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2 years ago

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What is the torque τb about axis b due to the force f⃗ ? (b is the point at cartesian coordinates (0,b), located a distance b fr

om the origin along the y axis.) express the torque about axis b in terms of f, θ, ϕ, π, and/or other given coordinate data?

1 answer:

yKpoI14uk [10]2 years ago

8 0

Check the attached file for the solution.

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If a 3-kg rabbit's leg muscles act as imperfectly elastic springs, how much energy will they hold if the rabbit lands from a hei

DiKsa [7]

Answer;

- 15 J

Explanation;

-Potential energy is defined as mechanical energy, stored energy, or energy caused by its position.

-For the gravitational force the formula is P.E. = mgh, where m is the mass in kilograms, g is the acceleration due to gravity (9.8 m /s² at the surface of the earth) and h is the height in meters.

Potential energy of the rabbit at the peak of its height is

PE = (3)(10)(0.5) = 15 J

(around 14.7 but because energy is lost, it is less than that)

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2 years ago

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A segment of wire of total length 2.0 m is formed into a circular loop having 5.0 turns. If the wire carries a 1.2-A current, de

docker41 [41]

Answer:

Magnetic field at the center of the loop $B=5.89\times 10^{-5}\ T.$

Explanation:

It is given that total length of wire is 2 m and number of circular loop is 5 turns.

Therefore ,

$5\times ( 2\pi r)=2 \ m .\\\\r=\dfrac{1}{5 \pi}=0.064\ m.$

We know , magnetic field at the center of loop is given by :

$B=N\dfrac{\mu_o i}{2r}$

Putting all values in above equation we get :

$B=5\times \dfrac{4\pi\times 10^{-7}\times 1.2}{2\times 0.064}\\\\B=5.89\times 10^{-5}\ T.$

Hence , this is the required solution.

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2 years ago

An electric dipole with dipole moment p⃗ is in a uniform electric field E⃗ . A. Find all the orientation angles of the dipole me

Tasya [4]

Answer:

Explanation:

A) When a dipole is placed in an electric field , it experiences a torque equal to the following

torque = p x E = p E sinθ , where θ is angle between direction of p and E .

It will be zero if θ = 0

or if both p and E are oriented in the same direction.

It is the stable orientation of dipole.

If θ = 180° ,

Torque = 0

In this case both p and E are oriented in opposite direction .

It is the unstable orientation of the dipole because if we deflect the dipole by even small angle , it goes back to most stable orientation due to torque acting on it by electric field.

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2 years ago

A car with speed v and an identical car with speed 2v both travel the same circular section of an unbanked road. If the friction

yawa3891 [41]

Answer:

$F'=\dfrac{F}{4}$

Explanation:

Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.

For faster car on the road,

$F=\dfrac{mv^2}{r}$

v = 2v

$F=\dfrac{m(2v)^2}{r}$

$F=4\dfrac{m(v)^2}{r}$ ..........(1)

For the slower car on the road,

$F'=\dfrac{mv^2}{r}$ ............(2)

Equation (1) becomes,

$F=4F'$

$F'=\dfrac{F}{4}$

So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.

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2 years ago

The critical angle for a substance is measured at 53.7 degrees. light enters from air at 45.0 degrees. at what angle it will con

faust18 [17]

When light travels from a medium with greater refractive index $n_1$ to a medium with smaller refractive index $n_2$ , there exists an angle (called critical angle) above which the light is totally reflected, and the value of this angle is given by
$\theta_c = \arcsin ( \frac{n_2}{n_1} )$
In this problem, we know that the critical angle is $\theta_c = 53.7^{\circ}$ , so we can find the ratio between the refractive indices of the two mediums:
$\frac{n_2}{n_1} = \sin \theta_c = \sin 53.7^{\circ} =0.81$
and since the second medium is air (n=1.00), the refractive index of the first medium is
$n_1= \frac{n_2}{0.81}= \frac{1.00}{0.81}=1.23$

In the second part of the problem, we have light entering from air ( $n_i = 1.00$ ) at angle of incidence of $\theta_i = 45.0 ^{\circ}$ , into the second medium with $n_r = 1.23$ . By using Snell's law, we can find the angle of refraction of the light inside the medium:
$n_i \sin \theta_i = n_r \sin \theta_r$
$\sin \theta_r = \frac{n_i}{n_r} \sin \theta_i = \frac{1.00}{1.23} \sin 45^{\circ}=0.574$
$\theta_r = \arcsin(0.574)=35.1^{\circ}$

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2 years ago