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horsena [70]
2 years ago
5

Which of the following statements cannot be supported by Kepler's laws of planetary motion?

Physics
1 answer:
Firdavs [7]2 years ago
7 0
The second
option is the answer
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A man starts walking north at 2 ft/s from a point p. five minutes later a woman starts walking south at 4 ft/s from a point 500
icang [17]
Refer to the diagram shown below.

After 5 minutes (300 seconds):
The man travels north by (2 ft/s)*(300 s) = 600 ft
The woman, located at q, 500 east of p, begins walking south at 4 ft/s.
The distance separating them is
d₁ = √(600² + 500²) = 781.025 ft

After 20 minutes:
The man has traveled for 20 minutes (1200 s).
The woman has traveled for 15 minutes (900 s).
The man has moved (2 ft/s)*(1200 s) = 2400 ft north of p.
The woman has moved (4 ft/s)*(900 s) = 3600 ft south of q.
The distance separating them is
d₂ = √(6000² + 500²) = 6020.8 ft

The separation from d₁ to d₂ occurs in 15 minutes (900s).
Therefore the rate of separation is
Rate = (d₂ - d₁ ft)/(900 s) = (6020.8 - 781.025)/900 = 5.822 ft/s
or
Rate = (5.822 ft/s)*(60 s/min) = 349.32 ft/min

Answer: 349.32 ft/min (or 5.82 ft/s)

4 0
3 years ago
What kind of friction exists between solid objects moving in water?
Stels [109]
The best and most correct answer among the choices provided by the question is  drag<span>.

</span><span>Drag </span>kind of friction exists between solid objects moving in water.

Hope my answer would be a great help for you.    
If you have more questions feel free to ask here at Brainly.  
6 0
2 years ago
A slender rod is 90.0 cm long and has mass 0.120 kg. A small 0.0200 kg sphere is welded to one end of the rod, and a small 0.080
deff fn [24]

Answer:

Speed of 0.08 kg mass when it will reach to the bottom position is 1.94 m/s

Explanation:

When rod is released from rest then due to unbalanced torque about the hinge the system will rotate

Now moment of inertia of the system is given as

I = \frac{ML^2}{12} + \frac{m_1L^2}{4} + \frac{m_2L^2}{4}

now we have

M = 0.120 kg

m_1 = 0.02 kg

m_3 = 0.08 kg

now we have

I = \frac{0.120(0.90)^2}{12} + \frac{0.02(0.90)^2}{4} + \frac{0.08(0.90)^2}{4}

so we have

I = 8.1 \times 10^[-3} + 4.05 \times 10^[-3} + 0.0162

I = 0.02835

now by energy conservation we can say work done by gravity must be equal to change in kinetic energy

so we have

\frac{1}{2}I\omega^2 = m_1g \frac{L}{2} - m_2 g\frac{L}{2}

\frac{1}{2}(0.02835)\omega^2 = (0.08 - 0.02)(9.81)(0.45)

\omega = 4.32 rad/s

Now speed of 0.08 kg mass when it reaches to bottom point is given as

v = \omega \frac{L}{2}

v = 4.32 (0.45)

v = 1.94 m/s

3 0
3 years ago
What slows down electricity?
tino4ka555 [31]

Answer:

D

Explanation:

4 0
2 years ago
Read 2 more answers
A uniform rod of length L rests on a frictionless horizontalsurface. The rod pivots about a fixed, frictionless axis at oneend.
coldgirl [10]

Answer: a ) 6/19 V/L

b ) 3/19

Explanation:

5 0
3 years ago
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