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2 years ago

Describe a theme-park ride that has constant speed but changing velocity. Thank you!

1 answer:

8 0

How about a carousel (merry go round).
For any one horse or rider, Speed is constant but direction keeps changing, so velocity does too.

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A satellite is in circular orbit at an altitude of 1500 km above the surface of a nonrotating planet with an orbital speed of 9.

Ksju [112]

To solve this problem we will use the Newtonian theory about the speed of a body in space for which the speed of a body in the orbit of a planet is summarized as:

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

Where,

G = Gravitational Universal Constant

M = Mass of Planet

r = Radius of the planet ('h' would be the orbit from the surface)

The escape velocity is

$v = 14.9km/h = 14900m/s$

Through this equation we can find the mass of the Planet in function of the distance, therefore

$M = \frac{v^2R}{2G}$

$M = \frac{14900^2R}{2(6.67*10^{-11})}$

$M = 16.64*10^{17}R$

The orbital velocity is

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

$9200^2 = \frac{(6.67*10^{-11})(16.64*10^{17})R}{R+1500*10^3}$

$11.1*10^7R = (R+15000*10^3)(9200)^2$

$2.64*10^7R = 12.69*10^{13}$

$R = 4.81*10^6m$

The time period of revolution is,

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

$T = \frac{2\pi(4.81*10^6+1.5*10^6)}{9200}$

$T = 4307s$

$T = 72min = 1hour12min$

Therefore the orbital period of the satellite is closes to 1 hour and 12 min

3 0

2 years ago

You go to an amusement park with your friend Betty, who wants to ride the 70-m-diameter Ferris wheel. She starts the ride at the

Ksenya-84 [330]

Answer:

Part a)

$d = 23.94 m$

Part b)

$\theta = 110 degree$

Explanation:

Part a)

As we know that initially the position vector is r

then the same magnitude position vector is rotated by 40 degree angle

so displacement magnitude is the magnitude of change in position vector

so it is given as

$d = \sqrt{r_1^2 + r_2^2 + 2r_1r_2cos(180-\theta)}$

$r_1 = r_2 = 35 m$

$d = \sqrt{35^2 + 35^2 -2(35)(35)cos40}$

$d = 23.94 m$

Part b)

now we need to find the direction of the displacement vector

so let say it makes an angle with x axis so we have

$tan\theta = \frac{rsin\phi}{r - rcos\phi}$

$tan\theta = \frac{sin 40}{1 - cos40}$

$\theta = 110 degree$

4 0

2 years ago

Which best describes what forms in nuclear fission?A. two smaller, more stable nucleiB. two larger, less stable nucleiC. one sma

Step2247 [10]

Answer:B. two larger, less stable nuclei

Explanation: They collied and don't combine

4 0

2 years ago

Read 2 more answers

The block in the diagram below is AT REST. However, the tension in the cable is not the only thing holding the block back. Stati

Vedmedyk [2.9K]

Answer:

The tension in the rope is 229.37 N.

Explanation:

Given:

Mass of the block is, $m=33.2\ kg$

Coefficient of static friction is, $\mu = 0.214$

Angle of inclination is, $\theta = 31.5$ °

Draw a free body diagram of the block.

From the free body diagram, consider the forces in the vertical direction perpendicular to inclined plane.

Forces acting are $mg\cos \theta$ and normal $N$ . Now, there is no motion in the direction perpendicular to the inclined plane. So,

$N=mg\cos \theta\\N=(33.2)(9.8)\cos (31.5)\\N=277.415\ N$

Consider the direction along the inclined plane.

The forces acting along the plane are $mg\sin \theta$ and frictional force, $f$ , down the plane and tension, $T$ , up the plane.

Now, as the block is at rest, so net force along the plane is also zero.

$T=mg\sin \theta+f\\T=mg\sin \theta +\mu N\\T= (33.2)(9.8)(\sin (31.5)+(0.214\times 277.415)\\T= 170+59.37\\T=229.37\ N$

Therefore, the tension in the rope is 229.37 N.

3 0

2 years ago

If it were possible to remove gravity and friction, think about what would happen to a football if it were tossed into the air.

elena-14-01-66 [18.8K]

Ignoring fluid resistance, football will <span>maintain a constant speed until other forces accelerate the football.</span>

6 0

2 years ago

Read 2 more answers