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satela [25.4K]
3 years ago
14

Can someone please help me with this question? ASAP thank you!

Physics
1 answer:
Assoli18 [71]3 years ago
8 0

did you get the answer

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On a hot summer day, 3.50 ✕ 106 J of heat transfer into a parked car takes place, increasing its temperature from 36.5°C to 44.4
anygoal [31]

Answer:

a) \Delta s=443037.9747\ J.K^{-1}

b) \Delta s=31868131.8681\ J.K^{-1}

Explanation:

a)

Given:

amount of heat transfer occurred,dQ=3.5\times 10^6\ J

initial temperature of car, T_i=36.5+273=309.5\ K

final temperature of car, T_f=44.4+273=317.4\ K

We know that the change in entropy is given by:

\Delta s=\frac{dQ}{T_f-T_i}

\Delta s=\frac{3.5\times 10^6}{(44.4-36.5)} (heat is transferred into the system of car)

\Delta s=443037.9747\ J.K^{-1}

b)

amount of heat transfer form the system of house, dQ=5.8\times 10^8\ J

initial temperature of house, T_i=23.5+273=296.5\ K

final temperature of house, T_f=5.3+273=278.3\ K

\Delta s=\frac{dQ}{T_f-T_i}

\Delta s=\frac{5.8\times 10^8}{278.3-296.5}

\Delta s=31868131.8681\ J.K^{-1}

6 0
3 years ago
A 27 kg bear slides, from rest, 14 m down a lodgepole pine tree, moving with a speed of 6.1 m/s just before hitting the ground.
Nadusha1986 [10]
<h2>Thus the force of friction is 235 N</h2>

Explanation:

When the bear was at the height of 14 m . Its potential energy = m g h

here m is the mass of bear , g is acceleration due to gravity and h is the height .

Thus P.E =  27 x 10 x 14 = 3780 J

The K.E of the bear just before hitting = \frac{1}{2} m v²

=   \frac{1}{2} x 27 x ( 6.1 )²  = 490 J

The force of friction f = P.E - K.E = 3290 J

Because the work done = Force x Distance

Thus frictional force = \frac{3290}{14} = 235 N

3 0
3 years ago
if you have corona and you fart does it spread the corona sense its airborn?? should i start putting a mask over my anus to stop
Taya2010 [7]

Answer:

its soul for that context not sole also rip and i have no idea

Explanation:

4 0
3 years ago
Read 2 more answers
Which statement about the total velocity of a projectile launched at an angle less than 90° above a flat surface is true?
marissa [1.9K]

The correct answer is B the total velocity is equal at both landing and launch because before your about launch you have 0 velocity then when you have landed you also have 0 velocity. Hope This Helps

3 0
3 years ago
Read 2 more answers
A girl is floating in a freshwater lake with her head just above the water. If she weighs 610 N, what is the volume of the subme
Elden [556K]

Answer:

The volume of the submerged part of her body is 0.0622m^{3}

Explanation:

Let's define the buoyant force acting on a submerged object.

In a submerged object acts a buoyant force which can be calculated as :

B=ρ.V.g

Where ''B'' is the buoyant force

Where ''ρ'' is the density of the fluid

Where ''V'' is the submerged volume of the object

Where ''g'' is the acceleration due to gravity

Because the girl is floating we can state that the weight of the girl is equal to the buoyant force.

We can write :

W_{girl}=B (I)

Where ''W'' is weight

⇒ If we consider ρ = 1000\frac{kg}{m^{3}} (water density) and g=9.81\frac{m}{s^{2}} and replacing this values in the equation (I) ⇒

B=W_{girl}

B=610N

ρ.V.g = 610N

1000\frac{kg}{m^{3}}.V.(9.81\frac{m}{s^{2}})=610N (II)

The force unit ''N'' (Newton) is defined as

N=kg.\frac{m}{s^{2}}

Using this in the equation (II) :

(9810\frac{N}{m^{3}}).V =610N

V=\frac{610N}{9810\frac{N}{m^{3}}}

V=0.0622m^{3}

We find that the volume of the submerged part of her body is 0.0622m^{3}

8 0
3 years ago
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