The required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ
First, we must understand that the component of the velocity along the vertical is due to maximum height achieved and expressed as usin
θ.
The component of the velocity along the horizontal is due to the range of the object and is expressed as ucosθ.
If the <u>air resistance is ignored</u>, the velocity of the object will be constant throughout the flight and the initial velocity will be equal to the final velocity.
Hence the required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ
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Answer:
6.44 × 10^10 N/C
Explanation:
Electric field due to the ring on its axis is given by
E = K q r / (r^2 + x^2)^3/2
Where r be the radius of ring and x be the distance of point from the centre of ring and q be the charge on ring.
r = 0.25 m, x = 0.5 m, q = 5 C
K = 9 × 10^9 Nm^2/C^2
E = 9 × 10^9 × 5 × 0.25 / (0.0625 + 0.25)^3/2
E = 6.44 × 10^10 N/C
Answer:
option (B)
Explanation:
Intensity of unpolarised light, I = 25 W/m^2
When it passes from first polarisr, the intensity of light becomes

Let the intensity of light as it passes from second polariser is I''.
According to the law of Malus

Where, θ be the angle between the axis first polariser and the second polariser.

I'' = 11.66 W/m^2
I'' = 11.7 W/m^2
Answer:
a)
b)
c)
d)
m
e)λ=∞
Explanation:
De Broglie discovered that an electron or other mass particles can have a wavelength associated, and that wavelength (λ) is:

with h the Plank's constant (
) and P the momentum of the object that is mass (m) times velocity (v).
a)

b)

c)

d)
m
e) 
λ=∞
Answer:
An apple in free fall accelerates toward the Earth with a free fall acceleration, g. The force of the apple on the Earth also causes the Earth to accelerate toward the falling apple. By Newton's Third Law, the force of the Earth on the apple is exactly equal and opposite to the force of the apple on the Earth. By Newton,s Second law, the force of the Earth on the apple is equal to the mass of the apple times g , the accelerations due to gravity. And, the force of the the apple on the Earth is equal to the mass of the Earth times the acceleration of the Earth toward the apple. In conclusion, the magnitude of the forces are equal, or
F ( apple on the Earth) = F( the Earth on the apple) or
M( mass of the earth) x a( the acceleration of the earth toward the apple) = m(mass of the apple) x g( the acceleration of the apple toward the Earth) or
a = (m/M) g
Explanation: