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3 years ago

In the first law of Thermodynamics ΔE = Q - W, what does ΔE stand for???

2 answers:

7 0

<span>Δ</span>E = q + w

q = heat (quantity of)

q and w can be positive or negative depending on if work/heat is being absorbed/done on the system or released/done by the system

Gnesinka [82]3 years ago

3 0

It is D. hope this helped!!!!

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What is massand its si unit

natulia [17]

A large body of matter with no definite shape.
The SI unit is-kilogram(kg).

6 0

3 years ago

Circle the larger unit:

DerKrebs [107]

1. Centimeter
2. Kilogram
3. Millisecond
4. DL
5. Kg
6. Mm
7. S
8. Mm
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4 0

3 years ago

Can someone explain how they got their answer or how I get the change in number? :(

enyata [817]

Answer:

See Below

Explanation:

Okay, I thinkkk what it is asking by what you summarzied for me issss:

They split the total time into four quarters. They then took (for the first quarter) the start time. Then when the first quarter ends and the second quarter starts is the "end" time.

They then subtract the start time of the second quarter from the end time of the first quarter.

I hope this helps, good luck! :D

5 0

2 years ago

A car travels initially at 24 m/s, until it enters the highway. If the car accelerates at 4 m/s^2 for a 96 meters, what is the c

marishachu [46]

initial velocity=u=24m/s
Acceleration=a=4m/s^2
Distance=s=96m
Final velocity=v

Using 3rd equation of kinematics

$\boxed{\Large{\sf v^2-u^2=2as}}$

$\\ \Large\sf\longmapsto v^2=u^2+2as$

$\\ \Large\sf\longmapsto v^2=24^2+2(4)(96)$

$\\ \Large\sf\longmapsto v^2=576+768$

$\\ \Large\sf\longmapsto v^2=1344$

$\\ \Large\sf\longmapsto v=\sqrt{1344}$

$\\ \Large\sf\longmapsto v=36.6m/s$

3 0

3 years ago

calculate earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the s

IceJOKER [234]

Answer:

Hello your question is incomplete below is the complete question

Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000

answer : V = 1.624* 10^-5 m/s

Explanation:

First we have to calculate the value of a

a = 93 * 10^6 mile/m * 1609.344 m

= 149.668 * 10^8 m

next we will express the distance between the earth and the sun

$r = \frac{a(1-E^2)}{1+Ecos\beta }$ --------- (1)

a = 149.668 * 10^8

E (eccentricity ) = ( 1/60 )^2

$\beta$ = 90°

input the given values into equation 1 above

r = 149.626 * 10^9 m

next calculate the Earths velocity of approach towards the sun using this equation

$v^2 = \frac{4\pi^2 }{r_{c} }$ ------ (2)

Note :

Rc = 149.626 * 10^9 m

equation 2 becomes

( $V^2 = (\frac{4\pi^{2} }{149.626*10^9})$

therefore : V = 1.624* 10^-5 m/s

4 0

3 years ago