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3 years ago

15

List 2 ways that Dopplar Radar is used in technology

1 answer:

Katyanochek1 [597]3 years ago

8 0

The doppler radar is used in technology in two ways;

<span>· </span>Continuous Doppler radar – it has the capability of receiving signals in means to provide output in velocity from the target

<span>· </span>It may be use as radar gun in which police use to detect speeding.

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A total charge of 4.70 is distributed on two metal spheres. When the spheres are 10 cm apart, they each feel a repulsive force o

Anna35 [415]

Answer:0.114 C

Explanation:

Given

Total 4.7 C is distributed in two spheres

Let $q_1$ and $q_2$ be the charges such that

$q_1+q_2=4.7$

and Force between charge particles is given by

$F=\frac{kq_1q_2}{r^2}$

$4.7\times 10^11=\frac{9\times 10^9\times q_1\cdot q_2}{0.1^2}$

$q_1\cdot q_2=0.522$

put the value of $q_1$

$q_2\left ( 4.7-q_2\right )=0.522$

$q_2^2-4.7q_2+0.522=0$

$q_2=\frac{4.7\pm \sqrt{4.7^2-4\times 1\times 0.522}}{2}$

$q_2=0.114 C$

thus $q_1=4.586 C$

3 0

3 years ago

Microwave ovens emit microwave energy with a wavelength of 12.2 cm. What is the energy of exactly one photon of this microwave r

Iteru [2.4K]

Answer:

$1.63\cdot 10^{-24} J$

Explanation:

The energy of a photon is given by:

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda$ is the wavelength of the photon

In this problem, we have a microwave photon with wavelength

$\lambda=12.2 cm=0.122 m$

Substituting into the equation, we find its energy:

$E=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{0.122 m}=1.63\cdot 10^{-24} J$

5 0

3 years ago

an object with a mass of 2.4 kg has a force of 12.6 N applied to it. What is the resulting acceleration of the object? WITH PROO

Scorpion4ik [409]

Newton’s Law: F = MA
A = F/M (change equation)
12.6 N/ 2.4 kg = 5.25
Answer: acceleration is 5.25 m/s^2

3 0

3 years ago

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p

igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

$B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

$B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

$B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

$F = qvB$

$F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

$F = 21.3 \times 10^{-16}$ N

From the newton's second law

$F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

$a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

7 0

2 years ago

13. You push with 56 N on a 15-kg box, and there is a 23-N force of friction. How fast will the box accelerate?

Flura [38]

Answer:

The acceleration is 2.2 m/s^2

Explanation:

In the attached image, we can see the free body diagram. And using the second law of Newton it will be possible to find the acceleration of the box.

4 0

3 years ago