Answer:
The time is 0.5 sec.
Explanation:
Given that,
Voltage V= 12.00 V
Inductance L= 1.20 H
Current = 3.00 A
Increases rate = 8.00 A
We need to calculate change in current
![\Delta A = 8.00-3.00= 5.00\ A](https://tex.z-dn.net/?f=%5CDelta%20A%20%3D%208.00-3.00%3D%205.00%5C%20A)
We need to calculate the time interval
Using formula of inductor
![V=L\dfrac{\Delta A}{\Delta t}](https://tex.z-dn.net/?f=V%3DL%5Cdfrac%7B%5CDelta%20A%7D%7B%5CDelta%20t%7D)
![\Delta t =\dfrac{L\Delta A}{V}](https://tex.z-dn.net/?f=%5CDelta%20t%20%3D%5Cdfrac%7BL%5CDelta%20A%7D%7BV%7D)
Where,
= change in current
V = voltage
L = inductance
Put the value into the formula
![\Delta t=\dfrac{1.20\times5.00}{12.00}](https://tex.z-dn.net/?f=%5CDelta%20t%3D%5Cdfrac%7B1.20%5Ctimes5.00%7D%7B12.00%7D)
![\Delta t=0.5\ sec](https://tex.z-dn.net/?f=%5CDelta%20t%3D0.5%5C%20sec)
Hence, The time is 0.5 sec.
The gravitational force between two objects is given by:
![F=G \frac{m_1 m_2}{r^2}](https://tex.z-dn.net/?f=F%3DG%20%5Cfrac%7Bm_1%20m_2%7D%7Br%5E2%7D%20)
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is the separation between the two objects
The distance of the telescope from the Earth's center is
![r=6940 km=6.94 \cdot 10^6 m](https://tex.z-dn.net/?f=r%3D6940%20km%3D6.94%20%5Ccdot%2010%5E6%20m)
, the gravitational force is
![F=9.21 \cdot 10^4 N](https://tex.z-dn.net/?f=F%3D9.21%20%5Ccdot%2010%5E4%20N)
and the mass of the Earth is
![m_1=5.98 \cdot 10^{24} kg](https://tex.z-dn.net/?f=m_1%3D5.98%20%5Ccdot%2010%5E%7B24%7D%20kg)
, therefore we can rearrange the previous equation to find m2, the mass of the telescope:
Supposing the runner is condensed to a point and moves upward at 2.2 m/s.
It takes a time = 2.2/g = 2.2/9.8 = 0.22 seconds to increase to max height.
Now looking at this condition in opposite - that is the runner is at max height and drops back to earth in 0.22 s (symmetry of this kind of motion).
From what height does any object take 0.22 s to fall to earth (supposing there is no air friction)?
d = 1/2gt²= (0.5)(9.8)(0.22)²= 0.24 m
Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.
At time t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is 7/3 . v1 = 7/3 m/s .
At time t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
v2 = zero .
At time t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is -3/5 . v3 = -0.6 m/s .
A wave is a result of the disturbance in the equilibrium state. There are two types of wave, transverse and longitudinal. Transverse wave affects amplitude while longitudinal wave affects the frequency of the wave. As for the transverse wave, the magnitude of the perpendicular disturbance of the wave is directly proportional to the amplitude of the wave. The higher the transverse disturbance the higher the amplitude.