Answer:
Because it can easily resist air resistance.
Explanation:
Since air resistance is not negligible, the crumpled paper will reach the ground first because it can easily resist air resistance surrounding it compare to the un-crumpled one that will be influenced by the air thereby causing the un-crumpled paper to spend more time in the air
For this case, let's
assume that the pot spends exactly half of its time going up, and half going
down, i.e. it is visible upward for 0.245 s and downward for 0.245 s. Let us take
the bottom of the window to be zero on a vertical axis pointing upward. All calculations
will be made in reference to this coordinate system. <span>
An initial condition has been supplied by the problem:
s=1.80m when t=0.245s
<span>This means that it takes the pot 0.245 seconds to travel
upward 1.8m. Knowing that the gravitational acceleration acts downward
constantly at 9.81m/s^2, and based on this information we can use the formula:
s=(v)(t)+(1/2)(a)(t^2)
to solve for v, the initial velocity of the pot as it enters
the cat's view through the window. Substituting and solving (note that
gravitational acceleration is negative since this is opposite our coordinate
orientation):
(1.8m)=(v)(0.245s)+(1/2)(-9.81m/s^2)(0.245s)^2
v=8.549m/s
<span>Now we know the initial velocity of the pot right when it
enters the view of the window. We know that at the apex of its flight, the
pot's velocity will be v=0, and using this piece of information we can use the
kinematic equation:
(v final)=(v initial)+(a)(t)
to solve for the time it will take for the pot to reach the
apex of its flight. Because (v final)=0, this equation will look like
0=(v)+(a)(t)
Substituting and solving for t:
0=(8.549m/s)+(-9.81m/s^2)(t)
t=0.8714s
<span>Using this information and the kinematic equation we can find
the total height of the pot’s flight:
s=(v)(t)+(1/2)(a)(t^2) </span></span></span></span>
s=8.549m/s (0.8714s)-0.5(9.81m/s^2)(0.8714s)^2
s=3.725m<span>
This distance is measured from the bottom of the window, and
so we will need to subtract 1.80m from it to find the distance from the top of
the window:
3.725m – 1.8m=1.925m</span>
Answer:
<span>1.925m</span>
Answer:
The coefficient of kinetic friction between the crate and the floor can be calculated using the formula μ = Ff / N, where Ff is the frictional force, N is the normal force, and μ is the coefficient of kinetic friction.
In this case, the normal force is equal to the weight of the crate, which is 24 kg * 9.8 m/s2 = 235.2 N. The frictional force can be calculated using the formula Ff = μ * N, where μ is the coefficient of kinetic friction and N is the normal force.
If we substitute the values for N and Ff into the formula for the coefficient of kinetic friction, we get:μ = 53 N / 235.2 N = 0.225
Therefore, the coefficient of kinetic friction between the crate and the floor is 0.225.
Answer:
v = 50.5 m/s
Explanation:
F = (m)(^v/^t)
115N = (0.04551kg)(v/(0.020s))
2,526.917161 m/s² = v/(0.020s)
v = 50.53834322 m/s
v = 50.5 m/s
Answer:
Capacitance, C = 26.1 picofarad
Explanation:
It is given that,
Side of square, x = 4.3546 cm = 0.043546 m
Distance between electrodes, d = 0.6408 mm = 0.0006408 m
Voltage, V = 73.68 V
Capacitance of parallel plates is given by :



or
C = 26.1 picofarad
So, the capacitance of the capacitor is 26.1 picofarad. Hence, this is the required solution.