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taurus [48]
3 years ago
13

You and a highway patrolman are driving at constant speeds in opposite directions on a straight highway. The patrolman is drivin

g at 60 mph and his radar gun determines your relative speed (the magnitude of the difference between your velocities) to be 135 mph. What is your speed at the time of measurement?
Physics
1 answer:
kompoz [17]3 years ago
3 0

Answer:  75 mph

Explanation:

The Relative Speed for a mobile is equal to the diference between the object and the observer:

Relative Speed (Rs) = Object's Velocity  - Observer's Velocity

Thinking on those terms, we would need to have a universal observer to do any understandable measurement on daily basics. This is why we all use earth as a Static Observer for every measurement we do everyday.

Using Earth as an observer, the Velocity for the Patrolman is:

Patrolman Velocity (Vp) = 60 mph

Because the radar gun does measure the Relative Speed for the object, which is 135 mph, we need to work with the equation to find the Velocity using Earth as a reference.

Object's Relative Velocity = Object's Velocity - Patrolman's Velocity

Object's Velocity = Object's Relative Velocity + Patrolman's Velocity

We need to keep in mind, the Patrolman is going on the opposite direction. Because of this the sign for his velocity should be negative.

Object's Velocity = 135 mph + ( -60 mph)

Object's Velocity = 135 mph - 60 mph

Object's Velocity = 75 mph

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Electromagnetic waves can travel through empty space true or false
Mariana [72]
False. Heat radiation from the sun cannot reach Earth, we cannot receive TV or GPS signals from satellites, and we cannot detect the light from distant stars.
Oh, wait . . .
6 0
3 years ago
. Draw a Cartesian coordinate system on a sheet of paper. On this Cartesian coordinate system, draw each vector to scale, starti
Triss [41]

Answer:

a) the first vector has magnitude 58 cm and the angle is 15 measured clockwise from the positive side of the x-axis

b) the second vector, the magnitude is 55.7 cm and the angle is 35 half from the negative side of the x-axis in a clockwise direction

c) the magnitude is 54.2 cm with an angle of 18 measured counterclockwise from the x-axis

Explanation:

For this exercise we draw a Cartesian coordinate system in this system: East coincides with the positive part of the x-axis and North with the positive part of the y-axis.

a) the first vector has magnitude 58 cm and the angle is 15 measured clockwise from the positive side of the x-axis

b) the second vector, the magnitude is 55.7 cm and the angle is 35 half from the negative side of the x-axis in a clockwise direction

c) the magnitude is 54.2 cm with an angle of 18 measured counterclockwise from the x-axis

In the attachment we can see the representation of the three vectors

8 0
3 years ago
A stationary 500 kg tank fires a 20 kg miegile at 100 m/s. What is the velocity of the tank after the missile is fired? Assume t
dedylja [7]

Answer:

v₁ = 4 [m/s].

Explanation:

This problem can be solved by using the principle of conservation of linear momentum. Where momentum is preserved before and after the missile is fired.

P=m*v

where:

P = linear momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

(m_{1}*v_{1})=(m_{2}*v_{2})

where:

m₁ = mass of the tank = 500 [kg]

v₁ = velocity of the tank after firing the missile [m/s]

m₂ = mass of the missile = 20 [kg]

v₂ = velocity of the missile after firing = 100 [m/s]

(500*v_{1})=(20*100)\\v_{1}=2000/500\\v_{1}=4[m/s]

8 0
3 years ago
The Earth's magnetic field is approximately 0.5 Gauss. What is the magnetic energy density of this field
ser-zykov [4K]

Answer:

The  magnetic energy density  is  E_B  =  9.95 *10^{-4} \  J/m^3

Explanation:

From the question we are told that  

    The  earths magnetic field is  B  =  0.5 \  Gauss

Now

          1 \ Gauss  \to  1 *10^{-4} \ T

=>        0.5  \ Gauss  \to  x T

So  

          x =  \frac{0.5 *  1*10^{-4}}{1}

          x = B= 5.0 *10^{-5} \  T

Generally the magnetic energy density is  mathematically represented as

        E_B  =  \frac{B^2}{ 2 *  \mu_o  }

  Here \mu_o is the  permeability of  free space with a constant value

          \mu_o  =  4\pi * 10^{-7} N/A^2

substituting values for equation above  

       E_B  =  \frac{(5*10^{-5})^2}{ 2 *   4\pi * 10^{-7}  }

       E_B  =  9.95 *10^{-4} \  J/m^3

5 0
3 years ago
A stretched string has a mass per unit length of 5.12 g/cm and a tension of 19.3 N. A sinusoidal wave on this string has an ampl
Sati [7]

Answer:

a. 0.143 mm b. 77.6 rad/m c. 483.18 rad/s  d. +1

Explanation:

a. ym

Since the amplitude is 0.143 mm, ym = amplitude = 0.143 mm

b. k

We know k = wave number = 2π/λ where λ = wavelength.

Also, λ = v/f where v = speed of wave in string = √(T/μ) where T = tension in string = 19.3 N and μ = mass per unit length = 5.12 g/cm = 5.12 ÷ 1000 kg/(1 ÷ 100 m) = 0.512 kg/m and f = frequency = 76.9 Hz.

So, λ = v/f = √(T/μ)/f

substituting the values of the variables into the equation, we have

λ = √(T/μ)/f

= √(19.6 N/0.512 kg/m)/76.9 Hz

= √(38.28 Nkg/m)/76.9 Hz

= 6.187 m/s ÷ 76.9 Hz

= 0.081 m

= 81 mm

So, k = 2π/λ

= 2π/0.081 m

= 77.6 rad/m

c. ω

ω = angular frequency = 2πf where f = frequency of wave = 76.9 Hz

So, ω = 2πf

= 2π × 76.9 Hz

= 483.18 rad/s

d. The correct choice of sign in front of ω?

Since the wave is travelling in the negative x - direction, the sign in front of ω is positive. That is +1.

4 0
3 years ago
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