Answer:
Final speed of the train is 7.5 m/s
Explanation:
It is given that,
Uniform acceleration of the train is, a = 1.5 m/s²
It starts from rest and travels for 5.0 s. We have to find the final velocity of the train. By using first equation of motion as :

Here, train starts from rest so, u = 0
v = 7.5 m/s
So, the final velocity of the train is 7.5 m/s. Hence, this is the required solution.
Question: What is the frequency of a wave that has a wave speed of 120 m/s and a wavelength of 0.40 m?
Answer: The equation that relates frequency of a wave to a waves speed and wavelength is Speed of Wave= Frequency X Wavelength. Since you are given speed and wavelength, you plug those two known numbers into the equation, 120= Frequency X 0.40. You then divide 120 by .4 to get your frequency of 300.
Explanation: this might help for
Answer:
Explanation:
Kinetic energy= K.E = (1/2) × mv²
K.E = 0.5× 0.8kg× 100m²/s² = 40 N
Answer:
(a) 2.85 m
(b) 16.5 m
(c) 21.7 m
(d) 22.7 m
Explanation:
Given:
v₀ₓ = 19 cos 71° m/s
v₀ᵧ = 19 sin 71° m/s
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
(a) Find Δy when t = 3.5 s.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²
Δy = 2.85 m
(b) Find Δy when vᵧ = 0 m/s.
vᵧ² = v₀ᵧ² + 2 aᵧ Δy
(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy
Δy = 16.5 m
(c) Find Δx when t = 3.5 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²
Δx = 21.7 m
(d) Find Δx when Δy = 0 m.
First, find t when Δy = 0 m.
Δy = v₀ᵧ t + ½ aᵧ t²
(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²
0 = t (18.0 − 4.9 t)
t = 3.67
Next, find Δx when t = 3.67 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²
Δx = 22.7 m
In this problem,
Applied force(F) = 10 N
The object’s mass (m) is 5 kg.
Having said that,
An object’s force is equal to the product of its mass and the acceleration it experiences as a result of the applied force.
i.e., Mass + Acceleration = Force (a)
F= m×a
Therefore,
A= F÷m
A= (10÷5) m/sec²
A= 2 m/sec²
Consequently, the object’s acceleration,
A=2 m/sec²
Concept of force and acceleration:
This states that the rate of velocity change of an object is directly proportional to the applied force and moves in the direction of the applied force.
It can be expressed mathematically as force (N) = mass (kg) x acceleration (m/s2). Therefore, an object with constant mass will accelerate in direct proportion to the applied force.
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