Don't mention.

Are you an aspirant?

Are you an aspirant?

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

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

\( T_{½}=0.693÷\lambda \)

\( \lambda=0.693÷T_{½} \)

\( \lambda=0.693÷2 \)s

\( \lambda=0.3465 \)s

\( \lambda\approx 0.347 \)s

You're welcome.

I am an aspirant too.

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

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

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

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 \)

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

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

₦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

#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

#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

40m/s=Distance/20s

Distance=40m/s*20s

Distance=800m

At what temperature will the Celsius scale coincide with the Fahrenheit scale? | Physics Past Questions

⇒(∘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.