Radar use reflected microwaves to detect objects and measure their... what?

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t

givi [52]

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

$f = f_0 (\frac{v_0}{v_0-v})$

Here,

$f_0$ = Frequency of Source

$v_s$ = Speed of sound

f = Frequency heard before slowing down

f' = Frequency heard after slowing down

v = Speed of the train before slowing down

So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

$f = f_0 (\frac{v_0}{v_0-v})$

$300 = f_0 (\frac{343}{343-v})$

$(300*343) - 300v = 343f_0$

Now the second expression will be,

$f' = f_0 (\frac{v_0}{v_0-v/2})$

$290 = (343)(\frac{v_0}{343-v/2})$

$290*343-145v = 343f_0$

Dividing the two expression we have,

$\frac{(300*343) - 300v}{290*343-145v} = 1$

Solving for v, we have,

$v = 22.12m/s$

Therefore the speed of the train before and after slowing down is 22.12m/s

6 0

3 years ago

A piece of glass has a temperature of 72.0 degrees Celsius. The specific heat capacity of the glass is 840 J/kg/deg C. A liquid

Nezavi [6.7K]

Answer:

741 J/kg°C

Explanation:

Given that

Initial temperature of glass, T(g) = 72° C

Specific heat capacity of glass, c(g) = 840 J/kg°C

Temperature of liquid, T(l)= 40° C

Final temperature, T(2) = 57° C

Specific heat capacity of the liquid, c(l) = ?

Using the relation

Heat gained by the liquid = Heat lost by the glass

m(l).C(l).ΔT(l) = m(g).C(g).ΔT(g)

Since their mass are the same, then

C(l)ΔT(l) = C(g)ΔT(g)

C(l) = C(g)ΔT(g) / ΔT(l)

C(l) = 840 * (72 - 57) / (57 - 40)

C(l) = 12600 / 17

C(l) = 741 J/kg°C

5 0

3 years ago

A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.

34kurt

Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated ( $Q$ ), measured in joules, by the cylindrical resistor is the power multiplied by operation time ( $\Delta t$ ), measured in hours. That is:

$Q = \dot Q \cdot \Delta t$ (1)

If we know that $\dot Q = 1.2\,W$ and $\Delta t = 86400\,s$ , then the amount of heat dissipated by the resistor is:

$Q = (1.2\,W)\cdot (86400\,s)$

$Q = 103680\,J$

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux ( $Q'$ ), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder ( $A$ ), measured in square meters:

$Q' = \frac{\dot Q}{A}$ (2)

$Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h }$ (3)

Where:

$D$ - Diameter, measured in meters.

$h$ - Length, measured in meters.

If we know that $\dot Q = 1.2\,W$ , $D = 4\times 10^{-3}\,m$ and $h = 2\times 10^{-2}\,m$ , the heat flux of the resistor is:

$Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }$

$Q' \approx 4340.589\,\frac{W}{m^{2}}$

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces ( $r$ ), no unit, is the ratio of the top and bottom surfaces to total surface:

$r = \frac{\frac{\pi}{2}\cdot D^{2}}{A}$ (3)

If we know that $A \approx 2.765\times 10^{-4}\,m^{2}$ and $D = 4\times 10^{-3}\,m$ , then the fraction is:

$r = \frac{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2} }{2.765\times 10^{-4}\,m^{2}}$

$r = 0.045$

The fraction of heat dissipated from the top and bottom surfaces is 0.045.

7 0

3 years ago

An ant crawls 50 cm to the right, stops, and then crawls 30 cm to the left. Calculate the distance that the ant travels, and cal

borishaifa [10]

Answer and Explanation:

If the ant was to crawl 50cm to the right, then come back 30 cm, then the total distance walked would be <u>80cm</u>.

- Combine 50cm and 30cm to get 80 cm.

For displacement, the answer is <u>20 cm.</u>

- When calculating displacement, you use the initial (starting) distance. and subtract that from the final distance, giving you the displacement, or the amount traveled from the starting point to the final point if you were to make a straight line from the starting point to end point. (0 to 50, then back 30 the same direction, so subtract 30 from 50 to get 20)

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

<em><u>I hope this helps!</u></em>

8 0

2 years ago

Read 2 more answers

A 100 N force is applied to a 50 kg crate resting on a level floor. The coefficient of kinetic friction is 0.15. What is the acc

Nookie1986 [14]

The acceleration of the crate after it begins to move is 0.5 m/s²

We'll begin by calculating the the frictional force

Mass (m) = 50 Kg

Coefficient of kinetic friction (μ) = 0.15

Acceleration due to gravity (g) = 10 m/s²

Normal reaction (N) = mg = 50 × 10 = 500 N

<h3>Frictional force (Fբ) =?</h3>

Fբ = μN

Fբ = 0.15 × 500

Next, we shall determine the net force acting on the crate

Frictional force (Fբ) = 75 N

Force (F) = 100 N

<h3>Net force (Fₙ) =?</h3>

Fₙ = F – Fբ

Fₙ = 100 – 75

Finally, we shall determine the acceleration of the crate

Mass (m) = 50 Kg

Net force (Fₙ) = 25 N

<h3>Acceleration (a) =?</h3>

a = Fₙ / m

a = 25 / 50

Therefore, the acceleration of the crate is 0.5 m/s²

Learn more on friction: brainly.com/question/364384

8 0

2 years ago