Answer
given,
diameter of the pipe is = (14 ft)4.27 m
minimum speed of the skater must have at very top = ?
At the topmost point of the pipe the normal force will be equal to zero.
F = mg
centripetal force acting on the skateboard

equating both the force equation


r = d/2 = 14/ 2 = 7 ft
or
r = 4.27/2 = 2.135 m
g = 32 ft/s² or g = 9.8 m/s²

v = 14.96 ft/s
or

v = 4.57 m/s
Distance fallen = 1/2 ( V initial + V final ) *t
We know
a = -9.8 m/s2
t=120s
To find distance fallen, we need to find V final
Use the equation
V final = V initial + a*t
Substitute known values
V final = 0 + (-9.8)(120)
V final = -1176 m/s
Then plug known values to distance fallen equation
Distance fallen = 1/2 ( 0 + 1176 )(120)
= 1/2(1776)(120)
=106,560 m
This way plugging into distance equation is actually the long way. A faster way is to plug the values into
Distance fallen = V initial * t + 1/2(a*t)
We won't need to find V final using another equation.
But anyways, good luck!
Answer:
The temperature of the Earth depends on many factors, including the concentration of greenhouse gases such as water vapour, methane and carbon dioxide. The Earth's temperature also depends on the rates at which light radiation and infrared radiation are: absorbed by the Earth's surface and atmosphere.
Answer:
μ=0.151
Explanation:
Given that
m= 3.5 Kg
d= 0.96 m
F= 22 N
v= 1.36 m/s
Lets take coefficient of kinetic friction = μ
Friction force Fr=μ m g
Lets take acceleration of block is a m/s²
F- Fr = m a
22 - μ x 3.5 x 10 = 3.5 a ( take g =10 m/s²)
a= 6.28 - 35μ m/s²
The final speed of the block is v
v= 1.36 m/s
We know that
v²= u²+ 2 a d
u= 0 m/s given that
1.36² = 2 x a x 0.96
a= 0.963 m/s²
a= 6.28 - 35μ m/s²
6.28 - 35μ = 0.963
μ=0.151
Answer:
vf = 22.36[m/s]
Explanation:
First we must understand the data given in the problem:
m = mass = 800 [kg]
F = force = 20000[N]
dx = displacement = 10[m]
From newton's second we know that the sum of forces must be equal to the product of mass by acceleration.
![F = m*a\\20000 = 800*a\\a = 20000/800\\a = 25 [m/s^2]](https://tex.z-dn.net/?f=F%20%3D%20m%2Aa%5C%5C20000%20%3D%20800%2Aa%5C%5Ca%20%3D%2020000%2F800%5C%5Ca%20%3D%2025%20%5Bm%2Fs%5E2%5D)
With the calculated acceleration, we can use the kinematics equations.
![v_{f} ^{2} =v_{o} ^{2}+2*a*dx\\ v_{o} = initial velocity = 0\\a = acceleration = 25[m/s^2]\\dx= displacement = 10[m]\\](https://tex.z-dn.net/?f=v_%7Bf%7D%20%5E%7B2%7D%20%3Dv_%7Bo%7D%20%5E%7B2%7D%2B2%2Aa%2Adx%5C%5C%20v_%7Bo%7D%20%3D%20initial%20velocity%20%3D%200%5C%5Ca%20%3D%20acceleration%20%3D%2025%5Bm%2Fs%5E2%5D%5C%5Cdx%3D%20displacement%20%3D%2010%5Bm%5D%5C%5C)
The key to using this equation is to clarify that the initial velocity is zero since the body is at rest, otherwise the initial velocity would be an initial data.
![v_{f} =\sqrt{2*25*10} \\v_{f} =22.36[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3D%5Csqrt%7B2%2A25%2A10%7D%20%5C%5Cv_%7Bf%7D%20%3D22.36%5Bm%2Fs%5D)
Another way of solving this problem is by means of the definition of work and kinetic energy, where work is defined as the product of the force by the distance.
W =F*d
W = 20000*10
W = 200000[J]
Kinetic energy is equal to work, therefore the value calculated above is equal to:
![E_{k}=W =0.5*m*v_{f}^{2} \\200000=0.5*800*v_{f}^{2}\\v_{f}=\sqrt{\frac{200000}{0.5*800} } \\v_{f}=22.36[m/s]](https://tex.z-dn.net/?f=E_%7Bk%7D%3DW%20%3D0.5%2Am%2Av_%7Bf%7D%5E%7B2%7D%20%5C%5C200000%3D0.5%2A800%2Av_%7Bf%7D%5E%7B2%7D%5C%5Cv_%7Bf%7D%3D%5Csqrt%7B%5Cfrac%7B200000%7D%7B0.5%2A800%7D%20%7D%20%5C%5Cv_%7Bf%7D%3D22.36%5Bm%2Fs%5D)