Answer:
a) D_ total = 18.54 m, b) v = 6.55 m / s
Explanation:
In this exercise we must find the displacement of the player.
a) Let's start with the initial displacement, d = 8 m at a 45º angle, use trigonometry to find the components
sin 45 = y₁ / d
cos 45 = x₁ / d
y₁ = d sin 45
x₁ = d sin 45
y₁ = 8 sin 45 = 5,657 m
x₁ = 8 cos 45 = 5,657 m
The second offset is d₂ = 12m at 90 of the 50 yard
y₂ = 12 m
x₂ = 0
total displacement
y_total = y₁ + y₂
y_total = 5,657 + 12
y_total = 17,657 m
x_total = x₁ + x₂
x_total = 5,657 + 0
x_total = 5,657 m
D_total = 17.657 i^+ 5.657 j^ m
D_total = Ra (17.657 2 + 5.657 2)
D_ total = 18.54 m
b) the average speed is requested, which is the offset carried out in the time used
v = Δx /Δt
the distance traveled using the pythagorean theorem is
r = √ (d1² + d2²)
r = √ (8² + 12²)
r = 14.42 m
The time used for this shredding is
t = t1 + t2
t = 1 + 1.2
t = 2.2 s
let's calculate the average speed
v = 14.42 / 2.2
v = 6.55 m / s
Answer:
Explanation:
The power provided by a resistor (wire in this case) is given by:
.
The resistance of a wire is given by:
Where for the resistivity the one of the copper should be used: 
.
The area A is that of a circle, which written in terms of its diameter is:
Putting all together:
Which for our values is:
F=ma
F= 4x1.2
F= 4.8 N
F= 4gsin30 - Friction
Friction= 19.6 - 4.8 N
Friction= 14.8 N
Friction= u x 4gcos30
14.8 / 4gcos30 = u
u= 0.43596...
u= 0.44
coefficient is 0.44
Answer:
Simple harmonic motion is repetitive. The period T is the time it takes the object to complete one oscillation and return to the starting position. ... If at t = 0 the object has its maximum displacement in the positive x-direction, then φ = 0, if it has its maximum displacement in the negative x-direction, then φ = π.
Explanation:
Simple harmonic motion, in physics, repetitive movement back and forth through an equilibrium, or central, position, so that the maximum displacement on one side of this position is equal to the maximum displacement on the other side
Answer:
Kinetic energy in your palm
