Answer:
Long sight occurs when the eyeball is too short or the lens is too thin, or both. As a result, light rays from near objects are focused behind the retina because the light rays are not converged enough. The image formed on the retina is therefore out of focus.
To correct this problem, people can wear glasses with convex lenses. Light rays from near objects are converged by the convex lenses before entering the eyes, so that light can be focused on the retina to form a sharp image. Additionally, long sight can also be corrected by surgical methods such as LASIK.
Initial speed of the car (u) = 15 m/s
Final speed of the car (v) = 0 m/s (Car comes to a complete stop after driver applies the brake)
Distance travelled by the car before it comes to halt (s) = 63 m
By using equation of motion, we get:
Acceleration of the car (a) = -1.78 m/s²
Magnitude of the car's acceleration (|a|) = 1.78 m/s²
Answer:
5 m/s
Explanation:
Here we can see there is no external force acted on a two masses when we consider the motion. If there is no external forces then momentum is conserved.
Initial momentum = Final momentum
0.5 × 10 = 1 × V
V = 5 m/s
V₁ = (1/g)₁ = Way₁ = 20(9.81)(0) = 0
V₂ (Vg)₂ = -WAy₂ = -20(9.81)(0.5) = -98.1J
The kinetic energy because the pool rotates about a fixed axis
W = VA/rA = VA/0.2 5VA
Mass momen of inertila about fixed axis which passes through point 0
I₀ = mko² = 50(0.280)² = 3.92 kg. m²
∴ The kinetic energy of the system is
T = 1/2 I₀w² + 1/2mAVA²
= 1/2(3.92)(5VA)² + 1/2 (20) VA² = 59VA²
Now that the system is at rest then T₁ = 0
Energy conservation is
T₁ +V₁ = T₂ + V₂
0+ 0 = 59VA² + (-98.1)
VA = 1.289 m/s
= 1.29 m/s
Answer:
The electric field at origin is 3600 N/C
Solution:
As per the question:
Charge density of rod 1, 
Charge density of rod 2, 
Now,
To calculate the electric field at origin:
We know that the electric field due to a long rod is given by:
Also,
(1)
where
K = electrostatic constant = 
R = Distance
= linear charge density
Now,
In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.
At x = - 1 cm = - 0.01 m:
Using eqn (1):
(towards)
Now, at x = 1 cm = 0.01 m :
Using eqn (1):
(towards)
Now, the total field at the origin is the sum of both the fields:
