A boy shoves his stuffed toy zebra down a frictionless chute. It starts at a height of 1.69 m above the bottom of the chute with
an initial speed of 1.23 m/s . The toy animal emerges horizontally from the bottom of the chute and continues sliding along a horizontal surface with a coefficient of kinetic friction of 0.255 . How far from the bottom of the chute does the toy zebra come to rest
2 answers:
Answer:
x = 6.94 m
Explanation:
For this exercise we can find the speed at the bottom of the ramp using energy conservation
Starting point. Higher
Em₀ = K + U = ½ m v₀² + m g h
Final point. Lower
= K = ½ m v²
Em₀ = Em_{f}
½ m v₀² + m g h = ½ m v²
v² = v₀² + 2 g h
Let's calculate
v = √(1.23² + 2 9.8 1.69)
v = 5.89 m / s
In the horizontal part we can use the relationship between work and the variation of kinetic energy
W = ΔK
-fr x = 0- ½ m v²
Newton's second law
N- W = 0
The equation for the friction is
fr = μ N
fr = μ m g
We replace
μ m g x = ½ m v²
x = v² / 2μ g
Let's calculate
x = 5.89² / (2 0.255 9.8)
x = 6.94 m
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We need to find the magnitude of the resultant on both x- and y-axis.
x-axis) The resultant on the x-axis is
in the positive direction.
y-axis) The resultant on the y-axis is
in the positive direction.
Both Fx and Fy are positive, so the resultant is in the first quadrant. We can find the angle and so the direction using
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Answer:
The rate of heat conduction through the layer of still air is 517.4 W
Explanation:
Given:
Thickness of the still air layer (L) = 1 mm
Area of the still air = 1 m
Temperature of the still air ( T) = 20°C
Thermal conductivity of still air (K) at 20°C = 25.87mW/mK
Rate of heat conduction (Q) = ?
To determine the rate of heat conduction through the still air, we apply the formula below.
Q = 517.4 W
Therefore, the rate of heat conduction through the layer of still air is 517.4 W