The diagram shows a heating curve for a substance. Which segment shows the substance changing from a gas to a liquid?

solmaris [256]

Q9. R₁ only.

Since the upper branch is cut at x, the circuit in this branch is not closed. No current will flow through this branch. In this case, the circuit becomes a series circuit and current only flows through R₁.

Q10. 3 pathways.

3 different resistors are connected in parallel, which means there will be three branches of parallel circuit, each with one resistor. Current can flow through 3 different branches, until the parallel circuits combine together again.

4 0

3 years ago

What form of light uses Doppler radar use?

jeka94

Answer:

Explanation:

Doppler radar works by sending <u>a beam of electromagnetic radiation waves, </u>tuned to a precise frequency, at a moving object. You can use Doppler radar on a stationary object, of course, but it's fairly uninteresting unless the target is moving

4 0

3 years ago

Air at 30°C and 2MPa flows at steady state in a horizontal pipeline with a velocity of 25 m/s. It passes through a throttle valv

erastovalidia [21]

Answer:

$V_2=159.9\ m/s$

$T_2=290.6K$

Explanation:

At initial condition

P=2 MPa

T=30°C

V=25 m/s

At final condition

P=0.3 MPa

Now from first law for open system

$h_1+\dfrac{V_1^2}{2000}=h_2+\dfrac{V_2^2}{2000}$

We know that for air

h= 1.010 x T KJ/kg

$1.01\times 303+\dfrac{25^2}{2000}=1.01\times T+\dfrac{V_2^2}{2000}$ -----1

Now from mass balance

$\rho_1A_1V_1=\rho_2A_2V_2$

We also know that

$\rho=\dfrac{P}{RT}$

$\dfrac{P_1}{RT_1}V_1=\dfrac{P_2}{RT_2}V_2$

$\dfrac{2}{R\times 303}\times 25=\dfrac{0.3}{RT_2}V_2$

$T_2=1.81V_2$ ----------2

Now from equation 1 and 2

$V_2^2+3673.749V-612916.49=0$

So we can say that

$V_2=159.9\ m/s$

This is the outlet velocity.

Now by putting the values in equation 2

$T_2=1.81\times 159.9$

$T_2=290.6K$

This is the outlet temperature.

4 0

4 years ago

In a supermarket, you place a 22.3-N (around 5 lb) bag of oranges on a scale, and the scale starts to oscillate at 2.7 Hz. What

allsm [11]

Answer:

Force constant, k = 653.3 N/m

Explanation:

It is given that,

Weight of the bag of oranges on a scale, W = 22.3 N

Let m is the mass of the bag of oranges,

$m=\dfrac{W}{g}$

$m=\dfrac{22.3}{9.8}$

m = 2.27 kg

Frequency of the oscillation of the scale, f = 2.7 Hz

We need to find the force constant (spring constant) of the spring of the scale. We know that the formula of the frequency of oscillation of the spring is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$