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2 years ago

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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.

1 answer:

Ad libitum [116K]2 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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You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

8 0

1 year ago

Read 2 more answers

Why is pure oxygen stored as a liquid under pressure

yKpoI14uk [10]

<h2>Answer: It is highly flammable.</h2>

Explanation:

Liquid oxygen is created from oxygen atoms that have been forced to assume the liquid state due to <u>compression (change of pressure) and temperature modification. </u>

Specifically this is achieved by cooling the oxygen enough to change it to its liquid state. So,<u> as the temperature drops, the atoms move more slowly because they have less energy. </u>

In this sense, in the liquid state it is easier to store and mobilize oxygen, taking into account that it is a highly flammable gas.

3 0

3 years ago

What best describes the volcanic process of magma with dissolved water vapor and other gases as it nears the Earth surface?

Assoli18 [71]

Answer:

Pressure on the molten rock lessens and the gases dissolved in rock can bubble and expand rapidly causing violent eruptions.

Explanation:

that's just how it works lol. hope this helps :]

6 0

2 years ago

Why is the sky blue and doesn't change color

Softa [21]

The statement is <u>false</u> because the sky <u>can change </u>colors during sunsets, sun rises, etc. The sky is not always blue.

7 0

2 years ago

If a car is accelerating downhill under a net force 3674 N , what additional force would cause the car to have a constant veloci

Semenov [28]

Answer:

For the car to move with constant velocity the additional force required is $F__{dg} }= -3674 \ N$

Explanation:

From the question we are told that

The net force of the car is $F_{net} = 3674 \ N$

Generally the total force acting on the car is the net force plus the force due to gravity acting in direction of the car (Let denote it as $F__{dg}$ )

So the total force acting on the car is mathematically represented as

$F_{net} + F__{dg}} = F$

Here this F representing the total force can be mathematically represented as

$F = ma$

Now for constant velocity to be attained, the acceleration of the car will be zero

So at constant velocity

$F = m * 0$

=> $F = 0 \ N$

So

$F_{net} + F__{dg}} = 0$

=> $F__{dg} }= -F_{net}$

=> $F__{dg} }= -3674 \ N$

6 0

2 years ago