Answer:
226.2 m/sec
Explanation:
We have given ![\Delta x=510nm=510\times 106{-9}m](https://tex.z-dn.net/?f=%5CDelta%20x%3D510nm%3D510%5Ctimes%20106%7B-9%7Dm)
The plank's constant ![h=6.67\times 10^{-34}](https://tex.z-dn.net/?f=h%3D6.67%5Ctimes%2010%5E%7B-34%7D)
Mass of electron ![m=9.11\times 10^{-31}kg](https://tex.z-dn.net/?f=m%3D9.11%5Ctimes%2010%5E%7B-31%7Dkg)
Now according to Heisenberg uncertainty principle ![\Delta v\geq \frac{h}{2\pi m\Delta x}](https://tex.z-dn.net/?f=%5CDelta%20v%5Cgeq%20%5Cfrac%7Bh%7D%7B2%5Cpi%20m%5CDelta%20x%7D)
So
Answer:
Electric dipole
Explanation:
the axis of a dipole makes an angle with the electric field. depending on the direction (clockwise/anticlockwise) we can get the torque (positive/negative).
hope this helps :D
Answer:
Work is defined as the process of energy transfer to the motion of an object through the application of force. This is usually represented as the product of force and displacement. The SI unit of work is Joule.
Explanation:
Power is defined as the amount of energy transferred in unit time. The SI unit of power is the watt. One watt is equal to one joule per second. Power is a scalar quantity.
Answer:
Explanation:
A ) When gymnast is motionless , he is in equilibrium
T = mg
= 63 x 9.81
= 618.03 N
B )
When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.
T = mg
= 618.03 N
C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2
Net force on it = T - mg , acting in upward direction
T - mg = m a
T = mg + m a
= m ( g + a )
= 63 ( 9.81 + .6)
= 655.83 N
D ) If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2
Net force acting in downward direction
mg - T = ma
T = m ( g - a )
= 63 x ( 9.81 - .6 )
= 580.23 N
Answer:
At least
(3 significant figures.)
Assumption: there's no air resistance.
Explanation:
Let
be the minimum speed at which that salmon leaves the water with an elevation angle of
.
.
- Initial horizontal velocity:
; - Initial vertical velocity:
.
Why is the initial horizontal velocity important? If there's no air resistance, the horizontal velocity of the fish will stay the same during its flight.
By the time the fish reaches the upper rim of the waterfall, it would have traveled
horizontally. How much time after take-off will that occur?
.
For the fish to continue upstream, its height should be at least
by the time it reaches the upper rim of the waterfall. In other words, at
,
.
If there's no air resistance, on the vertical direction the fish would accelerates downwards at a constant
during the entire flight. Its height could be expressed as
.
That's the same as:
.
Let
, Solve this equation for ![v](https://tex.z-dn.net/?f=v)
.
![\displaystyle \frac{1}{2}(-9.81) \cdot \left(\frac{1.87}{v\; \cos{32.7^{\circ}}}\right)^{2} + 1.87= 0.37](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B1%7D%7B2%7D%28-9.81%29%20%5Ccdot%20%5Cleft%28%5Cfrac%7B1.87%7D%7Bv%5C%3B%20%5Ccos%7B32.7%5E%7B%5Ccirc%7D%7D%7D%5Cright%29%5E%7B2%7D%20%2B%201.87%3D%200.37)
Given that
,
.
In other words, the fish must leave the water at a speed of at least
to continue upstream.