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lara31 [8.8K]
4 years ago
8

A student holds one end of a thread, which is wrapped around a cylindrical spool, as shown above. The student then drops the spo

ol from a height h above the floor, and the thread unwinds as it falls. The spool has a mass M and a radius R, and the thread has negligible mass. The spool can be approximated as a solid cylinder of moment of inertia I = 1 MR2. Express your answers in terms of M, R, h, and fundamental constants. At time t = 0, the spinning spool lands on the floor without bouncing and comes free from the thread. It continues to spin, but slips on the floor's surface while doing so. Assume a constant coefficient of sliding friction m. (c) Calculate the angular velocity of the spool as a function of time t.
Physics
1 answer:
lesya [120]4 years ago
7 0

by energy conservation we know that

KE or rotation + KE of translation = gravitational PE

now we have

\frac{1}{2}I\omega^2 + \frac{1}{2}mv^2 = mgH

also we know that

v = R\omega

now we have

\frac{1}{2}(\frac{1}{2}mR^2)\omega^2 + \frac{1}{2}m(R\omega)^2 = mgH

\frac{3}{4}mR^2\omega^2 = mgH

\omega = \sqrt{\frac{4gH}{3}}/R

now when it is rolling on ground the torque acting on it due to friction force is given by

\tau = R F_f

\tau = \mu mg R

\alpha = \frac{\mu mg R}{\frac{1}{2}mR^2}

\alpha = \frac{2 \mu g}{R}

now angular speed at any time is given as

\omega = \omega_i + \alpha t

\omega = \sqrt{\frac{4gH}{3}}/R -\frac{2 \mu g}{R} t

so above is the angular speed in terms of time "t"

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The half-life of the radioactive element beryllium-13 is 5 × 10-10 seconds, and half-life of the radioactive element beryllium-1
telo118 [61]
<h2>Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>

Explanation:

The half-life h of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.

In this case, we are given the half life of two elements:

beryllium-13: h_{B-13}=5(10)^{-10}s=0.0000000005s

beryllium-15: h_{B-15}=2(10)^{-7}s=0.0000002s

As we can see, the half-life of beryllium-15 is greater than the half-life of beryllium-13, but how great?

We can find it out by the following expression:

h_{B-15}=X.h_{B-13}

Where X is the amount we want to find:

X=\frac{h_{B-15}}{h_{B-13}}

X=\frac{2(10)^{-7}s}{5(10)^{-10}s}

Finally:

X=400

Therefore:

The half-life of beryllium-15 is <u>400 times greater than</u> the half-life of beryllium-13.

8 0
3 years ago
A ship is travelling due east at 30 km/hr and a boy runs across the deck
dsp73

Answer:

Vr = 20 [km/h]

Explanation:

In order to solve this problem, we have to add the relative velocities. We must remember that velocity is a vector, therefore it has magnitude and direction. We will take the sea as the reference measurement level.

Let's take the direction of the ship as positive. Therefore the boy moves in the opposite direction (Negative) to the reference level (the sea).

V_{r}=30-10\\V_{r}=20 [km/h]

8 0
3 years ago
A light spring obeys Hooke's law. The spring's unstretched length is 34.0 cm. One end of the spring is attached to the top of a
sleet_krkn [62]

When the spring is extended by 44.5 cm - 34.0 cm = 10.5 cm = 0.105 m, it exerts a restoring force with magnitude R such that the net force on the mass is

∑ F = R - mg = 0

where mg = weight of the mass = (7.00 kg) g = 68.6 N.

It follows that R = 68.6 N, and by Hooke's law, the spring constant is k such that

k (0.105 m) = 68.6 N   ⇒   k = (68.6 N) / (0.105 m) ≈ 653 N/m

5 0
2 years ago
Is a force necessary to keep an object in motion?
weqwewe [10]
Well according to Newton’s first law of motion, a body will remain in the state of rest or linear motion provided that an *external force* has been applied. So no, a force doesn’t need to keep a body to remain in linear motion, because F=ma, during uniform linear motion velocity is constant, hence acceleration is zero, so F=0
5 0
3 years ago
Air bags are designed to deploy in 10 ms. Given that the air bags expand 20 cm as they deploy, estimate the acceleration of the
joja [24]

As it is given that the air bag deploy in time

t = 10 ms = 0.010 s

total distance moved by the front face of the bag

d = 20 cm = 0.20 m

Now we will use kinematics to find the acceleration

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0.20 = 0 + \frac{1}{2}a*0.010^2

0.20 = 5 * 10^{-5}* a

a = 4000 m/s^2

now as we know that

g = 10 m/s^2

so we have

a = 400g

so the acceleration is 400g for the front surface of balloon

3 0
3 years ago
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