Answer:
8.505 m
Explanation:
Let V1 and V2 be velocities of puck A and B respectively
Since A and B move in the same direction, so the relative velocity will be V1+V2=3.5+3.9=7.4m/s
Or
Vr=7.4 m/s
Distance=S= 18 m
Time =t=?
S=Vr×t
==> t=S/Vr
==> t= 18/7.4=2.43 sec
At this time both will strike together
<em><u>Distance by puck A</u></em>
<em>V1=3.5 m/s</em>
Time=t= 2.43 sec
Distance covered=d=?
d=V1×t=3.5×2.43=8.505 m
So, puck A will cover 8.505 meters before collision
<span>Answer:
First we need to find the acceleration.
torque on cylinder Ď„ = T * r where T is the string tension;
T = m(g - a) where a is the acceleration of the cylinder. Then
Ď„ = m(g - a)r
But also τ = Iα. For a solid cylinder, I = ½mr²,
and if the string doesn't slip, then α = a / r, so
τ = ½mr² * a/r = ½mra.
Since Ď„ = Ď„, we have
m(g - a)r = ½mra → m, r cancel, leaving
g - a = ½a
g = 3a/2
a = 2g/3 where g, of course, is gravitational acceleration.
We know that v(t) = a*t, so for our cylinder
v(t) = 2gt / 3 â—„ linear velocity
and ω = v(t) / r = 2gt / 3r ◄ angular velocity</span>
Answer:
the bigger the mass, the smaller the acceleration and the smaller speed
Explanation:
this can be proven from the formula F = ma
F = m×speed/time
Or a practical example is that of a heavy object and a light object if pushed with the same force, it'll be observed that the small object moves faster or with a greater speed than the heavy object
The centripetal acceleration = 236.63 m/s²
The force = 17.98 N
<h3>Further explanation</h3>
Given
mass = 76 g = 0.076 kg
r = 1.5 m
f = 2 rps = 2 rotation per second
Required
The centripetal acceleration
The Force tension
Solution
Centripetal force is a force acting on objects that move in a circle in the direction toward the center of the circle
F = centripetal force, N
m = mass, Kg
v = linear velocity, m / s
r = radius, m
The speed that is in the direction of the circle is called linear velocity
Can be formulated:
r = circle radius
f = rotation per second (RPS)
The linear velocity : 2 x 3.14 x 1.5 x 2 =18.84 m/s
The centripetal acceleration : ac = v²/R = 236.63 m/s²
The force : F = m x ac = 0.076 x 236.63 = 17.98 N
Do you remember the general equation for the distance covered
by a moving object ? There are not many perfect opportunities to
use it in all its glory, but I think this is one of them.
Position =
(starting distance) + (starting speed) (time) + (1/2) (acceleration) (time)²
H = starting position + (starting speed x t) + 1/2 A t²
Here's how we can use it, with some careful definitions:
-- Let's say the surface of the sea is zero height.
Then 'H' ... the position at the end ... is zero, when it plunks into the water, and
the starting, original position of the stone is +10 on the cliff in the man's hand.
-- Starting speed is +5 ... 5 m/s upward, when he tosses it.
-- Acceleration is 9.8 m/s² downward ... the acceleration of gravity.
I think this is going to work out just beautifully !
0 = (5) + 5t - 1/2 (9.8) t²
-4.9 t² + 5t + 5 = 0 That's the whole thing right there. Look how gorgeous that is !
Solve it for 't' with the quadratic equation,
A = -4.9
B = 5
C = 5
When you solve a quadratic with the formula, you always get two roots.
If it's a real-world situation, one of them might not make sense. That's
the result in this case.
The two roots are
t = - 0.622 second
and
t = + 1.642 second
The first one isn't useful, because it means 0.622 second <u>before</u> the man
tossed the stone up.
So our answer is: We hear the 'plunk' <em>1.642 second</em> after the upward toss.
