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Tju [1.3M]
3 years ago
11

Radar uses radio waves of a wavelength of 2.2 m . The time interval for one radiation pulse is 100 times larger than the time of

one oscillation; the time between pulses is 10 times larger than the time of one pulse. What is the shortest distance to an object that this radar can detect?
Physics
1 answer:
liq [111]3 years ago
8 0

Answer:

<em>The shortest distance to an object that this radar can detect would be </em>

<em>111 m</em>

Explanation:

The shortest distance is the minimum distance that would be detected by the radar, The time of oscillation can be obtained thus;

T = 1/f   ..........1

but f= v/λ substituting f into equation 1 we have;

T = λ/v...............2

Where λ is the wavelength = 2.2 m

v is the velocity of  light  since the radar is an electromagnetic wave

   = 3 x10^{8} m/s

T = 2.2 m / 3 x10^{8} m/s

T = 7.33 x10^{-9} s

<u>Calculating the time interval required by the pulse</u>

Since the interval of the pulse (t) is 100 times greater than the period of oscillation, the time of the pulse is expressed as;

t = 100 x T

t = 100 x 7.33 x10^{-9} s

t = 7.33 x 10^{-7} s

Therefor the time of the pulse is 7.33 x 10^{-7} s

<u>Calculating for the shortest distance of the radar</u>

The shortest distance of the radar can be obtained using the equation below;

λ_s = (t/2) x v ............3

Substituting into equation 3 we have

λ_s = (7.33 x 10^{-7} /2) x 3 x10^{8} m/s

λ_s  =  111 m

Therefore the shortest distance to an object that this radar can detect would be 111 m

   

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6. The two ends of an iron rod are maintained at different temperatures. The amount of heat thatflows through the rod by conduct
motikmotik

Answer:

20cal/s

Explanation:

Question:

There are two questions. The first one has been answered:

From the formular, Power = Q/t = (kA∆T)/l

the amount heat depends on the duration of time interval, length of the iron rod, the thermal conductivity of iron and the temperature difference between the ends of the rod.

The amount of heat that flows through the rod by conduction during a given time interval does not depend upon the mass of the iron rod (D).

Second question:

The ends of a cylindrical steel rod are maintained at two different temperatures. The rod conducts heat from one end to the other at a rate of 10 cal/s. At what rate would a steel rod twice as long and twice the diameter conduct heat between the same two temperatures?

Solution:

Power = 10cal/s

Power = energy per unit time = Q/t

Where Q = energy

Power = (kA∆T)/l

k = thermal conductivity of iron

A = area

Area = πr^2

r = radius

Diameter = d = 2r

r = d/2

Area = (πd^2)/4

Length = l

∆T = change in temperature

10 = (kA∆T)/l

For a steel rod with length doubled and diameter doubled:

Let Length (L) = 2l

Diameter (D)= 2d

Area = π [(2d)^2]/4 = (π4d^2)/4

Area = 4(πd^2)/4

Using the formula Power = (kA∆T)/l, insert the new values for A and l

Power = [k × 4(πd^2)/4 × ∆T]/2l

Power = [4k((πd^2)/4) ∆T]/2l

Power = [(4/2)×k((πd^2)/4) ∆T]/l

Power = [2k(A) ×∆T]/l = 2(kA∆T)/l

Power of a steel that has its length doubled and diameter doubled = 2(kA∆T)/l

Recall initial Power = (kA∆T)/l = 10cal/s

And ∆T is the same

2[(kA∆T)/l] = 2 × 10

Power of a steel that has its length doubled and diameter doubled = 20cal/s

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3 years ago
A car is traveling at a speed of 45 km/h into town. It takes the car 2 hours to get there. How far has the car traveled?
hjlf

Answer:

90 km

Explanation:

45 km per hr

2 hrs

45 x 2 = 90

90 km/2 hrs

Hope this helps!

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3 years ago
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Which part of a laser printer applies a positive charge to the paper that attracts the toner particles to it
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The part of laser printer that applies a positive charge to the paper in order to attract the toner particles is known as transfer roller.

<h3 />

What is a laser printer:

A laser printer is a kind of printer that uses the electrostatic digital printing process to perform printing. It makes use of the static electricity and toner powder in place of liquid ink.

The toner is applied to specific areas which are dependent on the charge difference created or on the static electricity.

Following are the components of a laser printer:

  • Scanning unit:

        This unit of a laser printer generally consists of a laser diode, a

        scanning motor and a polygon mirror.
        It also consists of two-beam alignment lenses.

  • Cartridge unit:

        This unit of laser printer consists of three drums, namely primary

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        image transfer roller (ITR).
        The transfer roller is also present at a close vicinity of the  

        printer's  toner cartridge.

  • Fuser assembly unit:

        This unit of laser printer consists of a pressure roller and a fuser                roller, where the fuser roller assembly consists of a heating

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Therefore, the transfer roller unit of a laser printer applies a positive charge to the paper that attracts the toner particles to it.

Learn more about laser printers here:

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How are stars formed and classified?
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3 years ago
If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu
Mashcka [7]

3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}.

Volume of water:

V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}.

Mass of water:

m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}.

Amount of heat that the 15 L water absorbed:

Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}.

What's the mass of the hot steel tool?

The specific heat of carbon steel is 0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}.

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}.

\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}.

m = \dfrac{Q}{c\cdot \Delta T} = \dfrac{2.06910 \times 10^{6} \; \text{J}}{0.49\; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1152\; \textdegree{\text{C}}} = 3.6 \times 10^{3} \; \text{g} = 3.6 \; \text{kg}.

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3 years ago
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