Answer:
V = 0.896 m/s
Explanation:
This is a typical problem of momentum conservation, whic states the following:
m₁V₁ + m₂V₂ = m₁V₃ + m₂V₄ (1)
In this case V₃ and V₄ would be the final velocity of the trucks after the collision.
With the given data let's see what we have:
m₁ = 5.5x10⁵ kg
m₂ = 2.3x10⁵ kg
V₁ = 5 m/s
V₂ = -5 m/s because it's going to the left (-x axis)
V₄ = 9.1 m/s to the right (Meaning is positive)
V₃ = ??
So to calculate V₃ we just need to replace the data into (1) and solve for V₃:
(5.5x10⁵ * 5) - (2.3x10⁵ * 5) = 5.5x10⁵V₃ + (2.3x10⁵ * 9.1)
2.75x10⁶ + 1.15x10⁶ = 5.5x10⁵V₃ + 2.093x10⁶
V₃ = 2.75x10⁶ - 1.15x10⁶ - 2.093x10⁶ / 5.5x10⁵
V₃ = -0.493x10⁶ / 5.5x10⁵
V₃ = -0.896 m/s
With this sign, it means that is going in the same sense of the other truck, but it's going to the left so this would be positive:
<h2>
V₃ = 0.896 m/s</h2>
Hope this helps
Answer:
= 38.89 m/s
Explanation:
speed of first car (u) = 80 km/h = 22.22 m/s
time (t) = 30 s
distance (d) = 500 m
find the speed of second car (v)
let the speed of the second car be represented as 'v'
We can get the speed of the second car from the formula net speed = distance between cars / time interval
where
- distance between cars = 500 m
- time interval = 30 s
- net speed = v-22.22 (since both cars move in the same direction)
- substituting the above into the formula we have
v-22.22 = 500 / 30
v = 500/30 + 22.22
v = 38.89 m/s
Answer:
-39.2m/s
Explanation:
Using the equation of motion;
v = u + at
Since the ball is thrown upward, the acceleration due to gravity acting on it will be negative, hence a = -g
v = u - gt
Since g = 9.8m/s²
t = 4.0s
u = 0m/s
v = 0 + (-9.8)(4)
v = 0 + (-9.8)(4)
v = -39.2m/s
Hence the speed of the ball before release is -39.2m/s
Answer:
Explanation:
a) capacitive reactance Xc = 1/2πfC
f is the frequency of the AC source
C is the capacitance of thw capacitor
Given f = 60.0Hz
C = 12.2-µF
C = 12.2 × 10^-6Farads
Xc = 1/2π×60×12.2×10^-6
Xc = 1/0.004599
Xc = 217.44ohms
b) Using the formula Vrms = IrmsXc Vrms
Irms = Vrms/Xc
Irms = 27.0/217.44
Irms = 0.124Amperes
c) Since Irms = Im/√2
Where Im is the maximum current
Im = Irms×√2
Im = 0.124× √2
Im = 0.088A
d) No, the capacitor has no charge when the current is at maximum.
Answer:
Explanation:
Lets see the given parameters.
Length of wire (L) = 6.93
Linear Density of wire (m/L) = 0.019 kg/m
Tension Force (F) = 365 Newtons
Now, the fundamental frequency () of the wire is given by the formula:
Where
L is length of wire
F is the tension force
m/L is the linear density
Plugging in, we get:
We can say the fundamental frequency to be about 10 Hz
Now, we know that the lowest frequency humans can hear is 20 Hz.
The lowest harmonic number is gotten by dividing this by the fundamental frequency:
20/10 = 2
Thus,
n = 2