1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
borishaifa [10]
3 years ago
7

6. A water brake copupled to an engine on test absorbs 70 kW of power. Find the heat generated at the brake per minute and the m

ass flow of fresh water through the brake , in kg/min if the temperature increase of the water is 100 C .Assume all the heat generated is carried away by the cooling water
Physics
1 answer:
PtichkaEL [24]3 years ago
7 0

Answer:

4200 kJ

10 kg/min

Explanation:

70 kW means 70 kJ per second.  So in a minute, the amount of heat generated at the brake is:

70 kJ/s × 60 s/min = 4200 kJ

Heat = mass × specific heat capacity × increase in temperature

4200 kJ = m × 4.184 kJ/kg/C × 100 C

m = 10 kg

So every minute, 10 kg of water is heated.  So the mass flow is 10 kg/min.

You might be interested in
Determine the weight of an average physical science textbook whose mass is 3.1 kilograms. The acceleration due to gravity is 9.8
zvonat [6]
Weight equals mass*gravity
W = mg

Given m = 3.1 kg, g = 9.8 m/s^2

W = (3.1)(9.8)
W = 30.38
6 0
4 years ago
Exoplanets (planets outside our solar system) are an active area of modern research. Suppose astronomers find such a planet that
Leno4ka [110]

Answer:

8.829 m/s²

Explanation:

M = Mass of Earth

m = Mass of Exoplanet

g_e = Acceleration due to gravity on Earth = 9.81 m/s²

g = Acceleration due to gravity on Exoplanet

m=M-0.1M\\\Rightarrow m=0.9M

g_e=G\frac{M}{r^2}

g=G\frac{0.9M}{r^2}

Dividing the equations we get

\frac{g}{g_e}=\frac{G\frac{0.9M}{r^2}}{G\frac{M}{r^2}}\\\Rightarrow \frac{g}{g_e}=0.9\\\Rightarrow g=0.9g_e\\\Rightarrow g=0.9\times 9.81\\\Rightarrow g=8.829\ m/s^2

Acceleration due to gravity on the surface of the Exoplanet is 8.829 m/s²

3 0
3 years ago
To aid in the prevention of tooth decay, it is recommended that drinking water contain 0.800 ppm fluoride, F−. How many grams of
Leviafan [203]

Answer:

2.23 × 10^6 g of F- must be added to the cylindrical reservoir in order to obtain a drinking water with a concentration of 0.8ppm of F-

Explanation:

Here are the steps of how to arrive at the answer:

The volume of a cylinder = ((pi)D²/4) × H

Where D = diameter of the cylindrical reservoir = 2.02 × 10^2m

H = Height of the reservoir = 87.32m

Therefore volume of cylindrical reservoir = (3.142×202²/4)m² × 87.32m = 2798740.647m³

1ppm = 1g/m³

0.8ppm = 0.8 × 1g/m³

= 0.8g/m³

Therefore to obtain drinking water of concentration 0.8g/m³ in a reservoir of volume 2798740.647m³, F- of mass = 0.8g/m³ × 2798740.647m³ = 2.23 × 10^6 g must be added to the tank.

Thank you for reading.

4 0
3 years ago
A hammer of mass M is moving at speed v0 when it strikes a nail of negligible mass that is stuck in a wooden block. The hammer i
OleMash [197]

Answer:

i think it would be D

8 0
3 years ago
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
3 years ago
Other questions:
  • How can a small human retina detect objects larger than itself?
    12·2 answers
  • A series circuit has an impedance of 62.0 Ω and a power factor of 0.720 at a frequency of 47.0 Hz . The source voltage lags the
    7·1 answer
  • When the volcano Krakatoa erupted in 1883, it was heard 5000 km away. Which statement about the sound from the volcano is not co
    10·1 answer
  • The following angles are given in degrees and
    11·1 answer
  • In a nuclear reactor 1 g of mass is converted into energy. How much energy in Joules is produced?
    8·1 answer
  • 0.2 points
    8·1 answer
  • Sound cannot travel through a vacuum. Explain why.
    7·1 answer
  • When the sun and moon pull at right angles to the earth wat kinda tide can yu expect
    12·1 answer
  • Si dos aparatos eléctricos que tienen diferente potencia funcionan durante el mismo tiempo, ¿ cual transformara mayor cantidad d
    10·1 answer
  • 22 Newton force is working on a 1,901 gram object. What is the acceleration in<br> meter/s^2 unit
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!