8.16m is the required height, a 5kg stone need to be raised.
One sort of potential energy is gravitational potential energy, which is equal to the product of the object's mass (m), the gravitational acceleration (g), and the object's height (h) as measured in relation to the ground's surface (the body).
We obtain the formula by considering the work done in raising a mass m through a height h.
Work in elevating mass m through height h is equal to force times distance.
The force must be greater than the mass m's weight, hence F = mg.
Work done = mgh = gravitational potential energy
Energy = Mass of the object × gravitational acceleration × height.
Mass of the stone = 5kg
Equating ;
∴ 400 J = 5 kg × 9.8 m/s² × height
Height = 8.16 m
Therefore, 8.16m is the required height.
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Answer:
1. A force is a push or pull upon an object resulting from the object's interaction with another object.
Explanation:
Everything starts from spectroscopy. Astronomers only have concentrated information at wavelengths that are emitted from the stars. What they do with this information is to obtain the frequency range of the stars and through spectroscopes they are responsible for dividing the radiation beams and determining the coincidence with the emission of those same waves, of chemical elements. From these observation techniques it is possible to obtain the composition and according to the color, obtaining characteristics such as temperature. The spectrum of stars consists of dark and bright lines called Fraunhofer lines. This spectrum is compared to the spectrum of different elements to find the composition of the stars. This is possible because the elements emit or absorb only specific wavelengths.
Let <em>F</em> be the magnitude of the force applied to the cart, <em>m</em> the mass of the cart, and <em>a</em> the acceleration it undergoes. After time <em>t</em>, the cart accelerates from rest <em>v</em>₀ = 0 to a final velocity <em>v</em>. By Newton's second law, the first push applies an acceleration of
<em>F</em> = <em>m a</em> → <em>a</em> = <em>F </em>/ <em>m</em>
so that the cart's final speed is
<em>v</em> = <em>v</em>₀ + <em>a</em> <em>t</em>
<em>v</em> = (<em>F</em> / <em>m</em>) <em>t</em>
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If we force is halved, so is the accleration:
<em>a</em> = <em>F</em> / <em>m</em> → <em>a</em>/2 = <em>F</em> / (2<em>m</em>)
So, in order to get the cart up to the same speed <em>v</em> as before, you need to double the time interval <em>t</em> to 2<em>t</em>, since that would give
(<em>F</em> / (2<em>m</em>)) (2<em>t</em>) = (<em>F</em> / <em>m</em>) <em>t</em> = <em>v</em>
Answer:
the filling stops when the pressure of the pump equals the pressure of the interior air plus the pressure of the walls.
Explanation:
This exercise asks to describe the inflation situation of a spherical fultball.
Initially the balloon is deflated, therefore the internal pressure is equal to the pressure of the air outside, atmospheric pressure, when it begins to inflate the balloon with a pump this creates a pressure in the inlet valve and as it is greater than the pressure inside, the air enters it, this is repeated in each filling cycle, manual pump.
When the ball is full we have two forces, the one created by the external walls and the one aired by the pressure of the pump, these forces are directed towards the inside, but the air molecules exert a pressure towards the outside, which translates into a force. When these two forces are equal, the pump is no longer able to continue introducing air into the balloon.
Consequently the filling stops when the pressure of the pump equals the pressure of the interior air plus the pressure of the walls.
