PHYSICS COLLEGE

A child attaches a 0.6 kg toy to a 0.9 meter length of string and spins it around in uniform circular motion. Calculate the tension in the string if the period of rotation is 4.1 seconds.

Answers

Answer 1

Answer:

Answer:

1.26812 N

Explanation:

$m$ = Mass of toy = $0.6\ \text{kg}$

$r$ = Length of string = $0.9\ \text{m}$

$t$ = Period of rotation = $4.1\ \text{s}$

Time period is given by

$t=(2\pi r)/(v)\n\Rightarrow v=(2\pi r)/(t)\n\Rightarrow v=(2\pi 0.9)/(4.1)\n\Rightarrow v=1.3792\ \text{m/s}$

The rotational velocity is 1.3792 m/s

The tension in the rope will be the centripetal force acting on the toy

$T=(mv^2)/(r)\n\Rightarrow T=(0.6* 1.3792^2)/(0.9)\n\Rightarrow T=1.26812\ \text{N}$

The tension in the string is 1.26812 N.

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Two particles, one with charge -6.29 × 10^-6 C and one with charge 5.23 × 10^-6 C, are 0.0359 meters apart. What is the magnitude of the force that one particle exerts on the other?
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Look at the Kunoichi's of Naruto but as a Gang. Who do you think looks the baddest out of the group(but in a good way)?

My opinion is Hinata...just saying

Answers

Answer:

hinata for sure

Explanation:

seems reasonable

A uniform, 4.5 kg, square, solid wooden gate 2.0 m on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.2 kg raven flying horizontally at 4.5 m/s flies into this door at its center and bounces back at 1.5 m/s in the opposite direction. What is the angular speed of the gate just after it is struck by the unfortunate raven?

Answers

Answer:

Explanation:

Mass of the gate, $m_1 = 4.5 kg$

Mass of the raven, $m_2 = 1.2 kg$

Initial speed of raven, $v_1 = 4.5 m/s$

Final speed of raven, $v_2 = - 1.5 m/s$

Moment of Inertia of the gate about the axis passing through one end:

$I = (1)/(3) m_1 a^2\nI = (1)/(3) *4.5 * 2^2\nI = 6 kg m^2$

Angular momentum of the gate, $L = I \omega$

$L = 5.33 \omega$

Using the law of conservation of angular momentum:

$m_2 v_f (a/2) + I\omega = m_2v_i (a/2)\nI\omega = m_2 (a/2)(v_i - v_f)\n$

what is the necessary condition on a force the result the conservation of angular momentum for a particle affected by that force?

Answers

Answer:

The force must be applied on the axis of rotation

Explanation:

A rotating system conserves its angular momentum only if there are no external torques on the system. In other words, the external torque must be equal to zero.

T=0

T=Fxd

Torque is equal to the vector product of a force by the distance between the axis of rotation and where the force is applied.

For this product to be zero, the force must be applied on the axis of rotation (d=0).

An element emits light at two nearly equal wavelengths, 577 nm and 579 nm If the light is normally incident on a diffraction grating with 2000 lines/cm., what is the distance between the 3rd order fringes of the two wavelengths on a screen 1 m from the grating?

Answers

Answer:

Explanation:

d = width of slit = 1 / 2000 cm =5 x 10⁻⁶ m

Distance of screen D = 1 m.

wave length λ₁ and λ₂ are 577 x 10⁻⁹ and 579 x 10⁻⁹ m.respectively.

distance of third order bright fringe = 3.5 λ D/d

for 577 nm , this distance = 3.5 x 577 x 10⁻⁹ x 1 /5 x 10⁻⁶

= .403 m = 40.3 cm

For 579 nm , this distance = 3.5 x 579 x 1 / 5 x 10⁻⁶

= 40.5 cm

Distance between these two = 0.2 cm.

Your friend is trying to construct a clock for a craft show and asks you for some advice. She has decided to construct the clock with a pendulum. The pendulum will be a very thin, very light wooden bar with a thin, but heavy, brass ring fastened to one end. The length of the rod is 80 cm and the diameter of the ring is 10 cm. She is planning to drill a hole in the bar to place the axis of rotation 15 cm from one end. She wants you to tell her the period of this pendulum.

Answers

Answer:

The period of the pendulum is $T = 1.68 \ sec$

Explanation:

The diagram illustrating this setup is shown on the first uploaded image

From the question we are told that

The length of the rod is $L = 80 \ cm$

The diameter of the ring is $d = 10 \ cm$

The distance of the hole from the one end $D = 15cm$

From the diagram we see that point A is the center of the brass ring

So the length from the axis of rotation is mathematically evaluated as

$AP = 80 + 10 -5 -15$

$AP = 70 \ cm = (70)/(100) = 0.7 \ m$

Now the period of the pendulum is mathematically represented as

$T = 2 \pi \sqrt{(AP)/(g) }$

$T = 2 \pi \sqrt{(0.7)/(9.8 ) }$

$T = 1.68 \ sec$

At one instant, a 17.0-kg sled is moving over a horizontal surface of snow at 4.10 m/s. After 6.15 s has elapsed, the sled stops. Use a momentum approach to find the magnitude of the average friction force acting on the sled while it was moving.