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zmey [24]
3 years ago
5

A tank of water is 4m deep. How deep does it appear when seen normally?​

Physics
1 answer:
Dennis_Churaev [7]3 years ago
5 0

Answer:

The tank appears to be 3 m deep only. Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

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Liula [17]

Answer:A skier could benefit from a sports-specific training program because it could help them become better at skiing. ... Other activities such as squats and lunges will also improve skiing skills since it builds up lower body strength which is necessary to ski tough terrain.

Explanation:

7 0
3 years ago
I need help please I will give points
stira [4]

Answer:

-5 N of force

Explanation:

Hope it helps!

3 0
2 years ago
How many joules of electrical energy is transferred per second by 6V 0.5 A lamp?
Degger [83]

When you ask for "joules per second", you're asking for "watts".
The rate of energy "transfer" is 'power'.  In this case, the light bulb
transfers energy out of the electrical circuit and into the space
around it, in the form of light and heat radiation.

Electrical power = (voltage) x (current) =

                              (6 volts) x (0.5 ampere) =

                                          3 watts  =  3 joules per second.
 
4 0
3 years ago
On a bet, you try to remove water from a glass by blowing across the top of a vertical straw immersed in the water. What is the
MArishka [77]

Answer:

       v₂ = 0.56 m / s

Explanation:

This exercise can be done using Bernoulli's equation

        P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Where points 1 and 2 are on the surface of the glass and the top of the straw

The pressure at the two points is the same because they are open to the atmosphere, if we assume that the surface of the vessel is much sea that the area of ​​the straw the velocity of the surface of the vessel is almost zero v₁ = 0

The difference in height between the level of the glass and the straw is constant and equal to 1.6 cm = 1.6 10⁻² m

We substitute in the equation

         P_{atm} + ρ g y₁ = P_{atm} + ½ ρ v₂² + ρ g y₂

         ½ v₂² = g (y₂-y₁)

        v₂ = √ 2 g (y₂-y₁)

Let's calculate

        v₂ = √ (2 9.8 1.6 10⁻²)

       v₂ = 0.56 m / s

5 0
2 years ago
If a roller coaster cart, with a mass of 100 kg, traveled this coaster, how much kinetic energy would it have at point 'E'?
zzz [600]

Answer:

Explanation:

Assuming no friction between the roller coaster car and the hill, and neglecting air resistance, the kinetic energy the roller coaster car would have at the bottom of the hill would be equal to its gravitational potential energy at the top of the hill, by conservation of energy.

8 0
3 years ago
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