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GuDViN [60]
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?
Physics
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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Answer:

35.2m

Explanation:

S= V^2 - U^2 ÷ 2A

S= (0) - (18) ÷ 2(4.6)

S= 324 ÷ 92

S= 35.2m

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3 years ago
A cube at 101 °C radiates heat at a rate of 67 J/s. If its surface temperature is increased to 166 °C, the rate at which it will
igor_vitrenko [27]

Answer:

The rate at which it will radiate heat is closest to 127.2J.

Explanation:

According to Stefan's law, the heat radiated by a black body can be written as:

\frac{dH}{dt} is directly proprtional to the  T^{4}.

where H is the amount of heat radiated and T is the surface temperature .

<u>Example</u> : The heat is transferred by radiation and an example of heat transferred by radiation is the sunlight reaching earth from sun.

<u>Note</u> : Here the temperature is expressed in Kelvin.

  • Initial rate of heat , \frac{dH_{1}}{dt}=67J/s
  • Final rate of heat ,  \frac{dH_{2}}{dt}= unknown
  • Initial temperature , T_{1}=374K
  • Final temperature , T_{2}=439K

\frac{H_1}{T_{1}^4}=\frac{H_2}{T_{2}^4}

H_{2}=67\times1.898

H_{2}=127.2.

The rate at which it will radiate heat is closest to 127.2J.

3 0
3 years ago
An inductor with an inductance of 2.30H and a resistance of 8.00 Ω isconnected to the terminals of a battery with an emf of 6.00
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Answer:

(a). The initial rate is 2.60 A/s.

(b). The rate of current increases is 0.8658 A/s.

(c). The current is 0.435 A.

(d). The final steady-state current is 0.75 A.

Explanation:

Given that,

Inductance = 2.30 H

Resistance = 8.00 Ω

Voltage = 6.00 V

(a). We need to calculate the initial rate of increase of current in the circuit.

Using formula of initial rate

V=initial\ rate\times inductance

initial\ rate=\dfrac{V}{L}

Put the value into the formula

initial \rate=\dfrac{6.00}{2.30}

initial\ rate=2.60\ A/s

The initial rate is 2.60 A/s.

(b). We need to calculate the rate of increase of current at the instant when the current is 0.500

Using formula of rate of increase of current

rate\ of \ current\ increase=initial\ rate\times e^{\dfrac{-t}{T}}....(I)

Where, T=\dfrac{L}{R}

T=\dfrac{2.30}{8.00}

T=0.2875

Using formula of current

i=\dfrac{V}{R}(1-e^{\dfrac{-t}{T}})

e^{\dfrac{-t}{T}}=1-(i\times\dfrac{R}{V})

e^{\dfrac{-t}{T}}=1-(0.500\times\dfrac{8.00}{6.00})

e^{\dfrac{-t}{T}}=0.333

The rate of current increases is

Put the value in the equation (I)

rate\ of\ current\ increase=2.60\times0.333

rate\ of\ current\ increase=0.8658\ A/s

The rate of current increases is 0.8658 A/s.

(c). We need to calculate the current

Using formula of current

i=\dfrac{V}{R}(1-e^{\dfrac{-t}{T}})

Put the value into the formula

i=\dfrac{6.00}{8.00}\times(1-e^{\dfrac{-0.250}{0.2875}})

i=0.435\ A

The current is 0.435 A.

(d). We need to calculate the final steady-state current

Using formula of steady state

i=\dfrac{V}{R}

i=\dfrac{6.00}{8.00}

i=0.75\ A

The final steady-state current is 0.75 A.

Hence, This is the required solution.

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