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siniylev [52]
3 years ago
7

I throw a thin uniform circular disc (think of a frisbee) into the air so that it spins with angular velocity ω about an axis wh

ich makes an angle α with the axis of the disc. Show that the magnitude of ω is constant.
Physics
1 answer:
Ad libitum [116K]3 years ago
5 0

Answer:

| w | = w√ ( sin^2 ∝ + cos^2∝ ) shows that w is independent of time hence it is constant

Explanation:

prove that the magnitude of <em>w</em> is constant

attached below is a detailed solution of the question

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Two ships of equal mass are 110 m apart. What is the acceleration of either ship due to the gravitational attraction of the othe
dusya [7]

Answer:

Acceleration of the ship, a=2.14\times 10^{-7}\ m/s^2

Explanation:

It is given that,

Mass of both ships, m=39000\ metric\ tons=39\times 10^6\ kg

Distance between two ships, d = 110 m

The gravitational force between two ships is given by :

F=G\dfrac{m^2}{d^2}

F=6.67\times 10^{-11}\ Nm^2/kg^2\times \dfrac{(39\times 10^6\ kg)^2}{(110\ m)^2}

F = 8.38 N

Let a is the acceleration. Now, using second law of motion as :

a=\dfrac{F}{m}

a=\dfrac{8.38\ N}{39\times 10^6\ kg}

a=2.14\times 10^{-7}\ m/s^2

So, the acceleration of either ship due to the gravitational attraction of the other is 2.14\times 10^{-7}\ m/s^2. Hence, this is the required solution.

7 0
3 years ago
A car with a mass of 600 kg is traveling at a velocity of 10 m/s. How much kinetic energy does it have?
Viefleur [7K]

Answer:

KE = 30,000 J

Explanation:

KE = \frac{1}{2} mv²

KE = \frac{1}{2} (600)(10)²

KE = \frac{1}{2} (600)(100)

KE = \frac{1}{2} (60000)

KE = 30,000 J

5 0
4 years ago
The student makes a single pile of the 500 sheets of paper. Which a metre rule, she measured the height of the pile. The height
Cerrena [4.2K]

Answer:  please see attached work.

Explanation:  please see attached work.  Assuming 500 sheets of paper = 20 lb.  (typicical value).  

6 0
3 years ago
A 594 Ω resistor, an uncharged 1.3 μF capacitor, and a 6.53 V emf are connected in series. What is the current in milliamps afte
ivanzaharov [21]

Answer:

6.88 mA

Explanation:

Given:

Resistance, R = 594 Ω

Capacitance = 1.3 μF

emf, V = 6.53 V

Time, t = 1 time constant

Now,

The initial current, I₀ = \frac{\textup{V}}{\textup{R}}

or

I₀ = \frac{\textup{6.53}}{\textup{594}}

or

I₀ = 0.0109 A

also,

I = I_0[1-e^{-\frac{t}{\tau}}]

here,

τ = time constant

e = 2.717

on substituting the respective values, we get

I = 0.0109[1-e^{-\frac{\tau}{\tau}}]

or

I = 0.0109[1-2.717^{-1}]

or

I = 0.00688 A

or

I = 6.88 mA

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3 years ago
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