Together, two students exert a force of 825 N in pushing a car a distance of 35 m.

Uest 1. State Newton's law of cooling.

garik1379 [7]

Answer:

Newton's law of cooling states that the rate of heat loss of a body is directly proportional to the difference in the temperatures between the body and its surroundings. The law is frequently qualified to include the condition that the temperature difference is small and the nature of heat transfer mechanism remains the same. As such, it is equivalent to a statement that the heat transfer coefficient, which mediates between heat losses and temperature differences, is a constant. This condition is generally met in heat conduction (where it is guaranteed by Fourier's law) as the thermal conductivity of most materials is only weakly dependent on temperature. In convective heat transfer, Newton's Law is followed for forced air or pumped fluid cooling, where the properties of the fluid do not vary strongly with temperature, but it is only approximately true for buoyancy-driven convection, where the velocity of the flow increases with temperature difference. Finally, in the case of heat transfer by thermal radiation, Newton's law of cooling holds only for very small temperature differences.

When stated in terms of temperature differences, Newton's law (with several further simplifying assumptions, such as a low Biot number and a temperature-independent heat capacity) results in a simple differential equation expressing temperature-difference as a function of time. The solution to that equation describes an exponential decrease of temperature-difference over time. This characteristic decay of the temperature-difference is also associated with Newton's law of cooling

6 0

3 years ago

Mercury’s natural state is where the atoms are close to each other but are still free to pass by each other in witch states coul

Stels [109]

Mercury's natural state is where the atoms are close to each other but are still free to pass by each other. In which state(s) could mercury naturally exist?

Liquid is the answer

5 0

3 years ago

Read 2 more answers

2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate

adoni [48]

Answer:

 » Image distance :

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

v is image distance
u is object distance, u is 10 cm
f is focal length, f is 5 cm

${ \tt{ \frac{1}{v} + \frac{1}{10} = \frac{1}{5} }} \\ \\ { \tt{ \frac{1}{v} = \frac{1}{10} }} \\ \\ { \tt{v = 10}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \: \: }}}}}$

 » Magnification :

• Let's derive this formula from the lens formula:

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

» Multiply throughout by fv

${ \tt{fv( \frac{1}{v} + \frac{1}{u} ) = fv( \frac{1}{f} )}} \\ \\ { \tt{ \frac{fv}{v} + \frac{fv}{u} = \frac{fv}{f} }} \\ \\ { \tt{f + f( \frac{v}{u} ) = v}}$

• But we know that, v/u is M

${ \tt{f + fM = v}} \\ { \tt{f(1 +M) = v }} \\ { \tt{1 +M = \frac{v}{f} }} \\ \\ { \boxed{ \mathfrak{formular : } \: { \tt{ M = \frac{v}{f} - 1 }}}}$

v is image distance, v is 10 cm
f is focal length, f is 5 cm
M is magnification.

${ \tt{M = \frac{10}{5} - 1 }} \\ \\ { \tt{M = 5 - 1}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}$

 » Nature of Image :

Image is magnified
Image is erect or upright
Image is inverted
Image distance is identical to object distance.

4 0

2 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

4 years ago

Calculate two stars appear to you to have the same brightness. if star a is 7.00 light-years away from you, while star b is 15.0

vagabundo [1.1K]

It is eight times more than the star A.

<h3>What is luminosity and on which it depends?</h3>

The luminosity of an object is a measure of its intrinsic brightness and is defined as the amount of energy the object emits in a fixed time.

luminousity depends upon the two factors are:

1) The star's actual brightness

Some stars are naturally more luminous than others ,so the brightness level from one star to next star is significantly different.

2) The star distance from us

The more distance of an object the dimmer it appears.

Energy emitted = sAT⁴

where s is stefan constant

A is surface area and T is temperature .

to learn more about Luminosity click here brainly.com/question/14140223

#SPJ4

7 0

1 year ago