Answer:
A) F = - 8.5 10² N, B) I = 21 N s
Explanation:
A) We can solve this problem using the relationship of momentum and momentum
I = Δp
in this case they indicate that the body rebounds, therefore the exit speed is the same in modulus, but with the opposite direction
v₀ = 8.50 m / s
v_f = -8.50 m / s
F t = m v_f -m v₀
F = 
let's calculate
F = 
F = - 8.5 10² N
B) let's start by calculating the speed with which the ball reaches the ground, let's use the kinematic relations
v² = v₀² - 2g (y- y₀)
as the ball falls its initial velocity is zero (vo = 0) and the height upon reaching the ground is y = 0
v = 
calculate
v = 
v = 14 m / s
to calculate the momentum we use
I = Δp
I = m v_f - mv₀
when it hits the ground its speed drops to zero
we substitute
I = 1.50 (0-14)
I = -21 N s
the negative sign is for the momentum that the ground on the ball, the momentum of the ball on the ground is
I = 21 N s
Answer:
457.81 Hz
Explanation:
From the question, it is stated that it is a question under Doppler effect.
As a result, we use this form
fo = (c + vo) / (c - vs) × fs
fo = observed frequency by observer =?
c = speed of sound = 332 m/s
vo = velocity of observer relative to source = 45 m/s
vs = velocity of source relative to observer = - 46 m/s ( it is taking a negative sign because the velocity of the source is in opposite direction to the observer).
fs = frequency of sound wave by source = 459 Hz
By substituting the the values to the equation, we have
fo = (332 + 45) / (332 - (-46)) × 459
fo = (377/ 332 + 46) × 459
fo = (377/ 378) × 459
fo = 0.9974 × 459
fo = 457.81 Hz
B. velocity at position x, velocity at position x=0, position x, and the original position
In the equation
= 
+2 a x (x - x₀)
= velocity at position "x"
= velocity at position "x = 0 "
x = final position
= initial position of the object at the start of the motion
Answer with Explanation:
We are given that
Restoring force,
We have to find the work must you do to compress this spring 15 cm.
Using 1 m=100 cm
Work done=
W=
![W=k[\frac{(\Delta s)^2}{2}]^{0.15}_{0}+q[\frac{(\Delta s)^4}{4}]^{0.15}_{0}](https://tex.z-dn.net/?f=W%3Dk%5B%5Cfrac%7B%28%5CDelta%20s%29%5E2%7D%7B2%7D%5D%5E%7B0.15%7D_%7B0%7D%2Bq%5B%5Cfrac%7B%28%5CDelta%20s%29%5E4%7D%7B4%7D%5D%5E%7B0.15%7D_%7B0%7D)
Ideal spring work=
Percentage increase in work=
%
Answer:
-78.96 J
Explanation:
The workdone by the torque in stopping the wheel = rotational kinetic energy change of wheel.
So W = 1/2I(ω₁² - ω₀²) where I = rotational inertia of wheel = 0.04 kgm², r = radius of wheel = 0.02 m, ω₀ = initial rotational speed = 10 rev/s × 2π = 62.83 rad/s, ω₁ = final rotational speed = 0 rad/s (since the wheel stops)
W = 1/2I(ω₁² - ω₀²) = 1/2 0.04 kgm² (0² - (62.83 rad/s)²) = -78.96 J
