Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the Sun relative to the center of mass of the system during the time Saturn completes half an orbit. Assume the mass of the Sun is 5.68 x10^29 kg, the mass of Saturn is 5.68 x10^26 kg, its period is 9.29 x10^8 s, and the radius of its orbit is 1.43 x 10^12 m. Ignore the influence of other celestial objects.?
Answers
Answer:
Explanation:
If the sun considered as x=0 on the axis to put the center of the mass as a:
solve to r1
Now convert to coordinates centered on the center of mass. call the new coordinates x' and y' (we won't need y'). Now since in the sun centered coordinates the angular momentum was
where T = orbital period
then L'(x',y') = L(x) by conservation of angular momentum. So that means
Since
then
Final answer:
In a two-body system such as the Sun-Saturn system, both bodies orbit around their mutual center of mass, or barycenter. Given the Sun's significantly larger mass, this barycenter is near the center of the Sun, and hence the Sun's change in velocity relative to the center of mass of the system as Saturn completes half an orbit is effectively zero.
Explanation:
The problem here is asking for the change in velocity of the Sun relative to the center of mass of the Sun-Saturn system as Saturn completes half an orbit. This is a situation involving orbital physics and center of mass systems.
However, in an isolated two-body orbit system like this, the center of mass does not change velocity - it would remain constant, not unless acted upon by an outside force, which the problem instructs us to ignore.
Saturn and the Sun both orbit around their common center of mass (their barycenter). Given that the Sun is immensely more massive than Saturn, this center of mass is located very close to the center of the Sun.
So, while the Sun does indeed move a little due to Saturn's influence, the change in velocity of Sun relative to the center of mass of the system during the time Saturn completes half an orbit, for all intents and purposes, is zero.
This is especially true unless the problem specifically mentions that the Sun is initially at rest with respect to the center of mass. In any other case, the relative velocity remains constant and hence the change is zero.
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A circular coil lies flat on a horizontal surface. A bar magnet is held above the center of the coil with its north pole pointing downward, and is released from rest. What is the direction of the induced current in the coil, as viewed from above, as the magnet approaches the coil in free fall?a. clockwiseb. counterclockwisec. There is no induced current in the coil.
Answers
The magnitude of the average force applied by the floor on the ball is 60 N
From the question given above, the following data were obtained:
Mass (m) = 0.15 Kg
Initial velocity (u) = 6.5 m/s
Final velocity (v) = 3.5 m/s
Time (t) = 0.025 s
Force (F) =?
The magnitude of the averageforce applied by the floor on the ball can be obtained as follow:
F = 60 N
Thus, the magnitude of the average force applied by the floor on the ball is 60 N
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Answer: Change in ball's momentum is 1.5 kg-m/s.
Explanation: It is given that,
Mass of the ball, m = 0.15 kg
Speed before the impact, u = 6.5 m/s
Speed after the impact, v = -3.5 m/s (as it will rebound)
We need to find the change in the magnitude of the ball's momentum. It is given by :
So, the change in the ball's momentum is 1.5 kg-m/s. Hence, this is the required solution.
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Answers
Answer:
T=12544 N*m
Explanation:
Given
L=4.0m
ms=500kg
mw=70kg
Torque is the force in a distance the relation is proportional so the torque of weight first is:
Ts = Fs*d
Ts = ms*g*L
Ts = 500kg*9.8m/s^2*2m
Ts = 9800 N*m
now torque of the worker
Tw = Fw*d
Tw = 70kg*9.8m/s^2*4m
Tw = 2744 N*m
Torque net is
Tnet = Tw+Ts
Tnet= 2744 + 9800 =12544 N*m
Final answer:
The total torque about the bolt due to the worker and the weight of the beam is 12544 Nm. This is found by adding the torque due to the beam and the worker which can be calculated using their weights and their distance from the pivot point (bolt).
Explanation:
The key to solving this question is understanding torque, which in physics represents the rotational effect of a force. Torque is calculated using the formula τ = r x F, where τ is the torque, r is the distance from the pivot point, and F is the force applied.
In this case, there are two forces to consider: the weight of the beam and the weight of the worker. Both of these can be calculated using the formula for weight (F = m*g), where m is mass and g is gravitational acceleration, which is approximately 9.8 m/s^2 on Earth. The weight of the beam is therefore 500 kg * 9.8 m/s^2 = 4900 N, and the weight of the worker is 70 kg * 9.8 m/s^2 = 686 N.
