Bebe Essien

The relation between the Fahrenheit and Celsius scale is given by the equation.
⇒(∘F−32)180=∘C100

Now, let us assume that Fahrenheit and Celsius scale coincide at x∘
Therefore,
⇒(∘x−32)180=∘x100
ON solving,
We get,
⇒59(∘x−32)=∘x
Therefore,
⇒x=−40∘
Therefore, Celsius and Fahrenheit scale coincide at -40 degrees.

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Mass of truck, m¹=500kg
Velocity of truck, v¹=6m/s
Mass of car, m²=100kg
Velocity of car, v²=0m/s
m¹v¹+m²v²=(m¹+m²)V
Where V is the common velocity after collision
(500*6)+(100*0)=(500+100)V
3000+0=600V
V=\( \cfrac{3000}{600} \)
V=5m/s

Half life, \( T_{½}=2 \)
\( T_{½}=0.693÷\lambda \)
\( \lambda=0.693÷T_{½} \)
\( \lambda=0.693÷2 \)s
\( \lambda=0.3465 \)s
\( \lambda\approx 0.347 \)s

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I am an aspirant too.

Volume, V¹=283cm³
Temperature, T¹=10°C
Converting 10°C To Kelvin =10+273=283K
Volume, V²=?
Temperature, T²=30°C
Converting 30°C To Kelvin=30+273=303K

\( \cfrac{V¹}{T¹}=\cfrac{V²}{T²} \)
\( \cfrac{283}{283}=\cfrac{V²}{303} \)
\( V²=\cfrac{283*303}{283} \)
V²=303cm³

The telescope operates at optical to mid-infrared wavelengths, but has an especially powerful capability to image large areas of the sky at optical wavelengths. Astronomers use these images to economically estimate the distances and properties of galaxies.

A vernier caliper is measuring device used to precisely measure linear dimensions. It is a very useful tool when measuring the diameter of round objects like cylinders because the measuring jaws can be secured on either side of the circumference.

Let the number of images formed be n
And the angle of inclination be \( \theta \)
n=360/\( \theta \)
If n is even, we subtract one
But if n is odd, we leave it
Now, n=360÷180
n=3
Now, n is even so we subtract one from it
n=3-1
n=2

Initial velocity, u=20m/s
Time, t=4s
Acceleration due to gravity, g=10m/s²
Height of the hill, h=?
h=ut\( \pm \)½gt²
h is positive because the body is falling down
h=ut+½gt²
h=(20\( \times \)4)+½\( \times \)10\( \times \)4²
=80+80
=160m

Coefficient of linear eapsnsivity,\( \alpha=4\times10^-⁵ \)
Change in temperature, \( \Delta\theta=95°C-15°C
=80° C
\)
Original length, l¹=20cm
Final length, l²=xcm
\( \alpha=(l²-l¹)÷l¹\Delta\theta
4\times10^–⁵=(x-20)÷(20\times80)
x=20.064cm \)

Initial velocity, u=40m/s
Time, t=20s
Final velocity, v=0m/s since the truck is coming to a halt
Acceleration, a=?
But,
v=u+at
0=40+a(20)
0=40+20a
20a=-40
a=-2m/s²
To find the distance traveled,
v²=u²+2as
0²=40²+2(-2)(s)
0=1600-4s
4s=1600
s=400m

The interest at the end of one year will be
₦1,000,000*4%=₦4,000
Total amount of money at the end of one year
=₦1,000,000+₦4,000
=₦1,040,000
The interest at the end of two years will be
₦1,040,000*4%=₦41,600
Total amount of money at the end of two years
=₦1,040,000+₦41,600
=₦1,081,600
The interest at the end of three years will be
₦1,081,600*40%=₦43,264
Total amount of money at the end of three years
=₦1,081,600+₦43,264
=₦1,124,864

In the first year the value of the machine depreciates by
#20,000*10%=#2,000
At the end of the first year,the value of the machine will be
#20,000-#2000=#18,000
In the second year, the value of the machine depreciates by
#18,000*10%=#1,800
At the end of the second year, the value of the machine will be
#18,000-#1,800=#16,200

In the first year the value depreciates by
#20,000*10%=#2,000
At the end of the first year,the value of the machine will be
#20,000-#2000=#18,000
In the second year, the value of the machine depreciates by
#18,000*10%=#1,800
At the end of the second year, the value of the machine will be
#18,000-#1,800=#16,200

Velocity=Distance/Time
40m/s=Distance/20s
Distance=40m/s*20s
Distance=800m