When does the nuclear membrane disappear?

A yoyo with a mass of m = 150 g is released from rest as shown in the figure.

avanturin [10]

(1) The linear acceleration of the yoyo is 3.21 m/s².

(2) The angular acceleration of the yoyo is 80.25 rad/s²

(3) The weight of the yoyo is 1.47 N

(4) The tension in the rope is 1.47 N.

(5) The angular speed of the yoyo is 71.385 rad/s.

<h3> Linear acceleration of the yoyo</h3>

The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.

∑τ = Iα

rT - Rf = Iα

where;

I is moment of inertia
α is angular acceleration
T is tension in the rope
r is inner radius
R is outer radius
f is frictional force

rT - Rf = Iα ----- (1)

T - f = Ma -------- (2)

a = Rα

where;

a is the linear acceleration of the yoyo

Torque equation for frictional force;

$f = (\frac{r}{R} T) - (\frac{I}{R^2} )a$

solve (1) and (2)

$a = \frac{TR(R - r)}{I + MR^2}$

since the yoyo is pulled in vertical direction, T = mg $a = \frac{mgR(R - r)}{I + MR^2} \\\\a = \frac{(0.15\times 9.8 \times 0.04)(0.04 - 0.0214)}{1.01 \times 10^{-4} \ + \ (0.15 \times 0.04^2)} \\\\a = 3.21 \ m/s^2$

<h3>Angular acceleration of the yoyo</h3>

α = a/R

α = 3.21/0.04

α = 80.25 rad/s²

<h3>Weight of the yoyo</h3>

W = mg

W = 0.15 x 9.8 = 1.47 N

<h3>Tension in the rope </h3>

T = mg = 1.47 N

<h3>Angular speed of the yoyo </h3>

v² = u² + 2as

v² = 0 + 2(3.21)(1.27)

v² = 8.1534

v = √8.1534

v = 2.855 m/s

ω = v/R

ω = 2.855/0.04

ω = 71.385 rad/s

Learn more about angular speed here: brainly.com/question/6860269

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3 0

2 years ago

A metal ball rolls from rest at Point A down the track to Point E as shown below.

Tju [1.3M]

Answer:

Explanation:

Velocity is at its greatest when kinetic energy is at its max which is when all the ball's energy has been transformed from potential energy to kinetic energy which is at the lowest point in its travels (assuming the ball is rolling down a ramp). You have no picture here so this answer is a general one, not a specific one.

8 0

2 years ago

What is the farthest distance parallaxes can be used to measure star distances from Earth?

JulsSmile [24]

Using current technology, useful parallax measurements can only be found for stars up to about 340 light years (100 parsecs) away.

8 0

3 years ago

A child throws a ball vertically upward to a friend on a balcony 28 m above him. The friend misses the ball on its upward flight

photoshop1234 [79]

Answer:

$t=1.9 sec$

Explanation:

From the question we are told that:

Height $h=28m$

Time $t=3s$

Generally the Newton's equation for Initial velocity upward is mathematically given by

$s=ut+\frtac{1}{2}at^2$

$28=3u-\frac{1}{2}*9.8*3^2$

$u=24.03m/s$

Generally the velocity at elevation and depression occurs as ball arrives and passes through S=28

$v=\sqrt{24.03-2*9.8*28}$

$v=5.35m/s and -5.35m/s$

Generally the Newton's equation for time to reach initial velocity is mathematically given by

$v=u+at$

$5.35=24.03-9.8t$

$t=\frac{28.03-5.35}{9.8}$

$t=1.9 sec$

4 0

2 years ago

Two college teams, A and B, are playing a game of tug of war. Team A applies a force of 103 newtons, but the rope does not move

Kipish [7]

Given that the rope is not moving (acceleration is zero), by the second Law of Newton (F=m*a), the net force acting on the rope is zero.

Then, the force applied by the team B equals the force applied by the tema A: 103 N.

7 0

3 years ago

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