The rotor in a certain electric motor is a flat, rectangular coil with 84 turns of wire and dimensions 2.61 cm by 3.64 cm. The r
1 answer:
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Answer:
Vr = 20 [km/h]
Explanation:
In order to solve this problem, we have to add the relative velocities. We must remember that velocity is a vector, therefore it has magnitude and direction. We will take the sea as the reference measurement level.
Let's take the direction of the ship as positive. Therefore the boy moves in the opposite direction (Negative) to the reference level (the sea).
Answer:
a) 52.915 m
b) The vertical velocity is approximately 21.092 m/s
The resultant velocity is approximately 26.5 m/s
Explanation:
a) The height of the window in the house from which the water was thrown = 15 m
The speed of the stream of water thrown = 20 m/s
The angle at which the water was thrown = 37° over the horizontal
The acceleration due to gravity, g = 10 m/s²
a) The distance from the base of the house at which the water will fall is given as follows;
y = y₀ + u·t·sin(θ) + 1/2·g·t²
Where;
y = The vertical height reached
u = The initial velocity
t = Time of flight
From the point the steam of water is thrown, we get;
y₀ = 15 m
Therefore;
y = 15 + 20 × t × sin(37°) - 1/2 × 10 × t²
y = 15 + 20 × t × sin(37°) - 5 × t²
When y = 0, Ground level, we get
0 = 15 + 20 × t × sin(37°) - 5 × t²
5·t² - 20×sin(37°)×t -15 = 0
∴ t = (20 ×sin(37°) ± √((-20 × ·sin(37°))² - 4 × (5) × (-15)))/(2 × 5)
t ≈ 3.3128302, or t ≈ 0.906
Therefore, the time of flight of the water, t ≈ 3.3128302 seconds
The distance from the base of the house at which the water will fall = The horizontal distance travelled by the water, x
x = u·cos(θ)×t
∴ x = 20 × cos(37°) × 3.3128302 ≈ 52.915
The distance from the base of the house at which the water will fall = x ≈ 52.915 m
b) The velocity at which the water will reach the ground, 'v', is given as follows;
The vertical velocity, = u·sin(θ)·t - g·t
At the ground, t ≈ 3.3128302 seconds
∴ = 20 × sin(37) - 10 × 3.3128302 ≈ -21.092
The vertical velocity at which the water will reach the ground, ≈ 21.092 m/s (downwards)
The resultant velocity, v = √(² + vₓ²)
∴ v = √(21.092² + (0 × cos(37°))²) ≈ 26.5
The resultant velocity at which the water will reach the ground, v ≈ 26.5 m/s.
Answer:
Push with force of 1N
Explanation:
I have explained in the paper.
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