Explanation:
A wave on a string is described is given by :
![D(x,t)=2\ cm\ sin[(12.57\ rad/m)-(638\ rad/s)t]](https://tex.z-dn.net/?f=D%28x%2Ct%29%3D2%5C%20cm%5C%20sin%5B%2812.57%5C%20rad%2Fm%29-%28638%5C%20rad%2Fs%29t%5D)
The linear density of the string is 5 g/m.
Where
x is in meters and t is in seconds
The general equation of a wave is given by :
(2) The speed of the wave in terms of tension is given by :
Also, 
So, 
T = 12.88 N
(3) The maximum displacement of a point on the string is equal to the amplitude of the wave. So, the maximum displacement is 2 cm.
(4) The maximum speed of a point on the string is given by :
v = 12.76 m/s
Hence, this is the required solution.
Answer:
hmax = 1/2 · v²/g
Explanation:
Hi there!
Due to the conservation of energy and since there is no dissipative force (like friction) all the kinetic energy (KE) of the ball has to be converted into gravitational potential energy (PE) when the ball comes to stop.
KE = PE
Where KE is the initial kinetic energy and PE is the final potential energy.
The kinetic energy of the ball is calculated as follows:
KE = 1/2 · m · v²
Where:
m = mass of the ball
v = velocity.
The potential energy is calculated as follows:
PE = m · g · h
Where:
m = mass of the ball.
g = acceleration due to gravity (known value: 9.81 m/s²).
h = height.
At the maximum height, the potential energy is equal to the initial kinetic energy because the energy is conserved, i.e, all the kinetic energy was converted into potential energy (there was no energy dissipation as heat because there was no friction). Then:
PE = KE
m · g · hmax = 1/2 · m · v²
Solving for hmax:
hmax = 1/2 · v² / g
At a point on the streamline, Bernoulli's equation is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = density of air, 0.075 lb/ft³ (standard conditions)
g = 32 ft/s²
Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s
Point 2 (stagnation):
At the stagnation point, the velocity is zero.
The density remains constant.
Let p₂ = pressure at the stagnation point.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
= 314.37 lb/ft²
= 314.37/144 = 2.18 lb/in²
Answer: 2.2 psi
Answer:
is this phsyics teacher, is he good i-
Explanation:
Answer:
6N
Explanation:
Given parameters:
Pressure applied by the woman = 300N/m²
Area = 0.02m²
Unknown:
Force applied = ?
Solution:
Pressure is the force per unit area on a body
Pressure = 
Force = Pressure x area
Force = 300 x 0.02 = 6N
