Answer:
Torque,
Explanation:
Given that,
The loop is positioned at an angle of 30 degrees.
Current in the loop, I = 0.5 A
The magnitude of the magnetic field is 0.300 T, B = 0.3 T
We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :
Let us assume that, 
is the angle between normal and the magnetic field, 
Torque is given by :
So, the net torque about the vertical axis is 
. Hence, this is the required solution.
Answer:
v₀ₓ = 63.5 m/s
v₀y = 54.2 m/s
Explanation:
First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:
K.E = (0.5)(mv₀²)
where,
K.E = initial kinetic energy of projectile = 1430 J
m = mass of projectile = 0.41 kg
v₀ = launch velocity of projectile = ?
Therefore,
1430 J = (0.5)(0.41)v₀²
v₀ = √(6975.6 m²/s²)
v₀ = 83.5 m/s
Now, we find the launching angle, by using formula for maximum height of projectile:
h = v₀² Sin²θ/2g
where,
h = height of projectile = 150 m
g = 9.8 m/s²
θ = launch angle
Therefore,
150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)
Sin θ = √(0.4216)
θ = Sin⁻¹ (0.6493)
θ = 40.5°
Now, we find the components of launch velocity:
x- component = v₀ₓ = v₀Cosθ = (83.5 m/s) Cos(40.5°)
<u>v₀ₓ = 63.5 m/s</u>
y- component = v₀y = v₀Sinθ = (83.5 m/s) Sin(40.5°)
<u>v₀y = 54.2 m/s</u>
The answer is X. I hope this helped
Answer:
Explanation:
mass of liquid m₁ = 161 g
temperature t₁ = 31.8
final temperature t₂ = 28.8
Let m g of ice melted to cool the liquid
heat gained = mass x latent heat of fusion + mass x loss of temp x s heat of water
= m x 80 + m x 1 x ( 31.8 - 28.8 ) ( latent heat of ice = 80 cals/g )
= 83 m
heat lost = 161 x 1 x ( 31.8 - 28.8 ) ( specific heat of water = 1 cal / g / k )
= 161 x 3
heat lost = heat gained
83 m = 161 x 3
m = 5.82 g
mass of remaining ice = 131 - 5.82
= 125.18 g
