Racing on a flat track, a 755 kg car going 32m/s rounds a
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Answer:
1) Please find attached, created with Microsoft Visio
2) The acceleration of the masses connected by the light string is 0.00735 m/s²
3) The tension in the cord is 0.147 N
4) The time it would take the block to go 1.2 m to the edge of the table is approximately 18.07 s
5) The velocity of the cart as soon as it gets to the edge of the table is 0.042 m/s
Explanation:
1) Please find attached, the required free body diagram, showing the tension, weight and frictional (zero friction) forces acting on the cart and the mass created with Microsoft Visio
2) The acceleration of the masses connected by the light string is given as follows;
F = Mass, m × Acceleration, a
The mass of the truck, M = 20.0 kg
The mass attached to the string, hanging rom the pulley, m = 0.0150 kg
The force, F acting on the system = The pulling force on the cart = The tension on the cable = The weight of the hanging mass = 0.0150 × 9.8 = 0.147 N
The pulling force acting on the cart, F = M × a
∴ F = 0.147 N = 20.0 kg × a
a = 0.147 N/(20.0 kg) = 0.00735 m/s²
The acceleration of the truck = a = 0.00735 m/s²
3) The tension in the cord = F = 0.147 N
4) The time, t, it would take the block to go 1.2 m to the edge of the table is given by the kinematic equation, s = u·t + 1/2·a·t²
Where;
s = The distance to the edge of the table = 1.2 m
u = The initial velocity = 0 m/s (The cart is assumed to be initially at rest)
a = The acceleration of the cart = 0.00735 m/s²
t = The time taken
Substituting the known values, gives;
s = u·t + 1/2·a·t²
1.2 = 0 × t + 1/2 ×0.00735 × t²
1.2 = 1/2 ×0.00735 × t²
t² = 1.2/(1/2 ×0.00735) ≈ 326.5306
t = √(1.2/(1/2 ×0.00735)) ≈ 18.07
The time it would take the block to go 1.2 m to the edge of the table = t ≈ 18.07 s
5) The velocity, v, of the cart as soon as it gets to the edge of the table is given by the kinematic equation, v² = u² + 2·a·s as follows;
v² = u² + 2·a·s
u = 0 m/s
v² = 0² + 2 × 0.00735 × 1.2 = 0.001764
v = √(0.001764) = 0.042
The velocity of the cart as soon as it gets to the edge of the table = v = 0.042 m/s.
Answer:
a) T = (281.47 i ^ + 714.56 j ^) N , b) F_net = (281.47 i ^ - 110.44 j ^) N ,
c) F = 281.70 N, d) θ = 338.58º , e) a = 3,588 m / s² , f) θ = 201.45º
Explanation:
For this exercise we will use Newton's second law on each axis
X axis
-Tₓ = m aₓ
Y Axisy
–W = m a_{y}
Let's use trigonometry to find the components of force
sin 21.5 = Tₓ / T
cos 21.5 = T_{y} / T
Tₓ = T sin 21.5
T_{y} = T cos 21.5
Tₓ = 768 sin 21.5 = 281.47 N
T_{y} = 768 cos 21.5 = 714.56 N
a) the force of the rope on Tarzan is
T = (281.47 i ^ + 714.56 j ^) N
b) The net force is the subtraction of the tension minus the weight of Tarzan
Y Axis F_net = 714.56 - 825 = -110.44 N
F_net = (281.47 i ^ - 110.44 j ^) N
c) Let's use Pythagoras' theorem
F = √ (Fₓ² + T_{y}²)
F = √ (281.47² + 110.44²)
F = 281.70 N
d) Let's use trigonometry
tan θ = F_{y} / Fₓ
θ = tan⁻¹ F_{y} / Fₓ
θ = tan⁻¹ (-110.44 / 281.47)
θ = -21.42º
This angle is average clockwise, for counterclockwise measurement
θ = 360 - 21.42
θ = 338.58º
Acceleration
X axis
Tₓ = m aₓ
aₓ = Tₓ / m
The mass of Tarzan is
m = W / g
m = 825 / 9.8 = 84.18 kg
aₓ = 281.47 / 84.18
aₓ = -3.34 m / s2
Y Axis
T_{y}-W = m a_{y}
a_{y} = (T_{y} -W) / m
a_{y} = (714.56-825) / 84.18
a_{y} = - 1,312 m / s²
Acceleration Module
a = √ aₓ² + a_{y}²
a = √ (3.34² +1.312²)
a = 3,588 m / s²
The angle
θ = tan⁻¹ a_{y} / aₓ
θ = tan⁻¹ (-1312 / -3.34)
θ = 21.45º
Notice that the two components of the acceleration are negative, so the angle is in the third quadrant, to measure from the x-axis
θ = 180 + 21.45
θ = 201.45º
Answer:
8 J and 2 J
Explanation:
Given that,
Mass of the rubber ball, m = 1 kg
Initial speed of the rubber ball, u = 4 m/s (in east)
Final speed of the rubber ball, v = -2 m/s (in west)
We need to find the kinetic energy of the ball before it hits the wall, the kinetic energy of the ball after it bounces off the wall.
Initial kinetic energy,
Final kinetic energy,
So, the initial kinetic energy is 8 J and the final kinetic energy is 2 J.
Answer:
The slower the train is moving, the less are the changes of the magnetic flux, thus the eddy currents become weaker.
Explanation:
A magnetic brakes is not a very efficient way of braking when a train is moving slowly because at low speeds, the changes in the magnetic flux are very less and so it causes the eddy current to become weaker.
Let us find the drag force which is proportional to the velocity of two conducting plates.
The EMF that is induced in the eddy currents are :
The force which is due to the induced magnetic field is,
Therefore,
Here, force is directly proportional to the velocity of the two conducting plates.
Therefore, we can say that when the speed of the train is low, the magnetic flux changes are less and thus the eddy currents are weaker.