Answer:
The magnitude of the magnetic force on the wire carrying greater current is is 4.85 x 10⁻⁴ N
Explanation:
Given;
distance of separation of the two wires, d = 18 cm
current in the first wire, I₁ = 8 A
current in the second wire, I₂ = 26 A
length of the wire, L = 2.1 m
permittivity of free space, μ₀ = 4π x 10⁻⁷ N·m/A = 1.257 × 10⁻⁶ N·m/A
The magnitude of the magnetic force on the wire carrying greater current is given as;
Therefore, the magnitude of the magnetic force on the wire carrying greater current is is 4.85 x 10⁻⁴ N
The kinetic energy of an object is given by:
KE = 0.5mv²
KE = kinetic energy, m = mass, v = speed
Given values:
KE = 0.161J, v = 2.33m/s
Plug in and solve for m:
0.161 = 0.5m(2.33)²
m = 0.059kg
Answer:
D. the linear velocity of the point of contact (relative to the inclined surface) is zero
Explanation:
The force of friction emerges only when there is relative velocity between two objects . In case of perfect rolling , there is no sliding so relative velocity between the surface and the point of contact is zero . In other words the velocity of point of contact becomes zero , even though , the whole body is in linear motion . It happens due point of contact having two velocities which are equal and opposite . One of the velocity is in forward direction and the other velocity which is due to rotation is in backward direction . So net velocity of point of contact becomes zero . Due to absence of sliding , displacement due to friction becomes zero . Hence work done by friction becomes zero.
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Answer:
Mass, m = 27g
Explanation:
<u>Given the following data;</u>
Initial Temperature, T1 = 25°C
Final temperature, T2 = 50°C
Quantity of heat = 2825J
Specific heat capacity of water = 4.184
Heat capacity is given by the formula;
Where,
- Q represents the heat capacity.
- m represents the mass of an object.
- c represents the specific heat capacity of water.
- dt represents the change in temperature.
Making mass, m the subject of formula, we have;
Change in temperature, dt = T2 - T1
Change in temperature, dt = 50-25 = 25°C
Substituting into the equation, we have;
Mass, m = 27g
Therefore, the mass of water that can be added is 27 grams.