Answer:
66.4 N
Explanation:
From Newton's second law, <em>F </em>=<em> ma</em>
where <em>F</em> is the force, <em>m</em> is the mass and <em>a</em> is the acceleration.
Because the object has acceleration in two directions and the mass is constant, the force will be in two directions. The component of the forces are:
The magnitude of the resultant force is given by
Answer:
66.26 m/s
Explanation:
Horizontal velocity, Vx = 55.3 m/s
Vertical velocity, Vy = 36.5 m/s
The value of the resultant velocity is given by the vector sum of the two velocities which are acting at 90°.
V = 66.26 m/s
Thus, the velocity of the vehicle is 66.26 m/s along its descent path.
Answer:
5.42 m/s
Explanation:
At minimum speed, the tension in the bar will be 0 when the ball is at the top of the arc, so the only force is gravity pulling down.
Sum of forces towards the center of the circle:
∑F = ma
mg = m v²/r
v = √(gr)
v = √(9.8 m/s² × 3.00 m)
v = 5.42 m/s
Answer:
F = -6472.9 N
F= -6.47 kN
Explanation:
First of all you have to convert the data to SI units
so for the velocity you have :
Vi = 43km/h *(1000m/1km)*(1h/3600s) ---> using conversion factors
Vi= 11.9444 m/s
dX : distance the passanger moves
dX = 54cm*(1m/100cm) --> using conversion factors
dX = 0.54 m
Now to calculate the force we are going to use the sum of focers equals to mass for acceleration:
Sum F = m*a
We have to find a so we are going to use the velocity's formula as follows to solve a:
Vf ^2 = Vi^2 +2*a*dX
Vf=0 --> the passenger does not move after the airbag inflates.
a= -(Vi^2)/(2*dX)
you solve de acceleration with the data you hae and you will find
a = -132.1 m/ s^2
Now you can solve the Sum F equation
Sum F = 49 Kg * (-132.1 m/s^2)
F = -6472.9 N
F= -6.47 kN
Answer:
the rate of heat transfer after the system achieves steady state is -3.36 kW
Explanation:
Given the data in the question;
mass of water m = 50 kg
N = 300 rpm
Torque T = 0.1 kNm
V = 110 V
I = 2 A
Electric work supplied W₁ = PV = 2 × 110 = 220 W = 0.22 kW
Now, work supplied by paddle wheel W₂ is;
W₂ = 2πNT/60
W₂ = (2π × 0.1 × 300) / 60
W₂ = 188.495559 / 60
W₂ = 3.14 kW
So the total work will be;
W = 0.22 + 3.14
W = 3.36 kW
Hence total work done on the system is 3.36 kW.
At steady state, the properties of the system does not change so the heat transfer will be 3.36 KW.
The heat will be rejected by the system so the sign of heat will be negative.
i.e Q = -3.36 kW
Therefore, the rate of heat transfer after the system achieves steady state is -3.36 kW
