A sprinter accelerates from rest to a velocity of 12m/s in the first 6 seconds of the 100 meter dash .
1 answer:
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Answer:
Column X. Tangential Speed
Column Y. radius
Explanation:
The equation for centripetal acceleration is
= v² / r
Where v is the tangential velocity of the body and the radius of curvature.
To analyze this equation you must place the tangential velocity in one column and in the other the turning radius
Let's check the answers
Column X. Tangential Speed
Column Y. radius
This is the correct answer.
Answer:
a) a = g / 3
b) x (3.0) = 14.7 m
c) m (3.0) = 29.4 g
Explanation:
Given:-
- The following differential equation for (x) the distance a rain drop has fallen has the form:
- Where, v = Speed of the raindrop
- Proposed solution to given ODE:
v = a*t
Where, a = acceleration of raindrop
Find:-
(a) Using the proposed solution for v find the acceleration a.
(b) Find the distance the raindrop has fallen in t = 3.00 s.
(c) Given that k = 2.00 g/m, find the mass of the raindrop at t = 3.00 s.
Solution:-
- We know that acceleration (a) is the first derivative of velocity (v):
a = dv / dt ... Eq 1
- Similarly, we know that velocity (v) is the first derivative of displacement (x):
v = dx / dt , v = a*t ... proposed solution (Eq 2)
v .dt = dx = a*t . dt
- integrate both sides:
∫a*t . dt = ∫dt
x = 0.5*a*t^2 ... Eq 3
- Substitute Eq1 , 2 , 3 into the given ODE:
0.5*a*t^2*g = 0.5*a^2 t^2 + a^2 t^2
= 1.5 a^2 t^2
a = g / 3
- Using the acceleration of raindrop (a) and t = 3.00 second and plug into Eq 3:
x (t) = 0.5*a*t^2
x (t = 3.0) = 0.5*9.81*3^2 / 3
x (3.0) = 14.7 m
- Using the relation of mass given, and k = 2.00 g/m, determine the mass of raindrop at time t = 3.0 s:
m (t) = k*x (t)
m (3.0) = 2.00*x(3.0)
m (3.0) = 2.00*14.7
m (3.0) = 29.4 g
Answer:
μ = 0.692
Explanation:
In order to solve this problem, we must make a free body diagram and include the respective forces acting on the body. Similarly, deduce the respective equations according to the conditions of the problem and the directions of the forces.
Attached is an image with the respective forces:
A summation of forces on the Y-axis is performed equal to zero, in order to determine the normal force N. this summation is equal to zero since there is no movement on the Y-axis.
Since the body moves at a constant speed, there is no acceleration so the sum of forces on the X-axis must be equal to zero.
The frictional force is defined as the product of the coefficient of friction by the normal force. In this way, we can calculate the coefficient of friction.
The process of solving this problem can be seen in the attached image.
Answer:
f3 = 102 Hz
Explanation:
To find the frequency of the sound produced by the pipe you use the following formula:
n: number of the harmonic = 3
vs: speed of sound = 340 m/s
L: length of the pipe = 2.5 m
You replace the values of n, L and vs in order to calculate the frequency:
hence, the frequency of the third overtone is 102 Hz