Answer:
The value is 24meters
Explanation:
Using
r= xi+yj
To get the magnitude of vector x
We say
/r/= √x²+y²
So
30²= √18² + y²
y= √576
Y= 24m
The answer is A. Force is equal to mass x acceleration. The acceleration due to gravity is a constant so multiplying the mass by 2 would double the force.
Answer:
a) C = 4,012 10⁻¹⁴ F, b) Q = 1.6 10⁻¹¹ C
, c) U = 3.21 10⁻¹¹ J
Explanation:
a) The capacitance of a capacitor is
C = k e₀ A / d
Let's calculate
C = 4 8.85 10⁻¹² 17 10⁻⁴ / 0.150 10⁻²
C = 4,012 10⁻¹⁴ F
b) let's look the charge
C = Q / ΔV
Q = C ΔV
Q = 4,012 10⁻¹⁴ 400
Q = 1.6 10⁻¹¹ C
c) The stored energy
U = ½ C ΔV²
U = ½ 4,012 10⁻¹⁴ 400²
U = 3.21 10⁻¹¹ J
Time taken by the package to reach the sea level= 13.7 s
height=h=925 m
initial velocity along vertical= vi=0
acceleration due to gravity=g=9.8 m/s^2
using the kinematic equation h= Vi*t + 1/2 gt^2
925=0(t)+1/2 (9.8)t^2
4.9 t^2=925
t= 13.7 s
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
