Answer:
75 kgm/s
Explanation:
Impulse: This can be defined as the product of mass and change in velocity. The S.I unit is kgm/s.
From the question,
I = m(v-u)................... Equation 1
Where I = impulse, m = mass, v = final velocity, u = initial velocity.
Let the direction of the initial velocity be the positive direction.
Given: m = 5 kg, v = -10 m/s (bounce off), u = 5 m/s.
Substitute into equation 1
I = 5(-10-5)
I = 5(-15)
I = -75 kgm/s.
The negative sign tells that the impulse act in the same direction as the final velocity of the ball
Hence,
I = 75 kgm/s
Answer:
A.)1.52cm
B.)1.18cm
Explanation:
angular speed of 120 rev/min.
cross sectional area=0.14cm²
mass=12kg
F=120±12ω²r
=120±12(120×2π/60)^2 ×0.50
=828N or 1068N
To calculate the elongation of the wire for lowest and highest point
δ=F/A
= 1068/0.5
δ=2136MPa
'E' which is the modulus of elasticity for alluminium is 70000MPa
δ=ξl=φl/E =2136×50/70000=1.52cm
δ=F/A=828/0.5
=1656MPa
δ=ξl=φl/E
=1656×50/70000=1.18cm

Answer:
a) the values of the angle α is 45.5°
b) the required magnitude of the vertical force, F is 41 lb
Explanation:
Applying the free equilibrium equation along x-direction
from the diagram
we say
∑Fₓ = 0
Pcosα - 425cos30° = 0
525cosα - 368.06 = 0
cosα = 368.06/525
cosα = 0.701
α = cos⁻¹ (0.701)
α = 45.5°
Also Applying the force equation of motion along y-direction
∑Fₓ = ma
Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)
525sin45.5° + F + 212.5 - 600 = 27.95
374.46 + F + 212.5 - 600 = 27.95
F - 13.04 = 27.95
F = 27.95 + 13.04
F = 40.99 ≈ 41 lb
Answer:
v = 38.73 m/s
Explanation:
Given
Extension of the bow, x = 50 cm = 0.5 m
Force of the arrow, F = 150 N
Mass of the arrow, m = 50 g = 0.05 kg
speed of arrow, v = ? m/s
We start by finding the spring constant
Remember, F = kx, so
k = F/x
k = 150 / 0.5
k = 300 N/m
the potential energy if the bow when pulled back is
E = 1/2kx²
E = 1/2 * 300 * 0.5²
E = 0.5 * 300 * 0.25
E = 37.5 J
The speed of the arrow will now be found by using the law of conservation of energy
1/2kx² = 1/2mv²
kx² = mv²
v² = kx²/m, on substituting, we have
v² = (300 * 0.5²) / 0.05
v² = 75 / 0.05
v² = 1500
v = √1500
v = 38.73 m/s
Answer:
0.22m/s
Explanation:
The total momentum of the System is conserved. Total momentum of the system before the collision is equal to the total momentum of the system after collision. The total momentum is the sum of individual momentum of all the objects in that system.
momentum of an object = mass* velocity
Total Momentum before collision = 0.2*0.3 + 0.1*0.1= 0.07 kg⋅m/s;
Total momentum after collision = 0.1*0.26 + 0.2*x = 0.07;
Solve for x.