Answer:
rate of fission =5.89*10^3 1\Year
Explanation:
we know that
rate of fission is given as
mass = 1.0 g
avogardo number = 6.02*10^23
molar mass of isotopes 235U =235
Putting all value to get rate of emission
rate of fission
rate of fission =5.89*10^3 1\Year
(a) north
We can treat the aircraft as a single point charge moving in a magnetic field. In this case, the magnetic force exerted on the plane is
where
is the charge on the plane
v = 660 m/s is the velocity
is the magnitude of the magnetic field
is the angle between the direction of motion of the jet and of the magnetic field
Substituting,
The direction can be found by using Fleming's left hand rule. We have:
- index finger: magnetic field direction (straight up)
- middle finger: velocity of the plane (due west)
- force: thumb --> north
(b) Not negligible
As we can see from part (a), the magnitude of the force is not really big, so the effects are negligible.
For instance, we can compare this force with the weight of a plane. If we take a Boeing 737, its mass is about 80,000 kg, so its weight is
As we can see, this is several orders of magnitude bigger than the magnetic force calculated at point (a), so the effects of the magnetic force are negligible.
Answer:
Explanation:
<u>Free Fall Motion</u>
A free-falling object refers to an object that is falling under the sole influence of gravity. If the object is dropped from a certain height h, it moves downwards until it reaches ground level.
The speed vf of the object when a time t has passed is given by:
Where
Similarly, the distance y the object has traveled is calculated as follows:
If we know the height h from which the object was dropped, we can solve the above equation for t:
The stadium is h=32 m high. A pair of glasses is dropped from the top and reaches the ground at a time:
The pen is dropped 2 seconds after the glasses. When the glasses hit the ground, the pen has been falling for:
Therefore, it has traveled down a distance:
Thus, the height of the pen is:
Actually Welcome to the concept of Efficiency.
Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%
The efficiency is => 22% => 22/100.
so we get as,
E = W(output) /W(input)
hence, W(output) = E x W(input)
so we get as,
W(output) = (22/100) x 2.2 x 10^7
=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7
hence, W(output) = 4.84 x 10^6 J
The useful work done on the mass is 4.84 x 10^6 J
Answer:395.6 m/s
Explanation:
Given
mass of bullet
mass of wood block
Length of string
Center of mass rises to an height of
initial velocity of bullet
let and
be the velocity of bullet and block after collision
Conserving momentum
-------------1
Now after the collision block rises to an height of 0.38 cm
Conserving Energy for block
kinetic energy of block at bottom=Gain in Potential Energy
substitute the value of in equation 1