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3 years ago

If a ball is tossed straight up into the air, at what position is its potential energy the greatest?

2 answers:

4 0

A. when it reaches the top of its flight

potential energy increases as the height of the object does, and the highest point of a ball's flight is when it reaches the top.

abruzzese [7]3 years ago

3 0

Answer:

A.when it reaches the top of its flight

Explanation:

The potential energy of an object is given by:

$U=mgh$

where

m is the mass of the object

g is the acceleration of gravity

h is the heigth at which the object is located above the ground

Therefore, we see that the higher the object, the larger its potential energy: this means that the ball will have the greatest potential energy at the top of its trajectory.

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When is the average velocity of an object equal to the instantaneous velocity?

atroni [7]

Answer:

Average velocity of an object is equal to the instantaneous velocity when it's acceleration is zero.

Explanation:

4 0

4 years ago

How much heat energy (in megajoules) is needed to convert 7 kg of ice at -9°C C to water at 0°C?

Molodets [167]

Answer:

hence option A is correct

Explanation:

heat required from -9°C to 0°C ice = mass × specific heat of ice ×change in temperature

heat required from -9°C to 0°C ice = 7×2100×9 =132300 J =0.1323 MJ

( HERE SPECIFIC HEAT OF ICE IS A CONSTANT VALUE OF 2100

J/(kg °C )

heat required from 0°C ice to 0°C water = mass× specific heat of fusion of ice

= 7×3.36×10^5

= 2.352 × 10^6 J

= 2.352 MJ

TOTAL HEAT ENERGY REQUIRED = 0.1323 MJ +2.352 MJ

= 2.4843 MJ

hence option A is correct

5 0

3 years ago

Help meh in this question plzzz <br>

iragen [17]

The Moment of Inertia of the Disc is represented by $I = \frac{15}{32}\cdot M\cdot R^{2}$ . (Correct answer: A)

Let suppose that the Disk is a Rigid Body whose mass is uniformly distributed. The Moment of Inertia of the element is equal to the Moment of Inertia of the entire Disk minus the Moment of Inertia of the Hole, that is to say:

$I = I_{D} - I_{H}$ (1)

Where:

$I_{D}$ - Moment of inertia of the Disk.

$I_{H}$ - Moment of inertia of the Hole.

Then, this formula is expanded as follows:

$I = \frac{1}{2}\cdot M\cdot R^{2} - \frac{1}{2}\cdot m\cdot \left(\frac{1}{2}\cdot R^{2} \right)$ (1b)

Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole ( $m$ ):