<u>Out of the given options, the following interactions are part of the greenhouse effect,
</u>
- Gases in the atmosphere absorb heat
- Earth’s surface radiates energy back into the atmosphere
- Gases in the atmosphere radiate heat back to the surface
Answers: Options A, D and E
<u>Explanation:
</u>
The greenhouse effect, basically a warming effect caused by the greenhouse gases such as Carbon-Di-oxide, Methane, nitrous oxides, water vapour etc. These gases usually trap the heat that Earth Absorbs by the Sun.
In the day time, the Earth absorbs the energy in the form of heat which is radiated by the Sun. In the evening, the process gets reversed and the Earth starts releasing that heat into the atmosphere.
Now, this heat gets absorbed by this gases before it leaves the Earth's atmosphere and gets trapped there only, resulting in the temperature raise of the Earth's environment.
So, the prime causes of the greenhouse effect remains as the heat radiation from the Sun, the absorption of that heat by the Earth surface and the further absorption of that heat produced by the greenhouse gases that present in the atmosphere.
Answer:
<h2>
<u>Joule</u><u>:</u></h2>
1 Joule of work is said to be done when a force of 1 Newton is applied to move/displace a body by 1 metre.
1 Joule= 1 Newton × 1 metre
1 Newton is the amount of force required to accelerate body of mass 1 kg by 1m/s²
So units of N is kgm/s²
So,
1 Joule
=1kgm/s² × m
=1kgm²/s²
<h2><u>Erg</u><u>:</u></h2>
1 erg is the amount of work done by a force of 1 dyne exerted for a distance of one centimetre.
1 Erg =1 Dyne × 1 cm
1 dyne is the force required to cause a mass of 1 gram to accelerate at a rate of 1cm/s².
1 Erg=1 gmcm/s² × cm
1 Erg=1 gmcm/s² × cm=1gmcm²/s²
this is what you need to convert 1gmcm²/s² to 1kgm²/s²
<h3><u>
what you need to know for conversion</u></h3>
[1gm=0.001kg
1cm²
=1cm ×1cm
=0.01 m × 0.01 m
=0.0001m²
second remains constant
]
So,
1gmcm²/s²
=0.001kg×0.0001m²/s²
=0.001kg×0.0001m²/s² =0.0000001kgm²/s²
Hence,
<h3>
<u>1 Erg</u><u>=</u><u>0.0000001</u><u> </u><u>Joule</u></h3><h3>
<u>1</u><u> </u><u>Joule</u><u>=</u><u>1</u><u>0</u><u>,</u><u>0</u><u>0</u><u>0</u><u>,</u><u>0</u><u>0</u><u>0</u><u> </u><u>Erg</u></h3>
<h2>⇒15 J=15×10000000 Erg</h2><h2> =150000000 Erg</h2><h2>
=1.5×10⁶ Erg</h2>
Answer:
pressure of the water = 3.3 × 
pa
Explanation:
given data
velocity v1 = 1.5 m/s
pressure P = 400,000 Pa
inside radius r1 = 1.00 cm
pipe radius r2 = 0.5 cm
h1 = 0 (datum at inlet)
h2 = 5.0 m (datum at inlet)
density of water ρ = 1000 kg/m³
to find out
pressure of the water
solution
we consider here flow speed in bathroom that is = v2 and Pressure in bathroom is = P2
here we will use both continuity and Bernoulli equations
because here we have more than one unknown so that
v1 × A1 = v2 × A2 × P1 + ρ g h1 + (0.5)ρ v1² = P2 + ρ g h2 + (0.5) ρ v2²
now we use here first continuity equation for get v2
v2 = 
v2 = 
v2 = 6 m/s
and now we use here bernoulli eqution for find here p2 that is
P2 = P1 - 0.5× ρ ×(v2² - v1²) - ρ g (h2- h1 )
P2 = 400000 - 0.5× 1000 ×(6² - 1.5²) - 1000 × 9.81 × (5-0 )
P2 = 3.3 × pa
Explanation:
We need convert the velocities first to m/s and we get the following:
v2 = 21 km/hr = 5.8 m/s
v1 = 11 km/hr = 3.1 m/s
We need to find the mass of the car also for later use do using the work-energy theorem:
6.0x10^3 J = (0.5) m [(5.8)^2 - (3.1)^2]
or
m = 499.4 kg
Now we determine work needed delta W to change its velocity from 21 km/hr to 33 km/hr
v2 = 33 km/hr = 9.2 m/s
v1 = 21 km/hr = 5.8 m/s
delta W = (0.5)(499.4)[(9.2)^2 - (5.8)^2]
= 1.3 x 10^4 J
Answer: the volume of the unknown mass
Explanation:
The other choices, molar masses of the reactants, the balanced chemical equation, and the molar masses of the product, are essential for the realization of stoichiometric calculations.
The balanced chemical equations tells the mole ratio of reactants (calcium carbonate) and products (calcium oxide and carbon dioxide).
The molar masses of reactants are used to convert between mass (grams) and moles.
The volume of the unknown mass is not needed when you are working only masses and moles since the relation is given by the molar masses. The volume is required when you are working with gas pressures, molarity contentrations or the information has to be worked from density measures.
