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3 years ago

Which of the following remains constant for a satellite in an elliptical orbit.

1 answer:

5 0

If the satellite doesn't have little rocket engines or other thrusters on it, AND it stays far enough from Earth that it doesn't have to plow through any air molecules, AND no pieces break off of it and drift away, AND there are no hamsters inside it running on treadmills connected to external thrusters, then there's no way for it to gain or lose energy, and its total energy remains constant.

Some of its energy is always changing, either from potential to kinetic or from kinetic to potential, as its distance from Earth changes. But the total stays constant.

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The scientist most often credited with the idea that matter can have wave-like properties is:

Flura [38]

I believe the correct answer from the choices listed above is option B. The scientist most often credited with the idea that matter can have wave-like properties is de Broglie. He related E=mc^2, E=hf and p=mv in order to acquire the de Broglie wavelength, whose formula λ=h/p where h is Planck's constant and p is momentum of the particle.

6 0

3 years ago

Read 2 more answers

In the study of sound, one version of the law of tensions is:

Phantasy [73]

Answer:

F1 = 80

Explanation:

f1= f2 √ (F1/F2)

Where f1 = 300, f2 = 260 and F2 = 60

Putting in the above formula

300 = 260√(F1/60)

Dividing both sides by 260

=> 1.15 = √(F1/60)

Squaring both sides

=> 1.33 = F1/60

Multiplying both sides by 60

=> F1 = 80

5 0

3 years ago

A hamster in it's ball starts at rest and accelerates to 3ms1 in 6 seconds.

taurus [48]

Answer:9m

Explanation:

Ball starts from rest . Time taken = 6 seconds. Distance travelled by ball. ∴Distance travelled = 9 m

Hope it helps you

Good luck

7 0

3 years ago

At time t=0 a grinding wheel has an angular velocity of 30.0 rad/s . It has a constant angular acceleration of 35.0 rad/s2 until

Rama09 [41]

Answer:

(A) 570 rad

(B) 10 s

Explanation:

The equations of motion for circular motions are used.

Initial angular velocity, $\omega_0 = 30.0 \text{ rad/s}$

Angular acceleration, $\alpha =35.0 \text{ rad/s}^2$

(A)

At t = 2.00 s, the angular displacement, θ, is given by

$\theta = \omega_0t+\frac{1}{2}\alpha t^2 = (30\times 2) + \frac{1}{2}\times35\times2^2=60+70 = 130\text{ rad}$

After this time, it decelerates through an angular displacement of 440 rad.

Total angular displacement = 130 + 440 rad = 570 rad

(B)

At the time the circuit breaker tips, the angular velocity is given by

$\omega = \omega_0+\alpha t = 30.0+(35.0\times 2) = 30.0+70.0 =100.0\ \text{rad/s}$

This becomes the initial angular velocity for the decelerating motion. Because it stops, the final angular velocity is 0 rad/s. The time for this part of the motion is calculated thus:

$\theta_2 = \left(\dfrac{\omega_i+\omega_f}{2}\right)t$