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AURORKA [14]
3 years ago
12

What happens to an object's gravitational potential energy as it falls from some height?​

Physics
1 answer:
saveliy_v [14]3 years ago
3 0

Answer:

Decreases

Explanation:

As an object falls from some height, the gravitational potential energy of the object decreases.

Potential energy is the energy due to the position of a body.

The potential energy is mathematically expressed as:

   Potential energy  = mass x acceleration due to gravity x height

So, potential energy is directly proportional to mass and height.

The higher the mass and height, the more the potential energy.

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Resistance training should be performed every day for maximum conditioning benefits.
marysya [2.9K]

Answer:

The answer is True

7 0
3 years ago
Read 2 more answers
As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th
NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek

Conservation of Momentum:

P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1

Energy Balance:

\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2}  ....... Eq 2

Substitute Eq 2 into Eq 1

0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V  \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}

Using Eq 1

0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m

7 0
3 years ago
what equastion do you use to solve Riders in a carnival ride stand with their backs against the wall of a circular room of diame
Hitman42 [59]

Answer:

μsmín = 0.1

Explanation:

  • There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
  • This friction force has a maximum value, that can be written as follows:

       F_{frmax} = \mu_{s} *F_{n} (1)

       where  μs is the coefficient of static friction, and Fn is the normal force,

       perpendicular to the wall and aiming to the center of rotation.

  • This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
  • This force has the following general expression:

       F_{c} =  m* \omega^{2} * r (2)

       where ω is the angular velocity of the riders, and r the distance to the

      center of rotation (the  radius of the circle), and m the mass of the

      riders.

      Since Fc is actually Fn, we can replace the right side of (2) in (1), as

      follows:

     F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)

  • When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

       m* g = m* \mu_{smin} * \omega^{2} * r (4)

  • (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
  • Cancelling the masses on both sides of (4), we get:

       g = \mu_{smin} * \omega^{2} * r (5)

  • Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

      60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)

  • Replacing by the givens in (5), we can solve for μsmín, as follows:

       \mu_{smin} = \frac{g}{\omega^{2} *r}  = \frac{9.8m/s2}{(6.28rad/sec)^{2} *2.5 m} =0.1 (7)

5 0
2 years ago
The magnetic field at 8 cm distance from a long straight wire, carrying is 0.2x10^-5 T. How much is the electric current in the
FrozenT [24]

Answer:

The electric current in the wire is 0.8 A

Explanation:

We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:

B=\frac{2\pi*a }{u*I}

B= Magnetic field due to a straight and long wire that carries current

u= Free space permeability

I= Electrical current passing through the wire

a  = Perpendicular distance from the wire to the point where the magnetic field is located

Magnetic Field Calculation

We cleared (I) of the formula (1):

I=\frac{2\pi*a*B }{u} Formula(2)

B=0.2*10^{-5}  T = 0.2*10^{-5} \frac{weber}{m^{2} }

a  =8cm=0.08m

u=4*\pi *10^{-7} \frac{Weber}{A*m}

We replace the known information in the formula (2)

I=\frac{2\pi*0.08*0.2*10^{-5}  }{4\pi *10x^{-7} }

I=0.8 A

Answer: The electric current in the wire is 0.8 A

4 0
3 years ago
A student weighing 120 lbs climbs a 12 ft flight of stairs in 9 seconds. how much power did the student create?
Alex Ar [27]
Power can be calculate through the equation,
                        Power = Force x velocity

It should be noted that velocity is calculated by dividing displacement by time. Thus, from the given in this item we can calculate for the power. 
                       Power = (120 lb) x (12 ft/9 s)
                         <em> </em><span><em>Power   = 160 lb.ft/s</em></span>
7 0
3 years ago
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