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:
ball clears the net
Explanation:
= initial speed of launch of the ball = 20 ms^{-1}
= angle of launch = 5 deg
Consider the motion of the ball along the horizontal direction
= initial velocity = 
= time of travel
= horizontal displacement of the ball to reach the net = 7 m
Since there is no acceleration along the horizontal direction, we have
Eq-1
Consider the motion of the ball along the vertical direction
= initial velocity = 
= time of travel
= Initial position of the ball at the time of launch = 2 m
= Final position of the ball at time "t"
= acceleration in down direction = - 9.8 ms⁻²
Along the vertical direction , position at any time is given as
Since Y > 1 m
hence the ball clears the net
Answer:
0.12 K
Explanation:
height, h = 51 m
let the mass of water is m.
Specific heat of water, c = 4190 J/kg K
According to the transformation of energy
Potential energy of water = thermal energy of water
m x g x h = m x c x ΔT
Where, ΔT is the rise in temperature
g x h = c x ΔT
9.8 x 51 = 4190 x ΔT
ΔT = 0.12 K
Thus, the rise in temperature is 0.12 K.
Answer:
592.92 x 10³ Pa
Explanation:
Mole of ammonia required = 10 g / 17 =0 .588 moles
We shall have to find pressure of .588 moles of ammonia at 30 degree having volume of 2.5 x 10⁻³ m³. We can calculate it as follows .
From the relation
PV = nRT
P x 2.5 x 10⁻³ = .588 x 8.32 x ( 273 + 30 )
P = 592.92 x 10³ Pa
