Ox:vₓ=v₀
x=v₀t
Oy:y=h-gt²/2
|vy|=gt
tgα=|vy|/vₓ=gt/v₀=>t=v₀tgα/g
y=0=>h=gt²/2=v₀²tg²α/2g=>tgα=√(2gh/v₀²)=√(2*10*20/24²)=√(400/576)=0.83=>α=tg⁻¹0.83=39°
cosα=vₓ/v=v₀/v=>v=v₀/cosα=24/cos39°=24/0,77=31.16 m/s
Ec=mv²/2=2*31.16²/2=971.47 J=>Ec≈0.97 kJ
Answer:
Kinematics is the study of motion, without any reference to the forces that cause the motion. It basically means studying how things are moving, not why they're moving. It includes concepts such as distance or displacement, speed or velocity, and acceleration, and it looks at how those values vary over time.
Explanation:
It is given that,
An electron is released from rest in a weak electric field of,
Vertical distance covered,
We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :
.............(1)
Electric force is and force of gravity is . As both forces are acting in downward direction. So, total force is:
Acceleration of the electron,
Put the value of a in equation (1) as :
v = 0.010 m/s
So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.
I believe your answer should be C. Speed.
Answer:
f1 = -3.50 m
Explanation:
For a nearsighted person an object at infinity must be made to appear to be at his far point which is 3.50 m away. The image of an object at infinity must be formed on the same side of the lens as the object.
∴ v = -3.5 m
Using mirror formula,
i/f1 = 1/v + 1/u
Where f1 = focal length of the contact lens, v = image distance = -3.5 m, u = object distance = at infinity(∞) = 1/0
∴ 1/f1 = (1/-3.5) + 1/infinity
Note that, 1/infinity = 1/(1/0) = 0/1 =0.
∴ 1/f1 = 1/(-3.5) + 0
1/f1 = 1/(-3.5)
Solving the equation by finding the inverse of both side of the equation.
∴ f1 = -3.50 m
Therefore a converging lens of focal length f1 = -3.50 m
would be needed by the person to see an object at infinity clearly