Answer:
x = 11.23 m
Explanation:
For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.
Let's reduce to SI system units
θ = 155 rev (2pi rad / rev) = 310π rad
α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²
Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)
w² = w₀² + 2 α θ
w =√ 2 α θ
w = √(2 4pi 310pi)
w = 156.45 rad / s
The relationship between angular and linear velocity
v = w r
v = 156.45 0.175
v = 27.38 m / s
In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive
y = 
t - ½ g t²
As it leaves the highest point its speed is horizontal
y = 0 - ½ g t²
t = √ (-2y / g)
t = √ (-2 (-0.820) /9.8)
t = 0.41 s
With this time we calculate the horizontal distance, because the constant horizontal speed
x = vox t
x = 27.38 0.41
x = 11.23 m
Answer:
A block of mass M = 5 kg is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force F = 40N is applied, the acceleration of the block will be then (g=10ms
2 ).
Mass of the block=5kg
Coeffecient of friction=0.2
external applied force, F=40N
The angle at which the force is applied=30degree
So the horizontal component of force=Fcos30=40×
23 =20 3 N
While the uertical component of the force acting in upward direction=Fsin30=40× 21
=20N
The normal reaction from the surface (N)=mg−Fsin30=50−20=30N
So the ualue of limiting friction=μN=0.2×30=6N
Hence the net horizontal force on the block=Fcos30=μN=20
3
N−6N=28.64N
The horizontal acceleration of the block=
m
Fcos30−μN = 528.64
=5.73m/s 2
Answer:
utrasonic
Explanation:
these are sounds beyond our hearing capacity range of 20-20kHz
Answer:
See below
Explanation:
Vertical position is given by
df = do + vo t - 1/2 a t^2 df = final position = 0 (on the ground)
do =original position = 2 m
vo = original <u>VERTICAL</u> velocity = 0
a = acceleration of gravity = 9.81 m/s^2
THIS BECOMES
0 = 2 + 0 * t - 1/2 ( 9.81)t^2
to show t =<u> .639 seconds to hit the ground </u>
During this .639 seconds it flies horizontally at 10 m/s for a distance of
10 m/s * .639 s =<u> 6.39 m </u>
The work done to stretch the spring will be 112 J.
<h3>What is spring force?</h3>
The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is
F = kx
The given data in the problem is;
F is the spring force =?
K is the spring constant= 8.5 N/m
x is the length by which spring got stretched = 1.2m
The work is done to stretch the spring is;
To learn more about the spring force refer to the link;
brainly.com/question/4291098
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