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Ann [662]
3 years ago
7

"A uniform cylinder of mass M and radius R is rolling without slipping. The velocity of its center of mass is v. What is the cyl

inder's total kinetic energy?"
Physics
1 answer:
Goshia [24]3 years ago
6 0

Answer:

Explanation:

Given

mass of cylinder is M

radius R

velocity of center of mass is v

As there is no slipping therefore cylinder will rotate as well as translate

Moment of inertia of cylinder I=\frac{MR^2}{2}

Kinetic Energy of cylinder K.E.=\frac{1}{2}Mv^2

Rotational energy R.E.=\frac{1}{2}I\omega ^2

for rolling

v=\omega \times r

where \omega =angular\ velocity\ of\ cylinder

R.E.=\frac{1}{2}\times \frac{MR^2}{2}\times (\frac{v}{R})^2=\frac{1}{4}Mv^2

Total kinetic Energy =\frac{1}{2}Mv^2+\frac{1}{4}Mv^2=\frac{3}{4}Mv^2

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25 PTs
Nuetrik [128]
Answer:
They are both wrong!

Liquid oxygen really is a pale blue color.
I’ve seen it.

And they cant say that liquid and solid oxygen is blue which makes the sky blue because they’re not and it doesn’t make up for the color of the sky.

The sky is actually blue because It reflects more light than It can absorb
aka Rayleigh Scattering.

-HOPE THAT HELPED
8 0
3 years ago
Please need help fast
iVinArrow [24]

(a) See graph in attachment

The appropriate graph to draw in this part is a graph of velocity vs time.

In this problem, we have a horse that accelerates from 0 m/s to 15 m/s in 10 s.

Assuming the acceleration of the horse is uniform, it means that the velocity (y-coordinate of the graph) must increase linearly with the time: therefore, the velocity-time graph will appear as a straight line, having the final point at

t = 10 s

v = 15 m/s

(b) 1.5 m/s^2

The average acceleration of the horse can be calculated as:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time interval

In this problem,

u = 0

v = 15 m/s

t = 10 s

Substituting,

a=\frac{15-0}{10}=1.5 m/s^2

(c) 75 m

For a uniformly accelerated motion, the distance travelled can be calculated by using the suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance travelled

u is the initial velocity

t is the time interval

a is the acceleration

In this problem,

u = 0

t = 10 s

a=1.5 m/s^2

Substituting,

s=0+\frac{1}{2}(1.5)(10)^2=75 m

(d) See attached graphs

In a uniformly accelerated motion:

- The distance travelled (x) follows the equation mentioned in part c,

x=ut+\frac{1}{2}at^2

So, we see that this has the form of a parabola: therefore, the graph x vs t will represents a parabola.

- The acceleration is constant during the motion, and its value is

a=1.5 m/s^2 (calculated in part b)

therefore, the graph acceleration vs time will show a flat line at a constant value of 1.5 m/s^2.

6 0
3 years ago
Two identical point charges in outer space are held apart at a distance D. As soon as the charges are released, each begins movi
AVprozaik [17]

Answer:

B. 4a

Explanation:

Force between the charges is inversely proportional to the square of the distance

=> Force will be 4 times and acceleration will be 4a  

=> Answer b).

6 0
3 years ago
Read 2 more answers
In designing buildings to be erected in an area prone to earthquakes, what are the things that you need to consider?
il63 [147K]

Answer:

In an earthquake prone area, the things that are needed to be considered for the buildings to stand constant and no cause of damages are as follows-

  • The foundation of the building must be made strong, using large amount of concretes so that it can hold the entire load of the building.
  • The pillars of the buildings are also should be firmed, rigid, strong and thick.
  • The buildings are needed to be construct in such a way that during any seismic event, the forces can be redistributed easily through the body of the building without causing any damage.
  • The buildings must be overall comprised of holding high strength and stiffness.
6 0
3 years ago
In a thundercloud there may be electric charges of 45.0 C near the top of the cloud and -45.0 C near the bottom of the cloud. Th
Gala2k [10]

Answer:

The electric force on the top charge is F=3.44\times 10^6\ N.

Explanation:

Given that,

Electric charges in a thundercloud, q_1=q_2=45\ C

The distance between charges, d = 2.3 km = 2300 m

Let F is the electric force on the top charge. The electric force is given by the formula as :

F=\dfrac{kq^2}{d^2}

F=\dfrac{9\times 10^9\times (45)^2}{(2300)^2}

F=3445179.58\ N

or

F=3.44\times 10^6\ N

So, the electric force on the top charge is F=3.44\times 10^6\ N. Hence, this is the required solution.                                                

7 0
3 years ago
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