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3 years ago

13

A satellite dish is in the shape of a parabolic surface. Signals coming from a satellite strike the surface of the dish and are

reflected to the focus, where the receiver is located. The satellite dish has a diameter of 14 feet and a depth of 2 feet. How far from the base of the dish should the receiver be placed?

1 answer:

RSB [31]3 years ago

3 0

Answer: 6.125 ft

Explanation:

If this dish has the form of a concave upward parabola and its vertex $p$ is at the origin, its corresponding equation is:

$x^{2}=4py$

Where:

$x$ is the radius, which can be found by dividing the diameter $d=14 ft$ by half. Hence $x=\frac{d}{2}=\frac{14 ft}{2}=7 ft$

$y=2 ft$ is the depth

$p$ is the vertex of the parabola, where its base is

Finding $p$ :

$p=\frac{x^{2}}{4y}$

$p=\frac{(7 ft)^{2}}{4(2 ft)}$

Finally:

$p=6.125 ft$ This is where the the receiver should be placed

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2 years ago

When the velocity of an object is doubled, by what factor is its momentum changed? By what factor is its kinetic energy changed?

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Answer:

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Explanation:

Kinetic Energy is given by 0.5*mass*velocity^2. Kinetic Energy is proportional to Velocity^2.

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If the velocity of an object is doubled, Kinetic energy would increase by a factor of 2^2 i.e 4 times. Momentum would increase by a factor of 2.

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3 years ago

Three packing crates of masses, M1 = 6 kg, M2 = 2 kg and M3 = 8 kg are connected by a light string of negligible mass that passe

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Answer:

39.81 N

Explanation:

I attached an image of the free body diagrams I drew of crate #1 and #2.

Using these diagram, we can set up a system of equations for the sum of forces in the x and y direction.

∑Fₓ = maₓ

∑Fᵧ = maᵧ

Let's start with the free body diagram for crate #2. Let's set the positive direction on top and the negative direction on the bottom. We can see that the forces acting on crate #2 are in the y-direction, so let's use Newton's 2nd Law to write this equation:

∑Fᵧ = maᵧ
T₁ - m₂g = m₂aᵧ

Note that the tension and acceleration are constant throughout the system since the string has a negligible mass. Therefore, we don't really need to write the subscripts under T and a, but I am doing so just so there is no confusion.

Let's solve for T in the equation...

T₁ = m₂aᵧ + m₂g
T₁ = m₂(a + g)

We'll come back to this equation later. Now let's go to the free body diagram for crate #1.

We want to solve for the forces in the x-direction now. Let's set the leftwards direction to be positive and the rightwards direction to be negative.

∑Fₓ = maₓ

F_f - F_g sinΘ = maₓ

The normal force is equal to the x-component of the force of gravity.

(F_n · μ_k) - m₁g sinΘ = m₁aₓ

(F_g cosΘ · μ_k) - m₁g sinΘ = m₁aₓ
[m₁g cos(30) · 0.28] - [m₁g sin(30)] = m₁aₓ
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Now let's go back to this equation:

T₁ = m₂(a + g)

We have 3 known variables and we can solve for the tension force.

T = 2(10.1059759 + 9.8)
T = 2(19.9059759)
T = 39.8119518 N

The tension force is the same throughout the string, therefore, the tension in the string connecting M2 and M3 is 39.81 N.

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2 years ago

A rotating flywheel has moment of inertia 18.0 kg⋅m^2 for an axis along the axle about which the wheel is rotating. Initially th

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Answer:

The rotational kinetic energy takes 0.430 seconds to become half its initial value.

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Where:

$\Delta E$ - Energy losses, measured in joules.

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By definition of rotational kinetic energy, we expand the equation above:

$\Delta E = \frac{1}{2}\cdot I\cdot (\omega_{1}^{2}-\omega_{2}^{2})$ (2)

Where:

$I$ - Moment of inertia of the flywheel, measured in kilograms per square meter.

$\omega_{1}$ , $\omega_{2}$ - Initial and final angular speed, measured in radians per second.

If we know that $K_{1} = 30\,J$ , $K_{2} = 15\,J$ and $I = 18\,kg\cdot m^{2}$ , then the initial angular speed is:

$K_{1} = \frac{1}{2}\cdot I \cdot \omega_{1}^{2}$ (3)

$\omega_{1}=\sqrt{\frac{2\cdot K_{1}}{I} }$

$\omega_{1} = \sqrt{\frac{2\cdot (30\,J)}{18\,kg\cdot m^{2}} }$

$\omega_{1} \approx 1.825\,\frac{rad}{s}$

$\omega_{1}\approx 0.291\,\frac{rev}{s}$

$K_{2} = \frac{1}{2}\cdot I \cdot \omega_{2}^{2}$ (4)

$\omega_{2}=\sqrt{\frac{2\cdot K_{2}}{I} }$

$\omega_{2} = \sqrt{\frac{2\cdot (15\,J)}{18\,kg\cdot m^{2}} }$

$\omega_{2} \approx 1.291\,\frac{rad}{s}$

$\omega_{2} \approx 0.205\,\frac{rev}{s}$

Under the assumption that flywheel is decelerating uniformly, we get that the time taken for the flywheel to slowdown is:

$t = \frac{\omega_{2}-\omega_{1}}{\alpha}$ (5)

If we know that $\omega_{1}\approx 0.291\,\frac{rev}{s}$ , $\omega_{2} \approx 0.205\,\frac{rev}{s}$ and $\alpha = -0.200\,\frac{rev}{s^{2}}$ , then the time needed is:

$t = \frac{0.205\,\frac{rev}{s}-0.291\,\frac{rev}{s}}{-0.200\,\frac{rev}{s^{2}} }$

$t = 0.43\,s$

The rotational kinetic energy takes 0.430 seconds to become half its initial value.

6 0

3 years ago