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2 years ago

7

Select all that apply. Greenhouse gases _____. absorb solar energy absorb carbon dioxide release carbon dioxide are released dur

ing the combustion of fossil fuels

1 answer:

dem82 [27]2 years ago

8 0

Gases that trap heat in atmosphere are called Greenhouse gases (GHGs). Examples are carbon dioxide, water vapour, nitrous oxides etc.

The selected choices are that Greenhouse gases-

1. absorb solar energy

2. are released during combustion of fossil fuels.

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A thin copper rod 1.0 m long has a mass of 0.050 kg and is in a magnetic field of 0.10 t. What minimum current in the rod is nee

slamgirl [31]

Answer:

$i = 4.9 A$

Explanation:

Force on a current carrying rod due to magnetic field is given as

$F = iLB$

here we know that

$i =$ current in the rod

$B = 0.10 T$

$L = 1.0 m$

now magnetic force is balanced by the weight of the rod

so we will have

$iLB = mg$

$i(1.0)(0.10) = 0.05 \times 9.8$

$i = 4.9 A$

8 0

2 years ago

A heat engine (Power Cycle) with a thermal efficiency of 35 percent efficiency produces 750 kJ of work. Heat transfer to the eng

frosja888 [35]

Answer:

a) The schematic illustrating is attached

b) The heat transfer to the heat engine is 2142.86 kJ, the heat transfer from the heat engine is 1392.86 kJ

c) The heat transfer to the heat engine is 1648.35 kJ, the heat transfer from the heat engine is 898.35 kJ

Explanation:

b) The heat transfer to the engine and the heat transfer from the engine to the air is:

$Q_{1} =\frac{W}{n}$

Where

W = 750 kJ

n = 35% = 0.25

Replacing:

$Q_{1} =\frac{750}{0.35} =2142.86kJ$

$Q_{2} =Q_{1} -W=2142.86-750=1392.86kJ$

c) The efficiency of Carnot engine is:

$n=1-\frac{300K}{550K} =0.455$

The heat transfer to the heat engine is:

$Q_{1c} =\frac{750}{0.455} =1648.35kJ$

The heat transfer from the heat engine is:

$Q_{2c} =1648.35-750=898.35kJ$

4 0

2 years ago

Two objects are placed in thermal contact and are allowed to come to equilibrium in isolation. the heat capacity of object a is

Harman [31]

Given:
Ca = 3Cb (1)
where
Ca = heat capacity of object A
Cb = heat capacity f object B

Also,
Ta = 2Tb (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.

Let
Tf = final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.

Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb) (3)

Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb

Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb

Answer:
$T_{f} =( \frac{1+6k}{1+3k} )T_{b}= \frac{1}{2}( \frac{1+6k}{1+3k})T_{a}$
where
$k= \frac{M_{a}}{M_{b}}$

7 0

2 years ago

Robin Hood wishes to split an arrow already in the bull's-eye of a target 40 m away.

tamaranim1 [39]

Answer:

5.843 m

Explanation:

suppose that the arrow leave the bow with a horizontal speed , towards he bull's eye.

lets consider that horizontal motion

distance = speed * time

time = 40/ 37 = 1.081 s

arrow doesnot have a initial vertical velocity component. but it has a vertical motion due to gravity , which may cause a miss of the target.

applying motion equation

(assume g = 10 m/s²)

$s=ut+\frac{1}{2}*gt^{2} \\= 0+\frac{1}{2}*10*1.081^{2}\\= 5.843 m$

Arrow misses the target by 5.843m ig the arrow us split horizontally

4 0

2 years ago

An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o

Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

$T=2\pi\sqrt{\dfrac{m}{k}}$

m is the mass of object

$m=\dfrac{W}{g}$

$m=\dfrac{2.45}{9.8}$

m = 0.25 kg

$k=\dfrac{4\pi^2m}{T^2}$

$k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}$

k = 24.09 N/m

or

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0

2 years ago