Answer:
x = 41.28 m
Explanation:
This is a projectile launching exercise, let's find the time it takes to get to the base of the cliff.
Let's start by using trigonometry to find the initial velocity
cos 25 = v₀ₓ / v₀
sin 25 = Iv_{oy} / v₀
v₀ₓ = v₀ cos 25
v_{oy} = v₀ sin 25
v₀ₓ = 22 cos 25 = 19.94 m / s
v_{oy} = 22 sin 25 = 0.0192 m / s
let's use movement on the vertical axis
y = y₀ + v_{oy} t - ½ g t²
when reaching the base of the cliff y = 0 and the initial height is y₀ = 21 m
0 = 21 + 0.0192 t - ½ 9.81 t²
4.905 t² - 0.0192 t - 21 = 0
t² - 0.003914 t - 4.2813 =0
we solve the quadratic equation
t =
t =
t₁ = 2.07 s
t₂ = -2.067 s
since time must be a positive scalar quantity, the correct result is
t = 2.07 s
now we can look up the distance traveled
x = v₀ₓ t
x = 19.94 2.07
x = 41.28 m
Torque = r x F
|F| = mg = 60 * 10 N = 600 N ( assuming g ~ 10m/s^2)
distance of fulcrum = torque / Force = 90/600 m = .15 m.
Answer:
Vd = 1.597 ×10⁻⁴ m/s
Explanation:
Given: A = 3.90×10⁻⁶ m², I = 6.00 A, ρ = 2.70 g/cm³
To find:
Drift Velocity Vd=?
Solution:
the formula is Vd = I/nqA (n is the number of charge per unit volume)
n = No. of electron in a mole ( Avogadro's No.) / Volume
Volume = Molar mass / density ( molar mass of Al =27 g)
V = 27 g / 2.70 g/cm³ = 10 cm³ = 1 × 10 ⁻⁵ m³
n= (6.02 × 10 ²³) / (1 × 10 ⁻⁵ m³)
n= 6.02 × 10 ²⁸
Now
Vd = (6A) / ( 6.02 × 10 ²⁸ × 1.6 × 10⁻¹⁹ C × 3.9×10⁻⁶ m²)
Vd = 1.597 ×10⁻⁴ m/s
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Given Information:
Magnetic field = B = 1×10⁻³ T
Frequency = f = 72.5 Hz
Diameter of cell = d = 7.60 µm = 7.60×10⁻⁶ m
Required Information:
Maximum Emf = ?
Answer:
Maximum Emf = 20.66×10⁻¹² volts
Explanation:
The maximum emf generated around the perimeter of a cell in a field is given by
Emf = BAωcos(ωt)
Where A is the area, B is the magnetic field and ω is frequency in rad/sec
For maximum emf cos(ωt) = 1
Emf = BAω
Area is given by
A = πr²
A = π(d/2)²
A = π(7.60×10⁻⁶/2)²
A = 45.36×10⁻¹² m²
We know that,
ω = 2πf
ω = 2π(72.5)
ω = 455.53 rad/sec
Finally, the emf is,
Emf = BAω
Emf = 1×10⁻³*45.36×10⁻¹²*455.53
Emf = 20.66×10⁻¹² volts
Therefore, the maximum emf generated around the perimeter of the cell is 20.66×10⁻¹² volts