Jennifer and Jamie have a class assignment to identify an example of climate change and an example of a change in weather. They

[A] Write an expression for the equivalent resistance of three resistors connected in parallel.[ no derivation needed]

anzhelika [568]

Answer:

1) R1 + ((R2 × R3)/(R2 + R3))

2) 0.5 A

3) 3.6 V

Explanation:

1) We can see that resistors R2 and R3 are in parallel.

Formula for sum of parallel resistors; 1/Rt = 1/R2 + 1/R3

Making Rt the subject gives;

Rt = (R2 × R3)/(R2 + R3)

Now, Resistor R1 is in series with this sum of R2 and R3. Thus;

Total resistance of circuit = R1 + ((R2 × R3)/(R2 + R3))

2) R_total = R1 + ((R2 × R3)/(R2 + R3))

We are given;

R1 = 7.2 Ω

R2 = 8 Ω

R3 = 12 Ω

R_total = 7.2 + ((8 × 12)/(8 + 12))

R_total = 7.2 + 4.8

R_total = 12 Ω

Formula for current is;

I = V/R

I = 6/12

I = 0.5 A

3) since current through the circuit is 0.5 and R1 is 7.2 Ω.

Thus, potential difference through R1 is;

V = IR = 0.5 × 7.2 = 3.6 V

4 0

3 years ago

You are 1.8 m tall and stand 2.8 m from a plane mirror that extends vertically upward from the floor. on the floor 1 m in front

velikii [3]

This problem must be solved using a sketch. I attached an illustration of the problem.
You must trace the ray that reflects from the top off the table to your eyes. This how eyesight works, light rays reflects off the objects into your eyes.
Law of reflection tells us that light ray reflects off the surface at the same angle in which it falls on it( i attached another illustration of this).
Now we can write tangens equations:
$tan(\theta)=\frac{h-0.8}{1}\\
tan(\theta)=\frac{1.8-h}{2.8}\\
\frac{h-0.8}{1}=\frac{1.8-h}{2.8}$
We solve for h:
$\frac{h-0.8}{1}=\frac{1.8-h}{2.8}\\
2.8h-0.8\cdot 2.8=1.8-h\\
3.8h=1.8+2.24\\
h=\frac{4.04}{3.8}\\
h=1.06m$

6 0

3 years ago

A rock is thrown horizontally from a building at 15 m/s. It hits the ground 45 m from the base of the building. How high was the

evablogger [386]

Answer:

Explanation:

hope this helps!

7 0

3 years ago

Two satellites are in circular orbits around the earth. the orbit for satellite a is at a height of 542 km above the earth's sur

Evgen [1.6K]

Let R be radius of Earth with the amount of 6378 km h = height of satellite above Earth m = mass of satellite v = tangential velocity of satellite
Since gravitational force varies contrariwise with the square of the distance of separation, the value of g at altitude h will be 9.8*{[R/(R+h)]^2} = g'
So now gravity acceleration is g' and gravity is balanced by centripetal force mv^2/(R+h):
m*v^2/(R+h) = m*g' v = sqrt[g'*(R + h)]
Satellite A: h = 542 km so R+h = 6738 km = 6.920 e6 m g' = 9.8*(6378/6920)^2 = 8.32 m/sec^2 so v = sqrt(8.32*6.920e6) = 7587.79 m/s = 7.59 km/sec
Satellite B: h = 838 km so R+h = 7216 km = 7.216 e6 m g' = 9.8*(6378/7216)^2 = 8.66 m/sec^2 so v = sqrt(8.32*7.216e6) = 7748.36 m/s = 7.79 km/sec

6 0

3 years ago

A refrigerator is used to cool water from 23°C to 5°C in a continuous manner. The heat rejected in the condenser is 570 kJ/min a

GenaCL600 [577]

Answer: Q=5.46 L/s

COP=2.58

Explanation:

Given that

Cp = 4.18 kJ/(kg.C

density = 1 kg/L

Heat rejected Qr= 570 kJ/min

Power in put W= 2.65 KW

From first law of thermodynamics

U = W+ q

q = Heat absorbed

U = internal energy

W = workdone

U = 570 kJ/min = 9.5 KW

9.5 = 2.65 + q

q = 6.85 KW

COP = q/W

COP = 6.58 / 2.65

COP=2.58

Lets take volume flow rate is Q

So mass flow rate of water m = ρ Q

q = m Cp ΔT

6.85 = 1 x Q x 4.18 ( 23-5)

Q=0.091 L/min

Q=5.46 L/s

7 0

3 years ago