Potential energy is stored energy meaning you are siting at the the top of the slide doing nothing when you are going down the slide you have both potential and kinetic energy because you are moving but youre not doing anything . Gravity and friction play apart by being gravity and pulling you to the ground and friction with the weight and water togeter going downward.
Answer:
Speed, v = 26.5 m/s
Explanation:
It is given that,
The ball hits the ball at an angle of 40 degrees.
Initial height of the ball, h = 1 m
Vertical distance, y = 3.4 m
Horizontal distance, x = 67 m
The trajectory of the particle in projectile motion is given by :
R is the range of projectile.
Putting all the values in above equation as :
R = 71.3 meters
Now using the formula of the range of the projectile as :
Substituting the values in above equation as :
u = 26.5 m/s
So, the speed with which Fred give to the ball is 26.5 m/s. Hence, this is the required solution.
Answer:
A. energy source
Explanation:
An energy source like a batter or emf can move electrons through the circuit of what it's connected to and provide a force field (the case of the battery)
Answer:
vB = 0.5418 m/s (→)
aB = - (0.3189/L) m/s²
ωcd = (0.2117/L) rad/s
Explanation:
a) Given:
vA = 0.23 m/s (↑) (constant value)
If
tan θ = vA/vB
For the instant when θ = 23° we have
vB = vA/ tan θ
⇒ vB = 0.23 m/s/tan 23°
⇒ vB = 0.5418 m/s (→)
b) If tan θ = vA/vB ⇒ vA = vB*tan θ
⇒ d(vA)/dt = d(vB*tan θ)/dt
⇒ 0 = tan θ*d(vB)/dt + vB*Sec²θ*dθ/dt
Knowing that
aB = d(vB)/dt
ωcd = dθ/dt
we have
⇒ 0 = tan θ*aB + vB*Sec²θ*ωcd
ωcd = - Sin (2θ)*aB/(2*vB)
If
v = ωcd*L
where v = vA*Cos θ ⇒ ωcd = v/L = vA*Cos θ/L
⇒ vA*Cos θ/L = - Sin (2θ)*aB/(2*vB)
⇒ aB = - vA*vB/((Sin θ)*L)
We plug the known values into the equation
aB = - (0.23 m/s)*(0.5418 m/s)/(L*Sin 23°)
⇒ aB = - (0.3189/L) m/s²
Finally we obtain the angular velocity of CD as follows
ωcd = vA*Cos θ/L
⇒ ωcd = 0.23 m/s*Cos 23°/L
⇒ ωcd = (0.2117/L) rad/s
Answer:
The value of leaking rate in the question is repeated. By searching on the web I could find the correct value wich is 0.002h^2 m^3 /min.
The depth of the water has to be equal to 7.07 m in order to have a stationary volume.
Explanation:
In order to have a stationary water level the flow of water that comes into the tank (0.1 m^3/min) must be equal to the flow of water that goes out of the tank (0.002*h^2 m^3/min), therefore:
0.002*h^2 = 0.1
h^2 = 0.1/0.002
h^2 = 50
h = sqrt(50) = 7.07 m