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Leviafan
4 days ago
5

Which new process was added to the nebular theory of solar system formation to account for the existence of Jupiter-mass planets

that orbit close to their stars (the "hot Jupiters")?
Physics
1 answer:
serg [2.5K]4 days ago
7 0
Intense activity and magnetic fields are manifested by features, with sunspots occurring that are vastly larger than those seen on the Sun. These cause disturbances in the star's spectrum that exceed those produced by planets, rendering them significantly more difficult to detect, even in cases of hot Jupiters.
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An ideal spring is mounted horizontally, with its left end fixed. The force constant of the spring is 170 N/m. A glider of mass
kicyunya [2264]

Answer:

A) The updated amplitude = 0.048 m

B) Period T = 0.6 seconds

Explanation: Please refer to the attached documents for the solution.

4 0
18 days ago
Most calculators operate on 6.0 V. If, instead of using batteries, you obtain 6.0 V from a transformer plugged into 110-V house
kicyunya [2264]
1/0.0545. The transformation ratio of primary coil turns to secondary coil turns is directly proportional to the voltage transformation occurring. With 6.0 V on the secondary side (output) and 110 V on the primary side (input), the voltage ratio is calculated as 6/110 = 0.0545. This means for each turn in the primary coil, there are 0.0545 turns in the secondary coil.
6 0
11 days ago
A communications satellite orbiting the earth has solar panels that completely absorb all sunlight incident upon them. The total
Ostrovityanka [2208]

Answer:

0.000047N

Explanation:

We know that

intensity (I) = P/ A

Where

P= power

A= Area

Thus, the power absorbed can be calculated as:

Power = Intensity x Area

This equals = 1.4 x 10^3 x(10)

Thus,

14000 Watts = 14 kWatt

However, the radiation pressure can be defined as

time-averaged intensity divided by the speed of light in a vacuum

So,

P = (1.4 x 1000)/c

Also,

F= P x A

Thus,

((1.4 x 1000)/(3 x10^8)) x 10

This results in

=0.000046666N

Rounded to two significant figures gives us

=0.000047 N

3 0
14 days ago
The iron ball shown is being swung in a vertical circle at the end of a 0.7-m string. how slowly can the ball go through its top
Keith_Richards [2263]
<span>A centripetal force maintains an object's circular motion. When the ball is at the highest point, we can assume that the ball's speed v is such that the weight of the ball matches the required centripetal force to keep it moving in a circle. Hence, the string will not become slack. centripetal force = weight of the ball m v^2 / r = m g v^2 / r = g v^2 = g r v = sqrt { g r } v = sqrt { (9.80~m/s^2) (0.7 m) } v = 2.62 m/s Thus, the minimum speed for the ball at the top position is 2.62 m/s.</span>
6 0
26 days ago
camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Usin
Sav [2230]

The question lacks details. Here is the full question.

The accompanying image was captured with a camera capable of shooting between one and two frames per second. A series of photos was merged into this single image, meaning the vehicles depicted are actually the same car, documented at different intervals.

Assuming the camera produced 1.3 frames per second for this image and that the length of the car is approximately 5.3 meters, based on this information and the photo, how fast was the car moving?

Answer: v = 6.5 m/s

Explanation: The problem requires calculating the car's velocity. Velocity can be computed using:

v=\frac{\Delta x}{\Delta t}

Since the camera captured 7 images of the car and its length is noted as 5.3, the car's displacement is:

Δx = 7(5.3)

Δx = 37.1 m

The camera operates at 1.3 frames per second and recorded 7 images, thus the time driven by the car is:

1.3 frames = 1 s

7 frames = Δt

Δt = 5.4 s

<pconsequently the="" car="" was="" driving="" at:="">

v=\frac{37.1}{5.4}

v = 6.87 m/s

<pthe car="" moved="" at="" an="" estimated="">velocity of 6.87 m/s.

</pthe></pconsequently>
7 0
17 days ago
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