A toddler pushes his 7.0 kg toy box at a relatively constant velocity across the tiled floor of the family room applying a horiz
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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.
As we know that when charge is released in electric field
It will have two forces on it
1. electrostatic force
2. gravitational force
now if the ball will accelerate upwards so we can say
net upward force = mass * acceleration
now we can find charge q on it by above equation
So above is the charge on the particle