Answer:
in what context?
Explanation:
thermal resistivity is the property of an object to resist transmission of heat.
electrical resistivity is the property of an object to resist transmission of electricity.
There are two forces at play:
- The gravitational force acting downward due to the mass of the bucket and the water that it contains.
- The upward force that your hand exerts on the bucket.
If the magnitude of the force your hand exerts on the bucket equals the magnitude of the gravitational force, the bucket is in static equilibrium. That means the bucket is not moving and the forces acting on it balance each other out, making the net force 0.
Having 0 net force means the bucket doesn't undergo any acceleration, or change in motion.
The potential difference across the parallel plate capacitor is 2.26 millivolts
<h3>Capacitance of a parallel plate capacitor</h3>
The capacitance of the parallel plate capacitor is given by C = ε₀A/d where
- ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
- A = area of plates and
- d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.
<h3>Charge on plates</h3>
Also, the surface charge on the capacitor Q = σA where
- σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
- a = area of plates.
<h3>
The potential difference across the parallel plate capacitor</h3>
The potential difference across the parallel plate capacitor is V = Q/C
= σA ÷ ε₀A/d
= σd/ε₀
Substituting the values of the variables into the equation, we have
V = σd/ε₀
V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m
V = 20.0 C/m × 10⁻³/8.854 F/m
V = 2.26 × 10⁻³ Volts
V = 2.26 millivolts
So, the potential difference across the parallel plate capacitor is 2.26 millivolts
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Answer:
Time, 
Explanation:
Given that,
When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel
before interacting.
Let t is the time interval required for the strong interaction to occur. It will move with the speed of light. So,

So, the time interval is 
Work is calculated by multiplying force by the distance that the object had moved. The applied force is 60 N, moving the object by 10 m. Thus, the work does is 600 J. For the friction force which is equal to,
100N x 0.250 = 25.0 N
the work done is,
W = (60 N - 25 N) x 10 m = 350 J
The kinetic energy of the box can be equated to this force. Thus, the answer is also 350 J.