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3 years ago

Which of these is a unit of heat? a. joule b. degree celsius c. kelvin d. tesla

1 answer:

3 0

-- Heat is a form of energy.
-- Joule is the SI unit of energy.
ergo
-- Joule is a unit of heat.

'Degree Celsius' and 'Kelvin' are units of temperature.
Heat and temperature are different things.
We won't go there right now.

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0.5-lbm of a saturated vapor is converted to asaturated liquid by being cooled in a weighted piston-cylinder device maintained a

klio [65]

Answer:

The boiling point temperature of this substance when its pressure is 60 psia is 480.275 R

Explanation:

Given the data in the question;

Using the Clapeyron equation

$(\frac{dP}{dT} )_{sat } = \frac{h_{fg}}{Tv_{fg}}$

$(\frac{dP}{dT} )_{sat } = \frac{\frac{H_{fg}}{m} }{T\frac{V_{fg}}{m} }$

where $h_{fg$ is the change in enthalpy of saturated vapor to saturated liquid ( 250 Btu

T is the temperature ( 15 + 460 )R

m is the mass of water ( 0.5 Ibm )

$V_{fg$ is specific volume ( 1.5 ft³ )

we substitute

$(\frac{dP}{dT} )_{sat } =( \frac{250Btu\frac{778Ibf-ft}{Btu} }{0.5})$ / $( (15+460)\frac{1.5}{0.5})$

$(\frac{dP}{dT} )_{sat } =$ 272.98 Ibf-ft²/R

Now,

$(\frac{dP}{dT} )_{sat } =$ $(\frac{P_2 - P_1}{T_2 - T_1})_{sat$

where P₁ is the initial pressure ( 50 psia )

P₂ is the final pressure ( 60 psia )

T₁ is the initial temperature ( 15 + 460 )R

T₂ is the final temperature = ?

we substitute;

$T_2$ $= ( 15 + 460 ) + \frac{(60-50)psia(\frac{144in^2}{ft^2}) }{272.98}$

$T_2 = 475 + 5.2751\\$

$T_2 =$ 480.275 R

Therefore, boiling point temperature of this substance when its pressure is 60 psia is 480.275 R

3 0

3 years ago

A boat is traveling at an initial velocity of 2.7 meters per second in the positive direction. It accelerates at a rate of 0.15

cupoosta [38]

Answer:

$\boxed {\boxed {\sf 4.5 \ m/s \ in \ the \ positive \ direction}}$

Explanation:

We are asked to find the final velocity of the boat.

We are given the initial velocity, acceleration, and time. Therefore, we will use the following kinematic equation.

$v_f= v_i + at$

The initial velocity is 2.7 meters per second. The acceleration is 0.15 meters per second squared. The time is 12 seconds.

$v_i$ = 2.7 m/s
a= 0.15 m/s²
t= 12 s

Substitute the values into the formula.

$v_f = 2.7 \ m/s + (0.15 \ m/s^2)(12 \ s)$

Multiply the numbers in parentheses.

$v_f= 2.7 \ m/s + (0.15 \ m/s/s * 12 \ s)$

$v_f = 2.7 \ m/s + (0.15 \ m/s *12)$

$\v_f=2.7 \ m/s + (1.8 \ m/s)$ $v_f=2.7 \ m/s + (1.8 \ m/s)$

Add.

$v_f=4.5 \ m/s$

The final velocity of the boat is <u>4.5 meters per second in the positive direction.</u>

5 0

2 years ago

Two cars are moving towards each other and sound emitted by first car with real frequency of 3000 hertz is detected by a person

sertanlavr [38]

Answer:

v ’= 21.44 m / s

Explanation:

This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s

f ’= f (v + v₀) / (v- $v_{s}$ )

where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer

in this exercise both the source and the observer are moving, we will assume that both have the same speed,

v₀ = v_{s} = v ’

we substitute

f ’= f (v + v’) / (v - v ’)

f ’/ f (v-v’) = v + v ’

v (f ’/ f -1) = v’ (1 + f ’/ f)

v ’= (f’ / f-1) / (1 + f ’/ f) v

v ’= (f’-f) / (f + f’) v

let's calculate

v ’= (3400 -3000) / (3000 +3400) 343

v ’= 400/6400 343

v ’= 21.44 m / s

3 0

3 years ago

What is acceleration?

Kisachek [45]

Acceleration is the force of something moving, or acceleration can be how fast an object is going!

7 0

3 years ago

Read 2 more answers

Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 9.0 km/h due east. Runner B is initia

sergejj [24]

Answer:

0.176m from the flagpole, westward.

Explanation:

Let the Eastward be the positive direction. So initially runner A is at position -6km, running with velocity of 9km/h while runner B is at position 5km running at a velocity of -8km/h. We can conduct the following equation for their distances over the same time t

$s_A = -6 + 9t$