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3 years ago

Catalytic ozone destruction occurs in the stratosphere by a two-step reaction:

1 answer:

7 0

Catalytic ozone destruction occurs in the stratosphere where the reactions involving bromine, chlorine, hydrogen, nitrogen and oxygen gases form compounds that destroy the ozone layer. The reactions uses a catalyst (speeds up the reaction) in a two step reaction. considering chlorine the reactions appears as follows;
step 1
Cl + O3 = ClO + O2
step 2
ClO + O = Cl + O2
Where by chlorine is released to destroy the ozone layer, this takes place many times even with the other elements (hydrogen, bromine, nitrogen) and the end result is a completely destroyed Ozone layer

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The resistance of a lamp that draws 0.5A of current when it is operated by a 12V battery will be Ω

lisabon 2012 [21]

According to Ohm's Law, the resistance, current and voltage are related as:

V = IR
⇒
R = V/I

V is given to be 12 Volts.
I is given to be 0.5 Ampere

So, resistance will be:

R = 12/0.5 = 24 ohms

5 0

3 years ago

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zhannawk [14.2K]

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During a braking test, a car is brought to rest beginning from an initial speed of 60 mi/hr in a distance of 120 ft. With the sa

maw [93]

Answer:

Explanation:

Given

Initial speed $u=60\ mi/hr\approx 88\ ft/s$

distance traveled before coming to rest $d_1=120\ ft$

using equation of motion

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$0-(88)^2=2\times a\times 120---1$

for $u_2=80\ mi/hr\approx 117.33\ ft/s$

using same relation we get

$0-(117.33)^2=2\times a\times (d_2)----2$

divide 1 and 2 we get

$(\frac{88}{117.33})^2=\frac{120}{d_2}$

$d_2=213.32\ ft$

So a distance if 213.32 ft is required to stop the vehicle with 80 mph speed

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3 years ago

The concrete slab of a basement is 11 m long, 8 m wide, and 0.20 m thick. During the winter, temperatures are nominally 17 C and

Lunna [17]

Answer:

Q = - 4312 W = - 4.312 KW

Explanation:

The rate of heat of the concrete slab can be calculated through Fourier's Law of heat conduction. The formula of the Fourier's Law of heat conduction is as follows:

Q = - kA dt/dx

Integrating from one side of the slab to other along the thickness dimension, we get:

Q = - kA(T₂ - T₁)/L

Q = kA(T₁ - T₂)/t

where,

Q = Rate of Heat Loss = ?

k = thermal conductivity = 1.4 W/m.k

A = Surface Area = (11 m)(8 m) = 88 m²

T₁ = Temperature of Bottom Surface = 10°C

T₂ = Temperature of Top Surface = 17° C

t = Thickness of Slab = 0.2 m

Therefore,