Answer:
4.47 m/s.
Explanation:
distance traveled, d = 10 miles
time, t = 1 hour
Speed of the runner, v = d / t
Speed of the runner = 10 miles / 1
Speed of the runner = 10 mph
1 mph ----------------------- 0.44704 m/s
10 mph -----------------------?
= 4.47 m/s
Thus, in 2 hours the distance traveled will change but the speed it still 10 mph or 4.47 m/s.
Answer:
b. The internal resistance must be much smaller than the other resistances in the circuit.
Explanation:
Ammeter is used to measure the current flowing through a circuit. It is connected in series configuration with the load. In such a scenario the resistance of the ammeter should be negligible so as to make sure that the voltage drop across the resistance of ammeter is zero and it shows the correct reading of the current in the circuit.
Answer:
60 m
Explanation:
After 3 seconds of travel at 20 m/s, the projectile is 3·20 = 60 meters horizontally from the cannon.
__
The vertical height after 3 seconds is 0.9 m, so the straight-line distance from cannon to target is √(60^2 +0.9^2) ≈ 60.007 meters.
This affirmative is false
Answer:
a) -2.516 × 10⁻⁴ V
b) -1.33 × 10⁻³ V
Explanation:
The electric field inside the sphere can be expressed as:
The potential at a distance can be represented as:
V(r) - V(0) =
V(r) - V(0) = ₀
V(r) = ₀
Given that:
q = +3.83 fc = 3.83 × 10⁻¹⁵ C
r = 0.56 cm
= 0.56 × 10⁻² m
R = 1.29 cm
= 1.29 × 10⁻² m
E₀ = 8.85 × 10⁻¹² F/m
Substituting our values; we have:
= -2.15 × 10⁻⁴ V
The difference between the radial distance and center can be expressed as:
V(r) - V(0) =
V(r) - V(0) =
V(r) =
V(r) =
V(r)
V(r) = -0.00133
V(r) = - 1.33 × 10⁻³ V