Exam AnswersNabteb

NABTEB GCE 2022 PHYSICS ANSWERS NOVEMBER 17TH 2022

NABTEB GCE 2022 PHYSICS ANSWERS NOVEMBER 17TH 2022

NABTEB GCE 2022 PHYSICS ANSWERS

 

(6a)

(i) Amplitude: The measure of the displacement of the wave from its rest position.

John AYO runs

 

(ii) Frequency: The frequency of a wave is the number of times per second that the wave cycles.

John AYO runs

 

(6b)

V=k√T

 

:. V₁/√T₁ = V₂/√T₂

 

V₂= V₁√T₂/√T₁

 

V₁= 330ms-¹, T₁ = 273k, V₂=?, T₂= 273+45 = 318k

 

V₂= 330√318/√273

V₂= 330×17.83/16.52

V₂= 5883.9/16.52

V₂= 356.17ms-¹

 

(1a)

{CHOOSE ANY TWO}

(i)  To give early warning signals of environmental disaster

(ii) To help detect and control desertification in the northern part of Nigeria

(iii) To assist in demographic planning

 

(1b)

-TABULATE-

COMPOUND MICROSCOPE

(i) It is an optical instrument which is used to view tiny objects.

(ii) The focal length of the objective lens is shorter than that of the eyepiece.

(iii) The distance between the objective lens and the eyepiece cannot be altered.

(iv) The image formed by the objective lens lies within the focus of the eyepiece.

(v) The final image is formed beyond the objective lens.

 

 

ASTRONOMICAL TELESCOPE

(i) It is an optical instrument which is used to view heavenly bodies.

(ii) The focal length of the objective lens is longer than that of the eyepiece.

(iii) The distance between the objective lens and the eyepiece can be altered.

(iv) The image formed by the objective lens lies exactly at the focus of the eyepiece.

(v) The final image is formed at infinity.

 

1c)

Area of land =100m×128m

=12800m²

1acre=4077m²

12800/4077

=3.16acre

Approximately 3acre

 

(3a)

Range is defined as the horizontal distance from the point of projection to the point where the projectile hits the projection plane again

 

(3b)

S = ut + ½at²

Where, S = Diatance

u = Initial Velocity

a = acceleration

t = time.

 

Acceleration = (Final velocity – Initial Velocity)/t

a = (v – u)/t

at = v – u————eq(i)

 

Now, We know,

Distance Traveled = Average Velocity × Time

S = (v + u)/2 × t

 

But From eq(i) , v = u + at

S = (u + at + u)/2 × t

S = [(2u + at)/2] × t

⇒ S = ut + (1/2)at²

 

(5a)

(i) Conduction: The double-walled glass vessel and vacuum prevents heat loss by conduction.

(ii) Radiation: The silver coating on the inner bottle of a thermos flask is used to prevent heat transfer by radiation

 

 

(5b)

P₂ = (P₁T₂)/T₁

Given, T₁= 50°C, P₁= 900mmHg, T₂= 100°C, P₂=?

P₂ = (900×100)/50

P₂ = 90,000/50

P₂ = 1,800mmHg

°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°

 

(3a)

Range is defined as the horizontal distance from the point of projection to the point where the projectile hits the projection plane again

 

(3b)

S = ut + ½at²

Where, S = Diatance

u = Initial Velocity

a = acceleration

t = time.

 

Acceleration = (Final velocity – Initial Velocity)/t

a = (v – u)/t

at = v – u————eq(i)

 

Now, We know,

Distance Traveled = Average Velocity × Time

S = (v + u)/2 × t

 

But From eq(i) , v = u + at

S = (u + at + u)/2 × t

S = [(2u + at)/2] × t

⇒ S = ut + (1/2)at²

°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°

 

(6a)

(i) Amplitude: The measure of the displacement of the wave from its rest position.

 

(ii) Frequency: The frequency of a wave is the number of times per second that the wave cycles.

 

 

(6b)

V=k√T

 

:. V₁/√T₁ = V₂/√T₂

 

V₂= V₁√T₂/√T₁

 

V₁= 330ms-¹, T₁ = 273k, V₂=?, T₂= 273+45 = 318k

 

V₂= 330√318/√273

V₂= 330×17.83/16.52

V₂= 5883.9/16.52

V₂= 356.17ms-¹

 

COMPLETED!!!

CONCLUSION:- Noted this not 100% verified but I will drop the question specimen for you guys to trace but we can called this 89% so just hold on with it thanks so much.

 

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