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: a) Case 2 b) Case 1
Explanation:
a) By definition, the magnitude of a torque, referred to a given point, is expressed as the product of the force that causes the torque, times the perpendicular distance to the reference point.
If we assume that the only force acting on the arm is the weight of the arm, and that this is concentrated in a point in the center of it (taking the arm as a solid bar with the center of mass at the mid-point), clearly the torque will be the greatest when the force be exactly perpendicular, which is the case of the arm placed straight out parallel to the ground (Case 2).
b) As the torque and the angular acceleration are directly proportional each other (being the rotational inertia the proportionality constant) the angular acceleration will be maximum when torque be maximum also, which is the case that the arm begins to swim, due to the perpendicular distance to the shoulder is the maximum possible (Case 1).
Answer:
a. 12,600 N
b. 1290 kg
Explanation:
a. Impulse = change in momentum
F Δt = m Δv
F (0.192 s) = (59 kg) (0 m/s − 41 m/s)
F = -12,600 N
b. F = mg
12,600 N = m (9.8 m/s²)
m = 1290 kg (or 2,830 lbs)
Complete Question
The complete question shown on the first uploaded image
Answer:
a)
The force on Q due to dipole is Attractive
b)
The charge Q exerts attractive force on the dipole
c)
Yes from the above parts, force depends on the sign of charge
d)
![F = kQq[\frac{d^{2}+2rd}{r^{2}(d+r)^{2}} ]](https://tex.z-dn.net/?f=F%20%3D%20kQq%5B%5Cfrac%7Bd%5E%7B2%7D%2B2rd%7D%7Br%5E%7B2%7D%28d%2Br%29%5E%7B2%7D%7D%20%5D)
e)
The magnitude o force decrease by a factor of 8.0 times
Explanation:
The explanation is shown on the second uploaded image
Province. Is what is located away from the capital.
