Can someone please help me on thisss

You set out to design a car that uses the energy stored in a flywheel consisting of a uniform 101-kg cylinder of radius r that h

Ket [755]

Ok, assuming "mj" in the question is Megajoules MJ) you need a total amount of rotational kinetic energy in the fly wheel at the beginning of the trip that equals
(2.4e6 J/km)x(300 km)=7.2e8 J
The expression for rotational kinetic energy is

E = (1/2)Iω²

where I is the moment of inertia of the fly wheel and ω is the angular velocity.
So this comes down to finding the value of I that gives the required energy. We know the mass is 101kg. The formula for a solid cylinder's moment of inertia is

I = (1/2)mR²

We want (1/2)Iω² = 7.2e8 J and we know ω is limited to 470 revs/sec. However, ω must be in radians per second so multiply it by 2π to get
ω = 2953.1 rad/s
Now let's use this to solve the energy equation, E = (1/2)Iω², for I:
I = 2(7.2e8 J)/(2953.1 rad/s)² = 165.12 kg·m²

Now find the radius R,

165.12 kg·m² = (1/2)(101)R²,
√(2·165/101) = 1.807m

R = 1.807m

8 0

3 years ago

PROJECTILE MOTION-Why isn't there any accelertion in the x direction while there is an acceleration of -g in the y direction? ..

Ahat [919]

There is no acceleration of g in the x direction because the gravitational acceleration points downward. Also, on most studies we ignore the tidal forces since we are dealing with small bodies compared to the size of the earth.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

5 0

4 years ago

A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr

VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

$rR:\\\\E=k\frac{Q}{r^2}$

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

$F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}$

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

$Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}$