Answer:
The latent heat of fusion of water is 334.88 Joules per gram of water.
Explanation:
Let the latent heat of ice be 'x' J/g
1) Thus heat absorbed by 100 gram of ice to get converted into water equals
2) heat energy required to raise the temperature of water from 0 to 25 degree Celsius equals
Thus total energy needed equals
3) Heat energy released by the decrease in the temperature of water from 25 to 11 degree Celsius is
Now by conservation of energy we have
Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45 )
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = - 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct
Answer:
The correct answer is 231 Mpa i.e option a.
Explanation:
using the equation of torsion we Have
where,
= shear stress at a distance 'r' from the center
T = is the applied torque
= polar moment of inertia of the section
r = radial distance from the center
Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.
We know that in case of a circular section the maximum shearing stresses due to a shear force occurs at the center and equals
Applying values we get