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ki77a [65]
3 years ago
12

An astronaut orbits planet Y in her spaceship. To remain in orbit at 410km above the planet's center, she maintains a speed of 6

8 m/s. What is the mass of planet Y? Note: the constant of universal gravity (G) equals 6.674 × 10-11 N ⋅ m2/kg2.
Physics
1 answer:
choli [55]3 years ago
3 0
Formula for orbital speed, v =  √(GM/R)

Where G is the universal gravitational constant, M = Central Mass,

R = Distance between centers of Mass.

Given.  v = 68 m/s, M = ? , R = 410 km = 410000 m., G = 6.674 * 10⁻¹¹ Nm²/kg²

68 = √(GM/R)

68 = √(6.674 * 10⁻¹¹ * M/410000)

68² =  (6.674 * 10⁻¹¹ * M)/410000

(68²  * 410000) / 6.674 * 10⁻¹¹    = M

2.84 × 10¹⁹    = M

Mass of Planet Y =  2.84 × 10¹⁹ kg
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Answer:

\frac{d^2x}{dt^2}+\frac{\beta}{m}\frac{dx}{dt}+\frac{k}{m}x=0

Explanation:

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m\frac{d^2x}{dt^2}=-kx-\beta \frac{dx}{dt}

where x(t) is the displacement from equilibrium position. The equation can be transformed into:

\frac{d^2x}{dt^2}+\frac{\beta}{m}\frac{dx}{dt}+\frac{k}{m}x=0  shich is the equation of motion.

7 0
3 years ago
The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8
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Missing figure and missing details can be found here:
<span>http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....
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Solution:
(a) The work done by the spring is given by
W= \frac{1}{2} k (\Delta x)^2 &#10;
where k is the elastic constant of the spring and \Delta x is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
W= \frac{1}{2} \cdot 500 N/m \cdot (0.127m-(-0.203m))^2=27.25 J

(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
W_W = -F_{//} (x_2 -x_1)
where  the negative sign is given by the fact that F_{//} points in the opposite direction of the displacement of the cart, and where
F_{//}=m g sin 15^{\circ}=6 kg \cdot 9.81m/s^2 \cdot sin 15^{\circ}=15.2 N
therefore, the work done by the weight is
W_W=-15.2 N \cdot (0.203m-(-0.127m))=-5.02 J

8 0
3 years ago
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Leni [432]

Answer:

Explanation:

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4 0
3 years ago
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A force of 20.0 N is applied to a 3.00 kg object for 4.00 seconds. Calculate the impulse experienced by the object.​
GenaCL600 [577]

Answer:

Impulse = 80Ns

Explanation:

Given the following data;

Mass = 3kg

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Time = 4 seconds

To find the impulse experienced by the object;

Impulse = force * time

Impulse = 20*4

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Therefore, the impulse experienced by the object is 80 Newton-seconds.

7 0
3 years ago
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