Answer:
1629 metres
Explanation:
Given that Carlos runs with velocity v = (5.6 m/s, 29o north of east) for 10 minutes.
To the north, the velocity will be:
V = 5.6 × sin 29
V = 2.715 m/s
Convert the time to second
Time = 10 × 60 = 600
Using the formula below
Velocity = distance/ time
Substitute all the parameters into the formula
2.715 = distance/600
Distance = 2.715 × 600
distance = 1629 m
The distance formula:d = v i · t + a · t² / 2Sir G.: ( v i = 0 )
d G = 0 + 0.3 · t² / 2Sir A. : d A = 0 + 0.2 t² / 288 m = 0.3 t² / 2 + 0.2 t² / 2 / · 2 ( multiple both sides by 2 )176 = 0.3 t² + 0.2 t²176 = 0.5 t²t² = 176 : 0.5t² = 352t = √352t = 18.76 sd G. = 0.3 · 18.76² / 2 = 0.3 · 352 / 2 = 52.8 mAnswer: The knights collide at 52.8 m relative to Sir George`s starting point.
Speed with which initially car is moving is 21 m/s
Reaction time = 0.50 s
distance traveled in the reaction time d = v t
d = 21 * 0.50 = 10.5 m
deceleration after this time = -10 m/s^2
now the distance traveled by the car after applying bakes
so total distance moved before it stop
d = 22.05 + 10.5 = 32.55 m
so the distance from deer is 35 - 32.55 = 2.45 m
now to find the maximum speed with we can move we will assume that we will just touch the deer when we stop
so our distance after brakes are applied is d = 35 - 10.5 = 24.5 m
again by kinematics
so maximum speed would be 22.1 m/s
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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