A car is traveling at 10 m/s. 10 seconds later the car is traveling 40 m/s. What is the car’s acceleration?
1 answer:
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Answer:
2a.) Wavelength = 1.8 m
2b.) F = 66.67 Hz
3a.) Find the attached file
3b.) Wavelength = 0.6 m
Explanation:
Given that the
Length L = 0.9m
Wavelength (λ) = 2L/n
Where n = number of harmonic
If n = 1, then
Wavelength (λ) = 2L = 2 × 0.9 = 1.8 m
b.)
If waves travel at a speed of 120m/s on this string, what is the frequency
associated with the longest wave (first harmonic)?
Given that V = 120 m/s
V = Fλ
But λ = 2L, therefore,
F = V/2L
F = 120/1.8
F = 66.67 Hz
3. b.) If there are two node, the position will be in 3rd position which is 3rd harmonic
Using the same formula,
Wavelength (λ) = 2L/n
Where n = 3
Wavelength (λ) = 2 × 0.9/3
Wavelength (λ) = 0.6 m
1) Time in the air: 0.78 s
The motion of the ball is a projectile motion, which consists of two independent motions:
- A horizontal motion with constant horizontal velocity
- A vertical motion with constant downward acceleration of
(acceleration of gravity)
The initial vertical velocity is
where the negative sign means the direction is downward.
The vertical position of the ball is given by
where
h = 4.50 m is the initial heigth of the ball when it starts falling down
The ball reaches the ground when y = 0, so we have:
This is a second-order equation; solving for t, we get
t = -1.18 s
t = 0.78 s
We discard the negative solution since it has no physical meaning, so we can say that the ball spent 0.78 s in the air.
2) Horizontal distance: 1.83 m
For this second part of the problem, we just have to consider the horizontal motion of the ball.
As we said previously, the motion of the ball along the horizontal direction is a uniform motion with constant velocity, which is given by
where u = 3.05 m/s is the initial speed and the angle of projection.
For a uniform motion, we can use the following relationship between distance covered and velocity:
and substituting t = 0.78 s, we find the total distance travelled along the horizontal direction by the ball before reaching the ground: