Hello Averyone Subs Chanel Me please "RezaDarmawangsa "<br>Thanks !!

!!!!!!!!!!!!!!PLEASE ANSWER THIS!!!!!!!!!!!!!!!!

dmitriy555 [2]

Answer:

age = int(input())

ticket = 100

total = 0

passengers = 1

while passengers <= 5:

if age < 3:

continue

else:

total += ticket

passengers += 1

print(total)

Explanation:

UuU

6 0

2 years ago

Ammonia enters the expansion valve of a refrigeration system at a pressure of 10 bar and a temperature of 20oC and exits at 3.0

icang [17]

Answer:

Ammonia enters the expansion valve of a refrigeration system at a pressure of 10 bar and a temperature of 20 o C and exits at 5.0 bar. The refrigerant undergoes a throttling process. Determine the temperature, in o C, and the quality of the refrigerant at the exit of the expansion valve.

Explanation:

3 0

3 years ago

You are preparing to work with Chemical A. You open the appropriate storage cabinet, and notice Chemical B, as well as Chemical

PtichkaEL [24]

The correct answer is; Stability and reactivity.

Further Explanation:

The stability and reactivity section of the SDS sheets is where to check for the possibility of hazardous reactions for the chemicals. This also lists the chemical stability of each chemical that people may be using. This can be found in section 10 of the OSHA Quick Card.

The SDS sheets has 16 sections for employees to use. Since 2015, the sections can be found in uniform format for easier and faster ways to find the section needed. The 16 sections for the SDS sheets are:

Identification
Hazard(s) identification
Composition/information on ingredients
First-aid measures
First-aid measures
First-aid measures
Handling and storage
Exposure controls/personal protection
Physical and chemical properties
Stability and reactivity
Toxicological information
Ecological information
Disposal considerations
Transport information
Regulatory information
Other information

Learn more about SDS sheets at brainly.com/question/9753408

#LearnwithBrainly

5 0

4 years ago

An electronic toy is powered by three 1.58-V alkaline cells, each with an internal resistance of 0.0205 Ω, and a 1.53-V carbon-z

mina [271]

Answer:

(a) The current in amperes that flows through the toy's circuit is 0.923A

(b) The power supplied to the toy is 5.78721W

(c) The internal resistance r2 of the failed dry cell is 72Ω

Explanation:

From the circuit diagram attached. We have the electric component:

B1 = 3* 1.58 = 4.74V

B2 = 1.53V

r1 = 3*0.0205 = 0.0615Ω

r2 = 0.105Ω

R = 6.625Ω

Since the internal resistances and the resistor R are connected in series, we can calculate the total resistance RT as

RT = r1 + r2 +R = 0.0615 + 0.105 + 6.625

= 6.7915Ω

Total Voltage supplied to the circuit by both batteries V = B1 + B2 = 4.74 + 1.53 = 6.27V

(a) CIRCUIT CURRENT

The current I, flowing through the circuit is i = $\frac{V}{R_{T}}$ = $\frac{6.27}{6.7915}$ =0.923A

The current in amperes that flows through the toy's circuit is 0.923A

(b) THE POWER SUPPLIED TO THE TOY

Power P = I*V =0.923*6.27 = 5.78721W

The power supplied to the toy is 5.78721W

(c) THE VALUE OF r2

Due to dry cell failure, the power supplied to the toy is reduced to 0.5W

Now Power P = $\frac{V^{2} }{R}$ . To calculate the new total resistance of the circuit we will make R the subject of the formular

R= $\frac{V^{2} }{P}$ = $\frac{6.27^{2} }{0.5}$ = $\frac{39.3129}{0.5}$ = 78.6258Ω

Remember that RT = r1 + r2 + R

r1 =RT- (R +r2)

r1 = 78.6258 - 6.6865 =71.9393Ω

The internal resistance r2 of the failed dry cell is 72Ω

6 0

3 years ago

Read 2 more answers

Q4. (20 points) For a bronze alloy, the stress at which plastic deformation begins is 271 MPa and the modulus of elasticity is 1

babunello [35]

Answer:

a) P = 86720 N

b) L = 131.2983 mm

Explanation:

σ = 271 MPa = 271*10⁶ Pa

E = 119 GPa = 119*10⁹ Pa

A = 320 mm² = (320 mm²)(1 m² / 10⁶ mm²) = 3.2*10⁻⁴ m²

a) P = ?

We can apply the equation

σ = P / A ⇒ P = σ*A = (271*10⁶ Pa)(3.2*10⁻⁴ m²) = 86720 N

b) L₀ = 131 mm = 0.131 m

We can get ΔL applying the following formula (Hooke's Law):

ΔL = (P*L₀) / (A*E) ⇒ ΔL = (86720 N*0.131 m) / (3.2*10⁻⁴ m²*119*10⁹ Pa)

⇒ ΔL = 2.9832*10⁻⁴ m = 0.2983 mm

Finally we obtain

L = L₀ + ΔL = 131 mm + 0.2983 mm = 131.2983 mm

3 0

4 years ago

Hello Averyone Subs Chanel Me please "RezaDarmawangsa "Thanks !!​

Hello Averyone Subs Chanel Me please "RezaDarmawangsa "Thanks !!