Answer:
How to find the maximum height of a projectile?
if α = 90°, then the formula simplifies to: hmax = h + V₀² / (2 * g) and the time of flight is the longest. ...
if α = 45°, then the equation may be written as: ...
if α = 0°, then vertical velocity is equal to 0 (Vy = 0), and that's the case of horizontal projectile motion.
Explanation:
It’s the gravity pull of the sun that makes them spin around 360 degrees
It’s all so why they say In a billion years the sun will smash into the sun
-facts :)
Answer:
The magnitude of the tension in the cable, T is 1,064.315 N
Explanation:
Here we have
Length of beam = 4.0 m
Weight = 200 N
Center of mass of uniform beam = mid-span = 2.0 m
Point of attachment of cable = Beam end = 4.0 m
Angle of cable = 53° with the horizontal
Tension in cable = T
Point at which person stands = 1.50 m from wall
Weight of person = 350 N
Therefore,
Taking moment about the wall, we have
∑Clockwise moments = ∑Anticlockwise moments
T×sin(53) = 350×1.5 + 200×2
T = 850/sin(53) = 1,064.315 N.
Answer:
11.6m
Explanation:
should be somewhere around 11.6 because
a²+b²=c²
so 6²+10²=136
now find the square root √136
=11.6
11.6²=136
Complete question:
The length of nylon rope from which a mountain climber is suspended has an effective force constant of 1.40 ×10⁴ N/m.
What is the frequency (in Hz) at which he bounces, given that his mass plus the mass of his equipment is 84.0 kg?
Answer:
The frequency (in Hz) at which he bounce is 2.054 Hz
Explanation:
Given;
effective force constant, K = 1.40 ×10⁴ N/m.
The total mass = his mass plus the mass of his equipment, m = 84 kg
The frequency (in Hz) at which he bounce is given by;
![f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\\\\f = \frac{1}{2\pi} \sqrt{\frac{1.4*10^4}{84}}\\\\f = 2.054 \ Hz](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7B1%7D%7B2%5Cpi%7D%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%7D%5C%5C%5C%5Cf%20%3D%20%20%5Cfrac%7B1%7D%7B2%5Cpi%7D%20%5Csqrt%7B%5Cfrac%7B1.4%2A10%5E4%7D%7B84%7D%7D%5C%5C%5C%5Cf%20%3D%202.054%20%5C%20Hz)
Therefore, the frequency (in Hz) at which he bounce is 2.054 Hz