The distance from the pivot (bolt) for the beam's weight is considered to be the midpoint of the beam, so it is 4.0 m / 2 = 2.0 m. For the worker, r equals the full length of the beam, which is 4.0 m. The total torque can be calculated by adding the torque due to the beam and the worker. Therefore, the total torque τ = (2.0 m * 4900 N) + (4.0 m * 686 N) = 9800 Nm + 2744 Nm = 12544 Nm.
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Answers
Answer:
The correct answer is d. tension pneumothorax.
Explanation:
The increasing build-up of air that is in the pleural space is what we call the tension pneumothorax and this happens due to the lung laceration that lets the air to flee inside the pleural space but it does not return.
O A. The Moon spinning on its axis
O B. The Sun spinning on its axis
C. Earth orbiting the Sun
D. A ballet dancer spinning in place
Answers
Answer:
Option C
Explanation:
Revolution: When an object moves around another object it is called revolution.
Rotation: When an object spins around its axis it's called rotation
Answer:C
Explanation:
Good luck!
Answers
The correct option is option (1)
The faster movement of air on the upper surface of the paper creates lower pressure above the paper.
Pressure difference:
The movement of air is always from a region of higher pressure to a region of lower pressure.
As we blow air above the paper strip a low pressure is created above the strip due to the fast movement or high speed of the air. And the pressure below the strip is higher in comparison to the pressure above since the air below is not moving.
So, due to the pressure difference, a force is generated on the paper strip by the air from the lower surface to the upper surface.
Learn more about pressure difference:
This is happened because "the air" above "moves faster" and "the pressure" is "lower".
Option: 1
Explanation:
Air movement take place from the region where air pressure is more than the region where the pressure is low. When we "blow" air above the "paper strip" paper rises because "low pressure" is created above the strip as high speed winds always travel with reduced air pressure. Hence due to higher air pressure below the strip, it is pushed upwards. This difference in pressure results into fast air moves. This happen because "speed" of the wind depends on "the difference between the pressures" of the air in the two regions.
Answers
Answer:
v = 4,244,699 m/s = (4.245 × 10⁶) m/s
Explanation:
The electric force on the proton is given by
F = qE
where q = charge on the proton = (1.602 × 10⁻¹⁹) C
E = Electric field = 720,000 N/C
F = (1.602 × 10⁻¹⁹ × 720000)
F = (1.153 × 10⁻¹³) N
But this force will accelerate the proton in this magnetic field in a form of trajectory motion.
We can obtain the acceleration using Newton's first law of motion relation
F = ma
m = mass of a proton = (1.673 × 10⁻²⁷) kg
a = (F/m)
a = (1.153 × 10⁻¹³)/(1.673 × 10⁻²⁷)
a = 68,944,411,237,298 m/s²
a = (6.894 × 10¹³) m/s²
This acceleration directs the proton from the positive plate to the negative plate, covering a distance of y = 0.006 m (the distance between the plates)
Using Equations of motion, we can obtain the time taken for the proton to move from the rest at the positive plate to the negative one.
u = initial velocity of the proton = 0 m/s
y = vertical distance covered by the proton = 0.006 m
a = acceleration of the proton in this direction = (6.894 × 10¹³) m/s²
t = time taken for the proton to complete this distance = ?
y = ut + (1/2) at²
0.006 = 0 + [(1/2)×(6.894 × 10¹³)×t²]
0.006 = (3.447 × 10¹³) t²
t² = (0.006)/(3.447 × 10¹³)
t² = 1.741 × 10⁻¹⁶
t = (1.32 × 10⁻⁸) s
Then we can then calculate the minimum speed to navigate the entire length of the plates without hitting the plates.
v = ?
x = 0.056 n
t = (1.32 × 10⁻⁸)
v = (x/t)
v = (0.056)/(1.32 × 10⁻⁸)
v = 4,244,699 m/s = (4.245 × 10⁶) m/s
Hope this Helps!!!
Answer:
v = 9.09×10⁵m/s
Explanation:
Given
d = the distance between plates = 0.6cm = 0.006
E = Electric field strength = 720000N/C
m =mass of the proton = 1.67 ×10-²⁷ kg
The
Electric potential energy of the field is converted into the the kinetic energy of the proton.
So
qV = 1/2mv²
But V = Ed
So q(Ed) = 1/2mv²
v² = 2qEd/m
v = √(2qEd/m)
v = √(2×1.6×10-¹⁹×720000×0.006/1.67×10-²⁷)
v = 9.09×10⁵m/s