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9 months ago

A ball is thrown vertically upward. As it rises, what happens to its total energy?.

1 answer:

3 0

The total energy of a ball stays constant as it is thrown upward because potential energy increases while kinetic energy decreases. When the ball reaches its maximum height, the velocity is zero. Therefore, only potential energy exists rather than kinetic energy.

The thrower's movement imparts kinetic energy to a ball thrown vertically. The maximum height that can be achieved after leaving the hand will depend on the actual velocity. Air resistance causes some of this energy to be lost to the air as frictional dissipation, which warms the air in the area as well as the ball's surface.

We can just talk about how the ball moves when it is in the gravitational field of the Earth if we ignore this for the purposes of this discussion. The ball's total energy as it is released is comprised of both its gravitational potential energy and its kinetic energy, which result from the ball's velocity (due to its position).

The gravitational potential energy begins to rise as the ball moves vertically upward at precisely the same pace as it loses kinetic energy. The ball experiences a steady downward acceleration of 9.81 m/s2, which causes it to initially decline until it briefly comes to a stop at its highest point.

Due to its current position in the Earth's gravitational field relative to its initial position, all of the energy at this point is gravitational potential energy. As the ball experiences constant downward acceleration, its motion immediately becomes apparent in that direction because the acceleration easily transforms gravitational potential energy back into kinetic energy.

As a result, at every point along the trajectory, the total of these interchangeable forms of energy remains constant.

To learn more about what happens when a ball is thrown vertically upward:

brainly.com/question/1121850

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How much heat do you need to raise the temperature of 150 g of oxygen from -30c to -15c?

Naddika [18.5K]

The amount of heat needed to raise the temperature of a substance by $\Delta T$ is given by
$Q=m C_s \Delta T$
where
m is the mass of the substance
Cs is its specific heat capacity
$\Delta T$ is the increase in temperature

For oxygen, the specific heat capacity is approximately
$C_s = 0.92 J/(g K)$
The variation of temperature for the sample in our problem is
$\Delta T= -15^{\circ}C-(-30^{\circ} C)=+15^{\circ}C=15 K$
while the mass is m=150 g, so the amount of heat needed is
$Q=m C_s \Delta T=(150 g)(0.92 J/g K)(15 K)=2070 J$

4 0

3 years ago

Read 2 more answers

If an object is thrown in an upward direction from the top of a building 160 ft. High at an initial speed of 21.82 mi/h what is

viktelen [127]

To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity: $V_{f}=V_{i}+gt$
where
$V_{f}$ is the final velocity
$V_{i}$ is the initial velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time
2. Kinematic equation for distance: $d=V_{i}t+ \frac{1}{2} gt^2$
where
$d$ is the distance
$V_{i}$ is the initial velocity
$V_{f}$ is the final velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time

First, we are going to convert 21.82 mi/h to ft/s:
$21.82 \frac{mi}{h} =31.21 \frac{ft}{s}$

Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore: $V_{f}=0$ . We also know that it initial speed is 31.21 ft/s, so $V_{i}=31.21$ . Lets replace those values in our formula to find $t$ :
$V_{f}=V_{i}+gt$
$0=31.21+(-32)t$
$-32t=-31.21$
$t= \frac{-31.21}{-32}$
$t=0.98$ seconds

Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
$d=V_{i}t+ \frac{1}{2} gt^2$
$d=31.21(0.98)+ \frac{1}{2} (-32)(0.98)^2$
$d=15.22$ ft

Now we can add the height of the building and the maximum height of the object:
$d=160+15.22=175.22$ ft

Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
$d=V_{i}t+ \frac{1}{2} gt^2$
$175.22=31.21t+ \frac{1}{2} (32)t^2$
$16t^2+31.21t-175.22=0$
$t=2.47$ or $t=-4.43$
Since time cannot be negative, $t=2.47$ is the time it takes the object to fall to the ground.

Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
$V_{f}=V_{i}+gt$
$V_{f}=31.21+(32)(2.47)$
$V_{f}=110.25$ ft/s

We can conclude that the speed of the object when it hits the ground is 110.25 ft/s

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2 years ago

Whats the net force

zavuch27 [327]

Answer:

13n pushing left

Explanation:

1563 - 1550

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3 years ago

The football player running towards the goal line has

blagie [28]

the football player has speed

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3 years ago

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Since you've determined that the power supply is a 700W dual rail, what does that make the maximum output power?

inysia [295]

700 makes the maximum output power.

<u>Explanation:</u>

In physics, power is the rate of doing work or of transferring heat, i.e. the amount of energy transferred or converted per unit time. The output power of a motor is the product of the torque that the motor generates and the angular velocity of its output shaft.

A joule is equal to one Newton-meter, which is the amount of work needed to move a 1 Newton force a distance of 1 meter. When you divide work by time, you get power, measured in units of joules per second. This is also called a Watt. 1 Watt = 1 Joule Sec. This is the formula to calculate output power.

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2 years ago