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3 years ago

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A planet orbits a star, in a year of length 2.35 x 107 s, in a nearly circular orbit of radius 3.49 x 1011 m. With respect to th

e star, determine (a) the angular speed of the planet, (b) the tangential speed of the planet, and (c) the magnitude of the planet's centripetal acceleration.

1 answer:

Naya [18.7K]3 years ago

7 0

Answer:

a) w = 9.599 10⁴ rad / s , b) v = 3.35 10¹⁶ m / s , c) a = 3.22 10²¹ m / s²

Explanation:

For this exercise we must use the relation of angular kinematics

a) angular velocity, the distance remembered in orbit between time (period)

w = 2π r / T

w = 2 π 3.59 10¹¹ / 2.35 10⁷

w = 9.599 10⁴ rad / s

b) linear and angular velocity are related by the equation

v = w r

v = 9,599 10⁴ 3.49 10¹¹

v = 3.35 10¹⁶ m / s

c) the centripetal acceleration is

a = v² / r = w² r

a = (9,599 10⁴)² 3.49 10¹¹

a = 3.22 10²¹ m / s²

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Answer:

9.4 m

Explanation:

We can use a moving frame of reference with the same speed as the car. From this frame of reference the car doesn't move. The origin is at the back of the car, the positive X axis points back and the positive Y axis points up.

If the ballon is launched at 9.7 m/s at 39 degrees of elevation.

Vx0 = 9.7 * cos(39) = 7.5 m/s

Vy0 = 9.7 * sin(39) = 6.1 m/s

If we ignore air drag, the baloon will be subject only to the acceleration of gravity. We can use the equation of position under constant acceleration.

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

Y0 = 0

a = -9.81 m/s^2

It will fall when Y(t) = 0

0 = 6.1 * t - 4.9 * t^2

0 = t * (6.1 - 4.9 * t)

t1 = 0 (this is when the balloon was launched)

0 = 6.1 - 4.9 * t2

4.9 * t2 = 6.1

t2 = 6.1 / 4.9 = 1.25 s

The distance from the car will be the horizonta distance it travelled in that time

X(t) = X0 + Vx0 * t

X(1.25) = 7.5 * 1.25 = 9.4 m

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3 years ago

What is the power of an electric with a current of 0.5 A and a voltage of 120 V ? The formula for power is P =IV

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Answer:

60Watts

Explanation:

Given parameters:

Current = 0.5A

Voltage = 120V

Unknown

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3 years ago

Read 2 more answers

Three identical train cars, coupled together, are rolling east at speed v0. A fourth car traveling east at 2v0 catches up with t

Bezzdna [24]

Answer:

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Explanation:

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P=mv where m is mass and v is velocity hence

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The final mass is 2m+3m=5m

From aw of conservation of momentum

$5mv_0=5mv$ hence $v=1v_o$

Therefore, the speed is $1v_0$

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3 years ago

The cue ball is deflected so that it makes an angle of 30.0° with its original direction of travel.

Alexus [3.1K]

Answer:

(a) θ= 43.89°

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Explanation:

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$u_{1} = 7.52\frac{m}{s}$

Ball 2:

$u_{2} = 0\frac{m}{s}$

As the collision is elastic, it means that kinetic energy and momentum are conserved. Following that, we apply the law of conservation of energy and momentum:

$\frac{1}{2} m_{1}u_{1}^{2} = \frac{1}{2} m_{1}v_{1}^{2} + \frac{1}{2} m_{2}v_{2}^{2}$

and

$m_{1}u_{1} = m_{1}v_{1} + m_{2}v_{2}$

Where u is the velocity before the collision, and v are the velocities after the collision. Both previous equations can be simplified as:

$u_{1}^{2} = v_{1}^{2} + v_{2}^{2}$

and

$u_{1} = v_{1} + v_{2}$

This because the two balls have the same mass. We know that the cue ball is deflected and makes an angle of 30°. From conservation in x-direction, we get::

$56.55 = v_{1} ^{2} + v_{2} ^{2}$

and

$u_{1} = v_{1}cos(30) + v_{2}cos(\theta)$

Solving we get:

$(\frac{2u_{1}-\sqrt{3}v_{1} }{2cos(\theta)})^{2} = 56.55-v_{1} ^{2}$

From conservation in y-direction, we get:

$0 = v_{1}sin(30) - v_{2}sin(\theta)$

From this, we solve this equation system and get the answers. Remember to add 30° to the angle obtained.

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3 years ago

A 2.2-m-long steel rod must not stretch more than 1.2 mm when it is subjected to a 8.5-kN tension force. Knowing that E = 200 GP

kodGreya [7K]

Answer:

a) 0.00996 m

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Explanation:

Unit conversions:

$E = 200GPa = 200\times10^9 Pa$

1.2 mm = 0.0012 m

8.5 kN = 8500 N

If the 2.2m rod cannot stretch more than 0.0012 m, its maximum strain is

$\epsilon = \frac{\Delta L}{L} = \frac{0.0012}{2.2} = 0.000545455$

With elastic modulus being E = 200 GPa, then its maximum stress must be

$\sigma = E\epsilon = 200\times10^9*0.000545455 = 109090909 Pa$

Knowing the tension force being F = 8500 N, we can calculate the appropriate cross section area

$A = \frac{F}{\sigma} = \frac{8500}{109090909} = 7.79\times10^{-5}m^2$

And its corresponding diameter is

$A = \pi d^2/4$

$7.79\times10^{-5} = \pi d^2/4$

$d^2 = \frac{4*7.79\times10^{-5}}{\pi} = 9.92\times10^{-5}$

$d = \sqrt{9.92\times10^{-5}} = 0.00996 m \approx 1 cm$

7 0

3 years ago