we are given in the problem the following dimensions or specifications
B = 0.000055 T r = 0.25 m constant mu0 = 4*pi*10-7
The formula that is applicable from physics is
B = mu0*I/(2*pi*r) I = 2*B*pi*r/mu0 I = 68.75 Amperes
Answer:
a. β = 8.23 K
b. β = 28.815 K
Explanation:
Heat pump can be find using the equation
β = TH / TH - TC
a.
TH = 15 ° C + 273.15 K = 288.15 K
TC = - 20 ° C + 273.15 K = 253.15 K
β = 288.15 K / 288.15 k - ( 253.15 K )
β = 8.23 K
b.
TH = 15 ° C + 273.15 K = 288.15 K
TC = 5 ° C + 273.15 K = 278.15 K
β = 288.15 K / 288.15 k - ( 278.15 K )
β = 28.815 K
1. a. longitudinal waves.
There are two types of waves:
- Transverse waves: in transverse waves, the oscillations of the wave occur in a direction perpendicular to the direction of propagation of the wave
- Longitudinal waves: in longitudinal waves, the oscillations of the waves occur parallel to the direction in which the waves are travelling.
So, these types of waves are called longitudinal waves.
2. d. a medium
There are two types of waves:
- Electromagnetic waves: these waves are produced by the oscillations of electric and magnetic field, and they can travel both in a medium and also in a vacuum (they do not need a medium to propagate)
- Mechanical waves: these waves are produced by the oscillations of the particles in a medium, so they need a medium to propagate - therefore, the correct choice is d. a medium
3. a. AM/FM radio
Analogue signals consist of continuous signals, which vary in a continuous range of values. On the contrary, digital signals consist of discrete signals, which can assume only some discrete values. For AM and FM radios, signals are transmitted by using analogue signals.
Answer: 49.92 m
Explanation:
In this situation the following equation will be useful:
Where:
is the final velocity of the car, when it finally stops
is the initial velocity of the car
is the constant acceleration of the car after the driver slams on the brakes
is the stopping distance
Isolating :
Answer:
σ₁ = 
C/m²
σ₂ = 
C/m²
Explanation:
The given data :-
i) The radius of smaller sphere ( r ) = 5 cm.
ii) The radius of larger sphere ( R ) = 12 cm.
iii) The electric field at of larger sphere ( E₁ ) = 358 kV/m. = 358 * 1000 v/m
Q₁ = 572.8 
C
Since the field inside a conductor is zero, therefore electric potential ( V ) is constant.
V = constant
∴
= 
C
Surface charge density ( σ₁ ) for large sphere.
Area ( A₁ ) = 4 * π * R² = 4 * 3.14 * 0.12 = 0.180864 m².
σ₁ = 
= 
= C/m².
Surface charge density ( σ₂ ) for smaller sphere.
Area ( A₂ ) = 4 * π * r² = 4 * 3.14 * 0.05² =0.0314 m².
σ₂ =
= 
= C/m²
