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3 years ago

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A majorette in the Rose Bowl Parade tosses a baton into the air with an initial angular velocity of 2.5 rev/s. If the baton unde

rgoes a constant acceleration while airborne of 0.2 rev/s2 and its angular velocity is 0.8 rev/s when the majorette catches it, how many revolutions does it make in the air?

1 answer:

oksian1 [2.3K]3 years ago

4 0

Answer:

14 rev

Explanation:

$w_{o}$ = initial angular velocity = 2.5 revs⁻¹

$w$ = final angular velocity = 0.8 revs⁻¹

$\alpha$ = Angular acceleration = - 0.2 revs⁻²

$\theta$ = Angular displacement

Using the equation

$w^{2} = w_{o}^{2} + 2 \alpha \theta\\0.8^{2} = 2.5^{2} + 2 (- 0.2) \theta\\ \theta = 14 rev$

So the number of revolutions are 14

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A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.

mash [69]

a) The acceleration of gravity is $4.96 m/s^2$

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

a)

The acceleration of gravity for an object near a planet is given by

$g=\frac{GM}{(R+h)^2}$

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2$

b)

The critical velocity for a satellite orbiting around a planet is given by

$v=\sqrt{\frac{GM}{R+h}}$

where we have again:

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s$

Converting into km/h,

$v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h$

c)

The period of the orbit is given by the circumference of the orbit divided by the velocity:

$T=\frac{2\pi (R+h)}{v}$

where

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

v = 6668 m/s

Substituting,

$T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s$

d)

One day consists of:

$t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s$

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

$n=\frac{t}{T}=\frac{86400}{8452}=10.2$

e)

The escape velocity for an object in the gravitational field of a planet is given by

$v=\sqrt{\frac{2GM}{R+h}}$

where here we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

Substituting, we find

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s$

f)

We can apply again the formula to find the escape velocity for the rocket:

$v=\sqrt{\frac{2GM}{R+h}}$

Where this time we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=0$ , because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

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3 years ago

Compare between chromosphere, corona and photosphere

gladu [14]

Answer:

Explained

Explanation:

Photosphere: The lowest layer of the sun is called photo sphere . It is about 300 miles thick from the surface. It is the source of solar flares. It is marked by bright bubbling granules of plasma.

chromosphere emits a reddish glow as the super heated hydrogen burns off but the red rim can only be seen during total solar eclipse.

The third layer of the sun atmosphere is Corona. It can also only be seen during during a total solar eclipse. Temperature in corona can reach as high as 3.5 million degree fahrenheit. As the gases cool they become solar winds.

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4 years ago

A proton traveling to the right enters a region of uniform magnetic field that points into the screen. When the proton enters th

muminat

Answer:

deflected toward bottom of the screen

Explanation:

When entering the region with magnetic field, a magnetic force is exerted on the proton. This force is perpendicular to both the direction of the magnetic field and the direction of the velocity of the proton.

The direction of the force can be determined by using the right-hand rule. We have:

- Index finger: direction of the velocity of the proton --> to the right

- Middle finger: direction of the magnetic field --> into the screen

- Thumb: direction of the magnetic force --> toward bottom of the screen

So, the correct answer is

deflected toward bottom of the screen

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3 years ago

A model airplane with a mass of 0.741kg is tethered by a wire so that it flies in a circle 30.9 m in radius. The airplane engine

liq [111]

(a) 24.6 Nm

The torque produced by the net thrust about the center of the circle is given by:

$\tau = F r$

where

F is the magnitude of the thrust

r is the radius of the wire

Here we have

F = 0.795 N

r = 30.9 m

Therefore, the torque produced is

$\tau = (0.795 N)(30.9 m)=24.6 N m$

(b) $0.035 rad/s^2$

The equivalent of Newton's second law for a rotational motion is

$\tau = I \alpha$

where

$\tau$ is the torque

I is the moment of inertia

$\alpha$ is the angular acceleration

If we consider the airplane as a point mass with mass m = 0.741 kg, then its moment of inertia is

$I=mr^2 = (0.741 kg)(30.9 m)^2=707.5 kg m^2$

And so we can solve the previous equation to find the angular acceleration:

$\alpha = \frac{\tau}{I}=\frac{24.6 Nm}{707.5 kg m^2}=0.035 rad/s^2$

(c) $1.08 m/s^2$

The linear acceleration (tangential acceleration) in a rotational motion is given by

$a=\alpha r$

where in this problem we have

$\alpha = 0.035 rad/s^2$ is the angular acceleration

r = 30.9 m is the radius

Substituting the values, we find

$a=(0.035 rad/s^2)(30.9 m)=1.08 m/s^2$

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3 years ago

What is a fuel?

Daniel [21]

Answer:

something that can be burned to produce heat

Explanation:

Combustion can be defined as an exothermic chemical reaction between physical substances, usually in the presence of oxygen and hydrocarbons to produce heat, light and carbon.

A fuel can be defined as something that can be burned to produce heat.

In the process of heat conduction, thermal energy is usually transferred from fast moving particles to slow moving particles during the collision of these particles. Also, thermal energy is typically transferred between objects that has different degrees of temperature and materials (particles) that are directly in contact with each other but differ in their ability to accept or give up electrons.

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3 years ago