Why do scientists believe the universe is still expanding?

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

2 years ago

A bug flying horizontally at 0.65 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, th

irina [24]

The angular momentum is defined as,

$L=I\omega$

Acording to this text we know for conservation of angular momentum that

$L_i=L_f$

Where $L_i$ is initial momentum

$L_f$ is the final momentum

How there is a difference between the stick mass and the bug mass, we define that

Mass of the bug= m

Mass of the stick=10m

At the point 0 we have that,

$L_i=mvl$

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

vector from the point of reference (O), and ve is the velocity

At the end with the collition we have

$L_f=(I_b+I_s)\omega$

Substituting

$L_f=(ml^2+\frac{10ml^2}{3})\omega$

$L_f=\frac{13}{10}ml^2w$

$m(0.65)l=\frac{13}{10}ml^2 \omega$

$\omega=\frac{1}{2l}$

Applying conservative energy equation we have

$\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'$

$\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)$

Replacing the values and solving

$l=\frac{13}{0.54g}$

Substituting

l=\frac{13}{0.54(9.8)}

$l=2.45cm$

7 0

2 years ago

A) what is the light source for a plant growing in the shade?

Ulleksa [173]

a.) Plants that thrive in the shade are often able to hold on to sunlight for extensive periods of time; they're in a sense like the camels of the plaNt WoRld.

b.) Though artificial lights are not nearly as beneficial as the sun, one could invest in one of them plant growing light thingies, but sun-loving plants might be sad if u do this instead of letting them soak in ePic rays from the sun.

6 0

2 years ago

A flat horizontal surface with an area of 518 cm^2 is inside a uniform electric field of 1.33 x 10^4 N/C. The angle between the

GarryVolchara [31]

Answer: 576.48 N*m^2/C

Explanation: In order to calculate the electric flux through the any surface we have to take into account the scalar product between the electric field vector and the normal vector to the surface.

So we have:

ФE= E*A= 1.33 * 10^4*0.0518* cos (33.2°)= 576.48 N*m^2/C

3 0

2 years ago

The equation P^xV^yT^z= constant is Boyle law for what is the values of x,y,z

chubhunter [2.5K]

I learned the equation as P•V=k•T .

So x=1, y=1, and z= -1 .

3 0

2 years ago