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ankoles [38]
3 years ago
6

Why do all objects above earth's surface have gravitational potential energy

Physics
1 answer:
gayaneshka [121]3 years ago
6 0
Gravitational potential energy<span> is </span>energy<span> an object possesses because of its position in a </span>gravitational<span> field. The most common use of </span>gravitational potential energy<span> is for an object near the surface of the Earth where the </span>gravitational<span> acceleration can be assumed to be constant at about 9.8 m/s</span>2<span>.</span>
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Weather balloons are filled with only a small amount of helium because the ____________________ of the balloon will increase as
Shkiper50 [21]
Weather balloons are filled with only a small amount of helium because the __Volume__. of the balloon will increase as the air pressure decreases at higher altitudes.
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3 years ago
In moving out of a dormitory at the end of the semester, a student does 1.82 x 104 J of work. In the process, his internal energ
emmainna [20.7K]

Answer:

Explanation:

(a) Work done, W = 1.82 x 10^4 J

(b) internal energy, U = - 4.07 x 10^4 J ( as it decreases)

(c) According to the first law of thermodynamics  

Q = W + U

Q = 1.82 x 10^4 - 4.07 x 10^4

Q = - 2.25 x 10^4 J

4 0
2 years ago
A playground merry-go-round has radius 2.10m and moment of inertia 2500kg*m^2 about a vertical axle through its center, and it t
vladimir2022 [97]

Answer

Radius of the wheel r = 2.1 m

Moment of inertia I = 2500 Kg m²

Tangential force applied F = 18 N

Time interval t = 16 s

Initial angular speed ω1 = 0

Final angular speed ω2 = ?

Let α be the angular acceleration.

Torque applied τ = Iα

                         F r = Iα

Angular acceleration α = F r/I

                                    = \dfrac{18\times 2.1}{2500}

                                    = 0.015 rad/s²

(a)From rotational kinematic relation

            Final angular speed ω₂ = ω₁ + αt

                                                 = 0 + (0.015 rad/s^2 * 16 s)

                                                 = 0.24 rad/s

(b) Work done W = 0.5 Iω₂² - (1/2)Iω₁²

                      = 0.5*( 2500 Kg m²)(0.24 rad/s)^2 - 0

                      =  72 J

(c) Average power supplied by the child P = W/t = \dfrac{72}{16}

                                                                        = 4.5 watt        

8 0
3 years ago
What is the best way to ensure that the results of your experiment are accurate?
Ainat [17]
Re do the experiment 3 times over again and if you get the same results then it should be correct
6 0
3 years ago
A bullet fired horizontally hits the ground in 0.5 sec. If it had been fired with a much higher speed in the same direction, and
stira [4]

Answer:

3. 0.5 sec.

Explanation:

A bullet fired horizontally follows a projectile motion, which consists of two independent motions:

- A horizontal motion with constant speed

- A vertical motion with  constant acceleration, g = 9.8 m/s^2, towards the ground

The time taken for the bullet to reach the ground can be calculated just by considering the vertical motion:

y(t) = h + v_{0y} t - \frac{1}{2}gt^2

where y is the vertical position at time t, h is the initial height, and v_{0y} is the initial vertical velocity of the bullet.

Since the bullet is fired horizontally, v_{0y}=0. So the equation becomes

y(t) = h - \frac{1}{2}gt^2

And the time that the bullet takes to reach the ground can be found by requiring y=0 and solving for t:

t=\sqrt{\frac{2h}{g}}

As we can see, in this equation there is no dependance on the initial speed of the bullet: therefore, if the bullet is fired still horizontally but with a different speed, it will still take the same time (0.5 s) to reach the ground.

4 0
2 years ago
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