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Ira Lisetskai [31]

3 years ago

15

A small uniform disk and a small uniform sphere are released simultaneously at the top of a high inclined plane, and they roll d

own without slipping. Which one will reach the bottom first?A) the one of smallest diameter B) the one of greatest mass C) the disk D) the sphere E) They will reach the bottom at the same time.

1 answer:

barxatty [35]3 years ago

3 0

Answer:

(D) the sphere

Explanation:

The bodies given are Disk and Solid sphere (uniform sphere)

Moment of inertia of the bodies are

I(disk) = $\frac{MR^2}{2}$

I(sphere) = $\frac{2MR^2}{5}$

Since the moment of inertia of sphere is less than that of disk, therefore sphere will reach the bottom first.

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Determine the kinetic energy of a 10 Kg roller coaster car that is moving with a speed of 2m/s.

skelet666 [1.2K]

In this question all required information's are already provided. Based on these details the answer to the question can be easily determined. Let us now write down all the information's that are already given.

Mass of the roller coaster = 1000 kg

Velocity of the roller coaster = 20.0 m/s

We know the formula for finding the kinetic energy is

Kinetic energy = 0.5 * mass * (velocity) ^2

= 0.5 * 1000 * (20)^2

= 0.5 * 1000 * 400

= 200000 Joules

So the Kinetic energy of the roller coaster is 200000 joules.

i hope this helps you friend good luck on your quiz or lesson

6 0

1 year ago

A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o

Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

$dE = k\dfrac{dQ}{r^2}cos(\theta )$ <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where $r$ is the radius of curvature.

Now

$dQ = \lambda rd\theta$ ,

where $\lambda$ is the charge per unit length, and it has the value

$\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.$

Thus, the electric field at the center of the curvature of the arc is:

$E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta }{r^2} cos(\theta)$

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

$fraction = \dfrac{3.0m}{2\pi (2.3m)} = 0.2076$

and this is

$0.2076*2\pi =1.304$ radians.

Therefore,

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.$

evaluating the integral, and putting in the numerical values we get:

$E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\$

$\boxed{ E = 31.329N/C.}$

4 0

3 years ago

The earth has a vertical electric field at the surface,pointing down, that averages 102 N/C. This field is maintained by various

Schach [20]

Answer:

$q = -461532.5 \ C$

Explanation:

From the question we are told that

The electric filed is $E = 102 \ N/C$

Generally according to Gauss law

=> $E A = \frac{q}{\epsilon_o }$

Given that the electric field is pointing downward , the equation become

$- E A = \frac{q}{\epsilon_o }$

Here $q$ is the excess charge on the surface of the earth

$A$ is the surface area of the of the earth which is mathematically represented as

$A = 4\pi r^2$

Where r is the radius of the earth which has a value $r = 6.3781*10^6 m$

substituting values

$A = 4 * 3.142 * (6.3781*10^6 \ m)^2$

$A =5.1128 *10^{14} \ m^2$

So

$q = -E * A * \epsilon _o$

Here $\epsilon_o$ s the permitivity of free space with value

$\epsilon_o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

substituting values

$q = -102 * 5.1128 *10^{14} * 8.85 *10^{-12}$

$q = -461532.5 \ C$

6 0

3 years ago

How much force is needed to accelerate a vehicle with a mass of 1000 kg at a rate<br> of 5 m/s2?

Mashcka [7]

The force needed to accelerate a vehicle with a mass of 1000kg at a rate of 5m/s2 would be 5000

8 0

2 years ago

If a sprinter accelerates from rest to 12 m/s north , what is their change in velocity ?

natima [27]

Answer: It would be 12 m/s.

Explanation: It would be this because If you go from rest to sprint it would be 12 m/s. Also, I did this the other day.

5 0

3 years ago

Read 2 more answers