Answer:
a) D_ total = 18.54 m, b) v = 6.55 m / s
Explanation:
In this exercise we must find the displacement of the player.
a) Let's start with the initial displacement, d = 8 m at a 45º angle, use trigonometry to find the components
sin 45 = y₁ / d
cos 45 = x₁ / d
y₁ = d sin 45
x₁ = d sin 45
y₁ = 8 sin 45 = 5,657 m
x₁ = 8 cos 45 = 5,657 m
The second offset is d₂ = 12m at 90 of the 50 yard
y₂ = 12 m
x₂ = 0
total displacement
y_total = y₁ + y₂
y_total = 5,657 + 12
y_total = 17,657 m
x_total = x₁ + x₂
x_total = 5,657 + 0
x_total = 5,657 m
D_total = 17.657 i^+ 5.657 j^ m
D_total = Ra (17.657 2 + 5.657 2)
D_ total = 18.54 m
b) the average speed is requested, which is the offset carried out in the time used
v = Δx /Δt
the distance traveled using the pythagorean theorem is
r = √ (d1² + d2²)
r = √ (8² + 12²)
r = 14.42 m
The time used for this shredding is
t = t1 + t2
t = 1 + 1.2
t = 2.2 s
let's calculate the average speed
v = 14.42 / 2.2
v = 6.55 m / s
I dont know but i know i dont lnow if this is true but the gravity is slowly going away every 4 year i dont know i think.
Answer:
1) No, the car does not travel at constant speed.
2) V = 9 ft/s
3) No, the car does not travel at constant speed.
4) V = 5.9 ft/s
Explanation:
In order to know if the car is traveling at constant speed we need to derive the given formula. That way we get speed as a function of time:
V(t) = 2*t + 2 Since the speed depends on time, the speed is not constant at any time.
For the average speed we evaluate the formula for t=2 and t=5:
d(2) = 8 ft and d(5) = 35 ft
Again, for the average speed we evaluate the formula for t=1.8 and t=2.1:
d(1.8) = 6.84 ft and d(2.1) = 8.61 ft
Answer: Option B
Explanation : When a negatively charged object A gets in contact with the neutral object B, the negative charge of object will induce the opposite charges on object B. Hence, there will be a positive charge on object B
When the capacitor is connected to the voltage, a charge Q is stored on its plates. Calling 
the capacitance of the capacitor in air, the charge Q, the capacitance and the voltage (
) are related by
(1)
when the source is disconnected the charge Q remains on the capacitor.
When the space between the plates is filled with mica, the capacitance of the capacitor increases by a factor 5.4 (the permittivity of the mica compared to that of the air):
this is the new capacitance. Since the charge Q on the plates remains the same, by using eq. (1) we can find the new voltage across the capacitor:
And since 
, substituting into the previous equation, we find:
