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1 year ago

What results when energy is transformed while juggling three bowling pins?

Physics

1 answer:

Nady [450]1 year ago

8 0

Answer:

his is an example of the transformation of gravitational potential energy into kinetic energy

Explanation:

The game of juggling bowling is a clear example of the conservation of mechanical energy,

when the bolus is in the upper part of the path mechanical energy is potential energy; As this energy descends, it becomes kinetic energy where the lowest part of the trajectory, just before touching the hand, is totally kinetic.

At the moment of touching the hand, a relationship is applied that reverses the value of the speed, that is, now it is ascending and the cycle repeats.

Therefore this is an example of the transformation of gravitational potential energy into kinetic energy

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Two objects are placed in thermal contact and are allowed to come to equilibrium in isolation. The heat capacity of Object A is

Oksi-84 [34.3K]

Answer:

Explanation:

Heat capacity A = 3 x heat capacity of B

initial temperature of A = 2 x initial temperature of B

TA = 2 TB

Let T be the final temperature of the system

Heat lost by A is equal to the heat gained by B

mass of A x specific heat of A x (TA - T) = mass of B x specific heat of B x ( T - TB)

heat capacity of A x ( TA - T) = heat capacity of B x ( T - TB)

3 x heat capacity of B x ( TA - T) = heat capacity of B x ( T - TB)

3 TA - 3 T = T - TB

6 TB + TB = 4 T

T = 1.75 TB

8 0

2 years ago

How much heat is released when 432 g of water cools down from 71'c to 18'c?

maria [59]

The heat released by the water when it cools down by a temperature difference $\Delta T$ is
$Q=mC_s \Delta T$
where
m=432 g is the mass of the water
$C_s = 4.18 J/g^{\circ}C$ is the specific heat capacity of water
$\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ}$ is the decrease of temperature of the water

Plugging the numbers into the equation, we find
$Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J$
and this is the amount of heat released by the water.

7 0

2 years ago

At a certain instant after jumping from the airplane A, a skydiver B is in the position shown and has reached a terminal (consta

Lubov Fominskaja [6]

Answer:

a=2330

b= 0.223secs

Explanation:

pb=2330m

t=0.223secs

6 0

2 years ago

A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c

netineya [11]

Answer:

$E=\frac{\lambda}{2\pi r\epsilon_0}$

Explanation:

We are given that

Linear charge density of wire= $\lambda$

Radius of hollow cylinder=R

Net linear charge density of cylinder= $2\lambda$

We have to find the expression for the magnitude of the electric field strength inside the cylinder r<R

By Gauss theorem

$\oint E.dS=\frac{q}{\epsilon_0}$

$q=\lambda L$

$E(2\pi rL)=\frac{L\lambda}{\epsilon_0}$

Where surface area of cylinder= $2\pi rL$

$E=\frac{\lambda}{2\pi r\epsilon_0}$

8 0

1 year ago

Planetary orbits... are spaced more closely together as they get further from the Sun. are evenly spaced throughout the solar sy

BaLLatris [955]

Answer:

E) are almost circular, with low eccentricities.

Explanation:

Kepler's laws establish that:

All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).

A planet describes equal areas in equal times (Kepler's second law).

The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).

$T^{2} = a^{3}$

Where T is the period of revolution and a is the semi-major axis.

Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.

However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)

In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).

Therefore, option A and B can not be true.

In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.

4 0

2 years ago