Answer:
The coefficient of static friction between the puppy and the floor is 0.7273.
Explanation:
The horizontal force applied to move the puppy from a steady state has to be greater than the force of static friction, after it is moving the force needs to be equal to be greater than the force of dynamic friction in order to maintain its movement. The force of static friction is given by:
Where is the static friction force,
is the coefficient of static friction and
is the normal force. Since there's no angle on the flor the normal force is equal to the weight of the puppy, therefore,
, to make the puppy moving we need to use a force of 80 N, therefore,
, so we can solve for the coefficient as shown below:
The coefficient of static friction between the puppy and the floor is 0.7273.
Answer:
Explanation:
Given
length of window
time Frame for which rock can be seen is
Suppose h is height above which rock is dropped
Time taken to cover
so using equation of motion
where y=displacement
u=initial velocity
a=acceleration
t=time
time taken to travel h is
Subtract 1 and 2 we get
and from equation
so
and
so
substitute the value of in equation 2
The initial potential energy of the wagon containing gold boxes will enable
it roll down the hill when cut loose.
The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.
Reasons:
Mass wagon and gold = 166 kg
Location of the wagon = 77 meters up the hill
Slope of the hill = 8°
Location of the rangers = 41 meters from the canyon
Mass of Lone Ranger, m₁ = 65 kg
Mass of Tonto m₂ = 66 kg
Solution;
Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m
Potential energy = m·g·h
Where;
g = Acceleration due to gravity ≈ 9.81 m/s²
Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912
Potential energy of wagon, P.E. ≈ 17457.0912 J
By energy conservation, P.E. = K.E.
Where;
v = The velocity of the wagon a the bottom of the cliff
Therefore;
Velocity of the wagon, v ≈ 14.5 m/s
Momentum = Mass, m × Velocity, v
Initial momentum of wagon = m·v
Final momentum of wagon and ranger = (m + m₁ + m₂)·v'
By conservation of momentum, we have;
m·v = (m + m₁ + m₂)·v'
Which gives;
The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s
The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the
gold and jump out of the wagon before the wagon heads over the cliff.
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