Answer:
The height of the image will be "1.16 mm".
Explanation:
The given values are:
Object distance, u = 25 cm
Focal distance, f = 1.8 cm
On applying the lens formula, we get
⇒ 
On putting estimate values, we get
⇒ 
⇒ 
⇒ 
As a result, the image would be established mostly on right side and would be true even though v is positive.
By magnification,
and
⇒ 
⇒ 
⇒ 
Answer:
Final speed of the train is 7.5 m/s
Explanation:
It is given that,
Uniform acceleration of the train is, a = 1.5 m/s²
It starts from rest and travels for 5.0 s. We have to find the final velocity of the train. By using first equation of motion as :

Here, train starts from rest so, u = 0
v = 7.5 m/s
So, the final velocity of the train is 7.5 m/s. Hence, this is the required solution.
Answer:
D
Explanation:
they can see this along with many other fish.
Answer:
A
B

C

D

Explanation:
Considering the first question
From the question we are told that
The spring constant is 
The potential energy is 
Generally the potential energy stored in spring is mathematically represented as 
=>
=>
=>
Considering the second question
From the question we are told that
The mass of the dart is m = 0.050 kg
Generally from the law of energy conservation

=> 
=> 
Considering the third question
The height at which the dart was fired horizontally is 
Generally from the law of energy conservation

Here KE is kinetic energy of the dart which is mathematical represented as

=> 
=> 
=> 
Considering the fourth question
Generally the total time of flight of the dart is mathematically represented as

=> 
=> 
Generally the horizontal distance from the equilibrium position to the ground is mathematically represented as

=> 
=> 