Complete Question
A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?
Answer:
Explanation:
From the question we are told that
Mass of ball 
Length of string 
Wind force 
Generally the equation for 
is mathematically given as
Max angle =
Generally the equation for max Height 
is mathematically given as
Generally the equation for Equilibrium Height 
is mathematically given as
The conservation of momentum states that the total momentum in a system is constant if there is no external force acting on the system. The total momentum in the gun bullet system is 0 so it must stay that way.
The momentum of the bullet is mv = 0.015*500=7.5
The momentum of the gun must be the same to keep the total momentum of the system equal to zero, so we know that p = 7.5 for the gun.
Substituting this in we get:
7.5=3.1x
x=7.5/3.1
x=2.42
So the speed of the gun is 2.4m/s.
Answer: a) 0.78 m/s b) 1.57 m/s
Explanation:
M = father's mass
m = son's mass = M/3
V = father's initial speed
v = son's initial speed
(1/2)MV^2 = (1/2)*(1/2)*m v^2
M*V^2 = (1/2)(M/3)v^2
V^2/v^2 = 1/4
V = v/2
Second equation:
(1/2)M*(V + 1.4)^2 = (1/2)m*v^2
= (1/2)*(M/3)*(3V)^2
cancel out the M's and (1/2)'s
(V + 1.4)^2 = 3V^2
V^2 + 2.8V + 1.96 = 3V^2
V^2 -1.4V -0.98 = 0
V^2 = 0.98/0.4 = 2.45
V = 1.57
Answer:
Average speed is total distance divided by total time.
v = d / t
Answer:
<h2> $1.50</h2>
Explanation:
Given data
power P= 2 kW
time t= 15 min to hours = 15/60= 1/4 h
cost of power consumption per kWh= 10 cent = $0.1
We are expected to compute the cost of operating the heater for 30 days
but let us computer the energy consumption for one day
Energy of heater for one day= 2* 1/4 = 0.5 kWh
the cost of operating the heater for 30 days= 0.5*0.1*30= $1.50
<u><em>Hence it will cost $1.50 for 30 days operation</em></u>
