Answer:
elastic partial width is 2.49 eV
Explanation:
given data
ER E = 250 eV
spin J = 0
cross-section magnitude σ = 1300 barns
peak P = 20ev
to find out
elastic partial width W
solution
we know here that
σ = λ²× W / ( E × π × P ) ...................1
put here all value
σ = (0.286)² × W × 
/ ( 250 × π × 20 )
1300 × 
= (0.286)² × W × / ( 250 × π × 20 )
solve it and we get W
W = 249.56 × 
so elastic partial width is 2.49 eV
Answer:
a) puck is subjected to both the forces of the hockey sticks in a horizontal direction,
b)the puck does not move since the sum of the forces is zero
c) changing the magnitude or direction of its applied force
Explanation:
a) The puck is subjected to both the forces of the hockey sticks in a horizontal direction, these forces are of equal magnitude and opposite direction since the puck is at rest.
In the direction of the y-axis (perpendicular to the ice) you have the weight of the disk and the normal to this weight that are also in equilibrium.
b) the puck does not move since the sum of the forces is zero, which implies that the forces of the hockey sticks are of equal magnitude and opposite direction.
c) the player has several ways to make the puck move
* slightly changing the angle of the club and therefore the direction of the force, in this case the disc comes out in the direction of this component
* inclined the stick slightly so that the force has a vertical component and the puck jumps in this direction
* Increasing the magnitude of the force so that the puck comes out in the opposite direction to the player
* The worst case, decreasing its force to zero and the disk comes out in its direction by the other force that had the same magnitude.
Answer:
Explanation:
Mass of first cart M1=2.4kg
Velocity of first cart U1=4.1m/s
Mass of second cart M2=1.7kg
Second cart is initially at rest U2=0
After an instant, the velocity of the second cart is U2=-2.8m/s
Now after collision the two cart move together with the same velocity I.e inelastic collision
Using conservation of momentum
Momentum before collision, = momentum after collision
M1U1 + M2U2 = (M1+M2)V
2.4×4.1 + 1.7× -2.8 =(2.4+1.7)V
9.84 - 4.76 = 4.1V
5.08=4.1V
V=5.08/4.1
V=1.24m/s
The momentum of the two cart at that instant is
M1U1+M2U2
2.4×4.1 + 1.7× -2.8
9.84 - 4.76
5.08kgm/s
So the momentum at the instant the velocity is 4.1m/s for cart 1 and -2.8m/s for cart 2 is 5.08kgm/s
As we can see that the boy is moving his hand up and down while producing the wave in the string.
Here we can see that the source of disturbance or source of wave here in this example is the boy who is holding the string at one end.
So here the motion of the hand of boy is representing the motion of medium molecules in the medium always
So here the disturbance of wave is travelling in the string to the right direction while if we see the direction of medium molecules then it is moving perpendicular to the string i.e. up and down
This type of wave is known as transverse waves in which medium molecules moves perpendicular to the direction of wave
We will find the mass from
mass = density x volume
We are told the density and must find the volume from the dimensions given
the volume of the washer will be the area x thickness (remembering to convert all measurements to meters)
if the washer had no hole, its area would be pi (0.0225m)^2 (remember to convert to meters and to use radius)
the area of the hole is pi(0.00625m)^2
so the area of the washer is pi[(0.0225m)^2 - (0.00625m)^2] = 1.5x10^-3 m
the volume of the washer is 1.5x10^-3 m x 1.5x10^-3 m = 2.25x10^-6 m^3 (the thickness of the washer is 1.5 mm = 1.5x10^-3m)
thus, the mass of the washer = 8598kg/m^3 x 2.25x10^-6m^3 = 0.0189kg = 18.9 grams
