The magnitude of the induced emf is given by:
ℰ = |Δφ/Δt|
ℰ = emf, Δφ = change in magnetic flux, Δt = elapsed time
The magnetic field is perpendicular to the loop, so the magnetic flux φ is given by:
φ = BA
B = magnetic field strength, A = loop area
The area of the loop A is given by:
A = πr²
r = loop radius
Make a substitution:
φ = B2πr²
Since the strength of the magnetic field is changing while the radius of the loop isn't changing, the change in magnetic flux Δφ is given by:
Δφ = ΔB2πr²
ΔB = change in magnetic field strength
Make another substitution:
ℰ = |ΔB2πr²/Δt|
Given values:
ΔB = 0.20T - 0.40T = -0.20T, r = 0.50m, Δt = 2.5s
Plug in and solve for ℰ:
ℰ = |(-0.20)(2π)(0.50)²/2.5|
ℰ = 0.13V
A is the answer for the problem
Answer:
1.92 kg of nitrogen.
Explanation:
The following data were obtained from the question:
Heat absorbed (Q) = 384000 J
Note: Heat of vaporisation (ΔHv) of nitrogen = 5600 J/mol
Next, we shall determine the number of mole of nitrogen that absorbed 384000 J.
This is illustrated below:
Q = mol·ΔHv
384000 = mole of N2 x 5600
Divide both side by 5600
Mole of N2 = 384000/5600
Mole of N2 = 68.57 moles
Next, we shall convert 68.57 moles of nitrogen, N2 to grams.
This can be obtained as follow:
Molar mass of N2 = 2 x 14 = 28 g/mol.
Mole of N2 = 68.57 moles.
Mass of N2 =..?
Mole = mass /molar mass
68.57 = mass of N2 /28
Cross multiply
Mass of N2 = 68.57 x 28
Mass of N2 = 1919.96 g
Finally, we shall convert 1919.96 g to kilograms.
This can be achieved as shown below:
1000g = 1 kg
Therefore,
1919.96 g = 1919.96/1000 = 1.92 kg.
Therefore, 1.92 kg of nitrogen were burned off.
Answer:
1.84 m
Explanation:
For the small lead ball to be balanced at the tip of the vertical circle just before it is released, the reaction force , N equal the weight of the lead ball W + the centripetal force, F. This normal reaction ,N also equals the tension T in the string.
So, T = mg + mrω² = ma where m = mass of small lead ball, g = acceleration due to gravity = 9.8 m/s², r = length of rope = 1.10 m and ω = angular speed of lead ball = 3 rev/s = 3 × 2π rad/s = 6π rad/s = 18.85 rad/s and a = acceleration of normal force. So,
a = g + rω²
= 9.8 m/s² + 1.10 m × (18.85 rad/s)²
= 9.8 m/s² + 390.85 m/s²
= 400.65 m/s²
Now, using v² = u² + 2a(h₂ - h₁) where u = initial velocity of ball = rω = 1.10 m × 18.85 rad/s = 20.74 m/s, v = final velocity of ball at maximum height = 0 m/s (since the ball is stationary at maximum height), a = acceleration of small lead ball = -400.65 m/s² (negative since it is in the downward direction of the tension), h₁ = initial position of lead ball above the ground = 1.3 m and h₂ = final position of lead ball above the ground = unknown.
v² = u² + 2a(h₂ - h₁)
So, v² - u² = 2a(h₂ - h₁)
h₂ - h₁ = (v² - u²)/2a
h₂ = h₁ + (v² - u²)/2a
substituting the values of the variables into the equation, we have
h₂ = 1.3 m + ((0 m/s)² - (20.74 m/s)²)/2(-400.65 m/s²)
h₂ = 1.3 m + [-430.15 (m/s)²]/-801.3 m/s²
h₂ = 1.3 m + 0.54 m
h₂ = 1.84 m
