Answer:
1170 m
Explanation:
Given:
a = 3.30 m/s²
v₀ = 0 m/s
v = 88.0 m/s
x₀ = 0 m
Find:
x
v² = v₀² + 2a(x - x₀)
(88.0 m/s)² = (0 m/s)² + 2 (3.30 m/s²) (x - 0 m)
x = 1173.33 m
Rounded to 3 sig-figs, the runway must be at least 1170 meters long.
<h3>
Answer:</h3>
225 meters
<h3>
Explanation:</h3>
Acceleration is the rate of change in velocity of an object in motion.
In our case we are given;
Acceleration, a = 2.0 m/s²
Time, t = 15 s
We are required to find the length of the slope;
Assuming the student started at rest, then the initial velocity, V₀ is Zero.
<h3>Step 1: Calculate the final velocity, Vf</h3>
Using the equation of linear motion;
Vf = V₀ + at
Therefore;
Vf = 0 + (2 × 15)
= 30 m/s
Thus, the final velocity of the student is 30 m/s
<h3>Step 2: Calculate the length (displacement) of the slope </h3>
Using the other equation of linear motion;
S = 0.5 at + V₀t
We can calculate the length, S of the slope
That is;
S = (0.5 × 2 × 15² ) - (0 × 15)
= 225 m
Therefore, the length of the slope is 225 m
Answer:
C. Supervising the game to make sure teams are playing fairly
Answer:
f= 4,186 10² Hz
Explanation:
El sistema descrito es un pendulo de torsión que oscila con con velocidad angular, que esta dada por
w = √ k/I
donde ka es constante de torsion de hilo e I es el momento de inercia del disco
El momento de inercia de indican que giran un eje que pasa por enronqueces
I= ½ M R2
reduzcamos las cantidades al sistema SI
R= 1,4 cm = 0,014 m
M= 430 g = 0,430 kg
substituimos
w= √ (2 k/M R2)
calculemos
w = RA ( 2 370 / (0,430 0,014 2)
w = 2,963 103 rad/s
la velocidad angular esta relacionada con la frecuencia por
w =2pi f
f= w/2π
f= 2,963 10³/ (2π)
f= 4,186 10² Hz
Answer:
to overcome the out of friction we must increase the angle of the plane
Explanation:
To answer this exercise, let's propose the solution of the problem, write Newton's second law. We define a coordinate system where the x axis is parallel to the plane and the other axis is perpendicular to the plane.
X axis
fr - Wₓ = m a (1)
Y axis
N- 
= 0
N = W_{y}
let's use trigonometry to find the components of the weight
sin θ = Wₓ / W
cos θ = W_{y} / W
Wₓ = W sin θ
W_{y} = W cos θ
the friction force has the formula
fr = μ N
fr = μ Wy
fr = μ mg cos θ
from equation 1
at the point where the force equals the maximum friction force
in this case the block is still still so a = 0
F = fr
F = (μ mg) cos θ
We can see that the quantities in parentheses with constants, so as the angle increases, the applied force must be less.
This is the force that balances the friction force, any force slightly greater than F initiates the movement.
Consequently, to overcome the out of friction we must increase the angle of the plane
the correct answer is to increase the angle of the plane
