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3 years ago

What can increase the gravitational potential energy of a object

Physics

1 answer:

love history [14]3 years ago

6 0

Answer:

Increase in mass and height

Explanation:

The gravitational potential energy of an object can increase if the mass and height of object is increased.

Gravitational potential energy is the energy due to the position of a body.

It is expressed as:

Gravitational potential energy = mass x acceleration due gravity x height

Increasing mass and height will cause an increase in gravitational potential energy.

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Why are the freezing points of water and the melting point the same

Sladkaya [172]

Because melting point and freezing point describe thesame transition of matter, in this case from liquid to solid (freezing) or equivalently, from solid to liquid (melting).

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3 years ago

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Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin

likoan [24]

Answer:

13.6 N

Explanation:

Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.

Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

Mg - T = Ma (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0 and T - mgcos57 - F = m(0) = 0

Mg - T = 0 (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

Mg - mgcos57 - F = 0

So, Mg - mgcos57 = F

F = Mg - mgcos57

F = (M - mcos57)g

Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

We find F, thus

F = (2.94 kg - 2.85 kgcos57)9.8 m/s²

F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

6 0

3 years ago

Guillaume puts a bottle of soft drink in a refrigerator and leaves it there until its temperature has dropped 15.1 K.

Zarrin [17]

Answer: (a) The magnitude of its temperature change in degrees Celsius is $15.1^{o}C$ .

(b) The magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is $27.2^{o}F$ .

Explanation:

(a) Expression for change in temperature is as follows.

$|\Delta T| = |x - y|K$

= 15.1 K

= $|(x - 273.15) - (y - 273.15)|^{o}C$

= $|x - y|^{o}C$

= $15.1^{o}C$

Therefore, the magnitude of its temperature change in degrees Celsius is $15.1^{o}C$ .

(b) Change in temperature from Celsius to Fahrenheit is as follows.

F = 1.8C + 32

C = $\frac{F - 32}{1.8}$

Since, K = C + 273

or, $\Delta K = \Delta C = \frac{\Delta F}{1.8}$

$\Delta F = 1.8 \Delta K$

= 1.8 (15.1)

= $27.18^{o}F$

or, = $27.2^{o}F$

Thus, we can conclude that the magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is $27.2^{o}F$ .

7 0

3 years ago

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A rifle is aimed horizontally at a target 49 m away. the bullet hits the target 2.3 cm below the aim point. part a what was the

Nezavi [6.7K]

We use the equation of motion for vertical component,

$s_{y} = u_{y} t+\frac{1}{2} gt^2.$

Here, $s_{y}$ is displacement of bullet, $u_{y}$ is vertical initial velocity of bullet which is equal to zero because bullet was fired horizontally, and t is time of flight.

Therefore,

$s_{y} =\frac{1}{2}gt^2$

Given, $s_{y} =2.3 \ cm = 2.3 \times 10^{-2} \ m$

Substituting the values, we get time of flight

$2.3 \times 10^{-2} \ m = \frac{1}{2} \times 9.8 \ m/s^2 \times t^2 \\\\ t =\sqrt{46.94 \times 10^{-4} \ s } = 0.069 \ s$

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3 years ago

Which scenario involves kinetic energy transforming into potential energy? A. a book lying on a shelf B. a train traveling at a

NeX [460]

The answer is C.

The Kinetic energy which was exerted and experience pulling the string of a bow is kept as a potential energy at the end of the arrow in contact with the string. Once release from aim at stationary position the potential energy is again transformed.

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4 years ago