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3 years ago

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An electric current is flowing through a long cylindrical conductor with radius a. The current density J is uniform in the cylin

der. In this problem we consider an imaginary cylinder with radius r around the axis AB where r

1 answer:

RUDIKE [14]3 years ago

4 0

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

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Amy has a mass of 30 kg, and she is riding a skateboard traveling 5 m/s. What is her momentum?

Flauer [41]

Hey there!

So we know that m*v=P.
And in this question m=30
v=5 m/s
P = 30*5 Kgm/s
P = 150 Kgm/s

So, your final answer is 150 Kg.m/s

Hope this helps! :)

5 0

3 years ago

Read 2 more answers

g initial angular velocity of 39.1 rad/s. It starts to slow down uniformly and comes to rest, making 76.8 revolutions during the

MrRa [10]

Answer:

Approximately $-1.58\; \rm rad \cdot s^{-2}$ .

Explanation:

This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.

This question does not provide any information about the time required for the rotation of this object to come to a stop. In linear motions with a constant acceleration, there's an SUVAT equation that does not involve time:

$v^2 - u^2 = 2\, a\, x$ ,

where

$v$ is the final velocity of the moving object,
$u$ is the initial velocity of the moving object,
$a$ is the (linear) acceleration of the moving object, and
$x$ is the (linear) displacement of the object while its velocity changed from $u$ to $v$ .

The angular analogue of that equation will be:

$(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ , where

$\omega(\text{final})$ and $\omega(\text{initial})$ are the initial and final angular velocity of the rotating object,
$\alpha$ is the angular acceleration of the moving object, and
$\theta$ is the angular displacement of the object while its angular velocity changed from $\omega(\text{initial})$ to $\omega(\text{final})$ .

For this object:

$\omega(\text{final}) = 0\; \rm rad\cdot s^{-1}$ , whereas
$\omega(\text{initial}) = 39.1\; \rm rad\cdot s^{-1}$ .

The question is asking for an angular acceleration with the unit $\rm rad \cdot s^{-1}$ . However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:

$\begin{aligned}\theta &= 76.8\; \rm \text{revolution} \\ &= 76.8\;\text{revolution} \times 2\pi\; \rm rad \cdot \text{revolution}^{-1} \\ &= 153.6\pi\; \rm rad\end{aligned}$ .

Rearrange the equation $(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ and solve for $\alpha$ :

$\begin{aligned}\alpha &= \frac{(\omega(\text{final}))^2 - (\omega(\text{initial}))^2}{2\, \theta} \\ &= \frac{-\left(39.1\; \rm rad \cdot s^{-1}\right)^2}{2\times 153.6\pi\; \rm rad} \approx -1.58\; \rm rad \cdot s^{-1}\end{aligned}$ .

7 0

3 years ago

The orbital period of a satellite is 2 × 106 s and its total radius is 2.5 × 1012 m. The tangential speed of the satellite, writ

LenaWriter [7]

The orbital period of the satellite[T] is given as $2*10^{6} S$ .

The radius of the satellite is given [R] $2.5*10^{12} m$ .

we are asked here to calculate the tangential speed of the satellite.

Before going to get the solution first we have understand the tangential speed.

The tangential speed of a satellite is given as the speed required to keep the satellite along the orbit. If satellite speed is less than tangential speed,there is the chance of it falling down towards earth. If it is more,then it will deviate from it orbit and can't stick to the orbit further.In a simple way the tangential speed is the linear speed of an object in a circular path.

Now we have to calculate the tangential speed [V].

Mathematically the tangential speed [V] written as -

V $=\frac{2\pi R}{T}$

where T is the time period of the satellite and R is the radius of the satellite.

$V=\frac{2*3.14*10^{12} }{2*10^{6} }$

$= 7.85*10^{6} m/s$

There is also another way through which we can get the solution as explained below-

We know that the tangential speed of a satellite V $=\sqrt{\frac{GM}{R^{2} } }$

where G is the gravitational constant and M is the mas of central object.

But we know that $g=\frac{GM}{R^{2} }$

⇒ $GM=gR^{2}$ where g is the acceleration due to gravity of that central object.

Hence $V=\sqrt{\frac{gR^{2} }{R} }$

⇒ $V=\sqrt{gR}$

By knowing the value of g due to that central object we can also calculate its tangential speed.

7 0

3 years ago

Read 2 more answers

A deciliter is how many times larger than a millimeter?

jarptica [38.1K]

Milliliter centiliter deciliter liter dekaliter hectoliter kiloliter

All related by 10's every move to the right is 10x larger than the one to its left

Since deciliter is two steps away from milliliter it is 10 x 10 or 100 times as large.

If the question is meant to be a trick then the answer is 99x larger.

It should read "A deciliter is how many milliliters" ofr "a milliliter is how much of a deciliter?"
Once you say larger than you could be confusing subtraction with multiplication. How much larger than 30 is 3? Answer is clearly 27. How many times as large as 3 is 30? Answer is clearly 20. How many times larger than 3 is 30? Hmmmm? Which one of the two does he mean?
I am sure your teacher meant you to consider multiplication, but just in case, I included the "trick" answer.

8 0

3 years ago

A golf ball is struck at 75 m/s at an angle of 60 degrees. What is the

Gnesinka [82]

Answer:

$v_x=37.50$

Explanation:

The rectangular components of a vector $\vec v$ having a magnitude v and angle θ are:

$v_x=v\cos\theta$

$v_y=v\sin\theta$

The golf ball has an initial speed of 75 m/s at an angle of 60 degrees.

The variables of the equations have the values:

v = 75 m/s

θ = 60°

Substituting into the formula:

$v_x=75\cos 60^\circ$

$v_x=75\cdot 0.5$

$v_x=37.5~m/s$

Without specifying units and with precision to the hundredths place:

$\mathbf{v_x=37.50}$

7 0

3 years ago