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irga5000 [103]
3 years ago
5

Starting from rest, a solid sphere rolls without slipping down an incline plane. At the bottom of the incline, what does the ang

ular velocity of the sphere depend upon?
Physics
1 answer:
Marrrta [24]3 years ago
3 0

Answer:

2/R*sqrt (g*s*sin(θ)) = w

Explanation:

Assume:

- The cylinder with mass m

- The radius of cylinder R

- Distance traveled down the slope is s

- The angular speed at bottom of slope w

- The slope of the plane θ

- Frictionless surface.

Solution:

- Using energy principle at top and bottom of the slope. The exchange of gravitational potential energy at height h, and kinetic energy at the bottom of slope.

                                         ΔPE = ΔKE

- The change in gravitational potential energy is given as m*g*h.

- The kinetic energy of the cylinder at the bottom is given as rotational motion: 0.5*I*w^2

- Where I is the moment of inertia of the cylinder I = 0.5*m*R^2

We have:

                              m*g*s*sin(θ) = 0.25*m*R^2*w^2

                              2/R*sqrt (g*s*sin(θ)) = w

- The angular velocity depends on plane geometry θ , distance travelled down slope s, Radius of the cylinder R , and gravitational acceleration g

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A steady current I flows through a wire of radius a. The current density in the wire varies with r as J = kr, where k is a const
grin007 [14]

Answer:

Explanation:

we can consider an element of radius r < a and thickness dr.  and Area of this element is

dA=2\pi r dr

since current density is given

J=kr

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di_{thru}=JdA=(kr)(2\pi\,r\,dr)=2\pi\,kr^2\,dr

integrating on both sides between the appropriate limits,

\int_0^Idi_{thru}=\int_0^a2\pi\,kr^2\,dr&#10;\\\\&#10;I=\frac{2\pi\,ka^3}{3} -------------------------------(1)

Magnetic field can be found by using Ampere's law

\oint{\vec{B}\cdot\,d\vec{l}}=\mu_0\,i_{enc}

for points inside the wire ( r<a)

now, consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius r.

by applying the Ampere's law, we can write

\oint{\vec{B}_{in}\cdot\,d\vec{l}}=\mu_0\,i_{enc}&#10;

by symmetry \vec{B} will be of uniform magnitude on this loop and it's direction will be tangential to the loop.

Hence,

B_{in}\times2\pi\,l=\mu_0\int_0^r(kr)(2\pi\,r\,dr)=&#10;\\\\2\pi\,B_{in} l=2\pi\mu_0k \frac{r^3}{3}&#10;\\\\B_{in}=\frac{\mu_0kl^2}{3}&#10;

now using equation 1, putting the value of k,

B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3}&#10;\\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3}&#10;

B)

now, for points outside the wire ( r>a)

consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius l.

applying the Ampere's law

\oint{\vec{B}_{out}\cdot\,d\vec{l}}=\mu_0\,i_{enc}&#10;

by symmetry \vec{B} will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence

B_{out}\times2\pi\,r=\mu_0\int_0^a(kr)(2\pi\,r\,dr)&#10;\\\\2\pi\,B_{out}r=2\pi\mu_0k\frac{a^3}{3}&#10;\\\\B_{out}=\frac{\mu_0ka^3}{3r}&#10;

again using,equaiton 1,

B_{out}= \mu_0 \frac{a^3}{3r} \times \frac{3 I}{2 \pi a^3}&#10;\\\\B_{out} = \frac{ \mu_{0} I}{2 \pi r}

8 0
3 years ago
What is the distance traversed by the particle between 0 seconds and 6 seconds?
ad-work [718]

Answer:

I dont know to be honest

Explanation:

i dont know to be honest

5 0
3 years ago
What is the full meaning of (i.p.s.m.n)​
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Answer:

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6 0
3 years ago
What is the equivalent resistance of a circuit that contains four 75.0 resistors connected in parallel with a 100.0 V battery?
zhannawk [14.2K]
To get the total resistance in a parallel circuit, you need to remember that unlike in a series, you do not just merely add the resistances. You need to get the reciprocal first of each resistance and add them together. 

\frac{1}{R_{T}} =  \frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}... +\frac{1}{R_{n}}

After adding them, you will get the reciprocal again and then compute for the value. The problem says that there are 4 resistors in the circuit that have a resistance of 75.

\frac{1}{R_{T}} = \frac{1}{75}+\frac{1}{75}+\frac{1}{75}+\frac{1}{75}

Add up the numerator and copy the denominator:

\frac{1}{R_{T}} = \frac{4}{75}

Then get the reciprocal to get the total resistance:

R _{T} = \frac{75}{4}  = 18.75

The answer to your question then is A. 18.8.
5 0
3 years ago
Read 2 more answers
8. two +1 C charges are separated by 3000m. What is the magnitude of the electric force between them?
Sidana [21]

Answer:

1000 N

Explanation:

The magnitude of the electrostatic force between two charged object is given by

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb constant

q1, q2 is the magnitude of the two charges

r is the distance between the two objects

Moreover, the force is:

- Attractive if the two forces have opposite sign

- Repulsive if the two forces have same sign

In this problem:

q_1=q_2=+1C are the two charges

r = 3000 m is their separation

Therefore, the electric force between the charges is:

F=(9\cdot 10^9)\frac{(1)(1)}{3000^2}=1000 N

8 0
3 years ago
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