.90 atm

Solution:

Now,

Atmospheric pressue, P_{atm} = 1 atm = 1.01\tiems 10^{5} Pa

At the top, the absolute pressure, P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa

Now, the pressure at the bottom will be equal to the atmopheric pressure, P_{b} = 1 atm = 1.01\times 10^{5} Pa

The velocity at the top, v_{top} = 0 m/s, l;et the bottom velocity, be v_{b}.

Now, by Bernoulli's eqn:

P_{t} + \frac{1}{2}\rho v_{t}^{2} + \rho g h_{t} = P_{b} + \frac{1}{2}\rho v_{b}^{2} + \rho g h_{b}

where

h_{t} -  h_{b} = 12.8 m

Density of sea water, \rho = 1030 kg/m^{3}

\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} =  v_{b}

\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} =  v_{b}

v_{b} = 28.63 m/s

5 0
3 years ago
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