A hot-air balloonist, rising vertically with a constant speed of 5.00 m/s releases a sandbag at the instant the balloon is 40.0
m above the ground. After it is released, the sandbag encounters no appreciable air drag. Compute the position of the sandbag at 0.250 s after its release.
40.9 m
39.7 m
41.3 m
40.3 m
40.9 m
39.7 m
41.3 m
40.3 m
1 answer:
Answer:
<em>First option</em>
<em>The position of the sandbag at 0.250 s after its release is </em><u><em>40.9 m</em></u>
Explanation:
We want to know the height of the balloon as a function of time, after 0.250s
Then we use the following equation of motion:
In this equation
is the speed
is gravitational acceleration
is the initial height
is the initial velocity.
If the sandbag is released 40 meters above the ground, then the initial height is 40 meters.
As the balloon rises at a constant speed, then the initial velocity of the balloon is 5 m/s
Now we substitute these values in the formula
Now we look for:
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Answer:
The value to be reported is 5.48V
Explanation:
The RMS (root mean square) is defined as the value of voltage that will produce the same heating effect, or power dissipation, in circuit, as this AC voltage.
The RMS voltage is also called effective voltage because it is just as effective as DC voltage in providing power to an element.
It is expressed as =
where Vm is the maximum or peak value of the voltage
In calculating the RMS of the voltage , we simply divide the peak voltage by square root of 2 (√2)
=
=
= 5.48 V
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2 /><h2>
</h2><h2>_____________________________________</h2><h2>DATA:</h2>
mass = m = 2kg
Distance = x = 6m
Force = 30N
TO FIND:
Work = W = ?
Velocity = V = ?
<h2>SOLUTION:</h2>According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.
To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,
Base is the x-axis of the graph which is Position i.e. 6m
Height is the y-axis of the graph which is Force i.e. 30N
So,
W = 90 J
The work done is 90 J.
According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.