Answer: It takes 12 N of force to pull the wagon.
Explanation:
Determine how would the frequency of the pendulum change if it was taken to the moon by finding the ratio of its frequency on the moon fM to its frequency on the earth fE. Suppose that gE is the free-fall acceleration on the earth and gM is the free-fall acceleration on the moon.
Express your answer in terms of some or all of the variables l, m, gE, gM.
fM/fE = ?
For the pendulum taken to the moon, The frequency change that would occur is mathematically given as
[tex]\frac{Fmoon}{Fearth}=0.408[/tex]
What frequency change would occur to the pendulum if it was taken to the moon?Generally, the equation for the Time period is mathematically given as
[tex]T=2\pi\sqrt{L/g}[/tex]
Therefore
[tex]\frac{Fmoon}{Fearth}=\frac{\sqrt{g/6L}}{\sqrt{g/6L}}\\\\\frac{Fmoon}{Fearth}=\sqrt{1/6}[/tex]
[tex]\frac{Fmoon}{Fearth}=0.408[/tex]
In conclusion, The frequency change
[tex]\frac{Fmoon}{Fearth}=0.408[/tex]
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Answer:
.408
Explanation:
Place the balloon in a bell jar. If available also add some shaving cream and fresh marshmallows. Ask the instructor for help if you are unfamiliar with this apparatus. The motor of the vacuum pump will remove air from the bell jar when it is turned on. Make a prediction about what you think will happen to the balloon as air is removed from the bell jar.
Answer:
The balloon will collapse
Explanation:
When air is removed from the bell jar, the balloon will collapse if the internal pressure from the balloon does not balance the atmospheric pressure from the surroundings.
Which diagram shows magnets that will attract each other? 2 bar magnets side by side with their long axes vertical, both red S on top and blue N on bottom. 2 bar magnets top to bottom with their long axes vertical, the top one with red S on top and blue N on bottom and the bottom magnet with blue N on top and red S on bottom. 2 bar magnets top to bottom with their long axes vertical, the top one with blue N on top and red S on bottom and the bottom magnet with red S on top and blue N on bottom. 2 bar magnets top to bottom with their long axes vertical, the top one with red S on top and blue N on bottom and the bottom magnet with red S on top and blue N on bottom.
2 bar magnets top to bottom with their long axes vertical, the top one with red S on top and blue N on bottom and the bottom magnet with red S on top and blue N on bottom. this diagram shows magnets that will attract each other. Hence option D is correct.
What is Magnet ?A permanent magnet is an item constructed of magnetised material that generates its own persistent magnetic field. A refrigerator magnet, for example, is commonly used to hold notes on a refrigerator door. Ferromagnetic (or ferrimagnetic) materials are those that can be magnetised and are strongly attracted to a magnet. These include the elements iron, nickel, and cobalt, as well as their alloys, some rare-earth metal alloys, and naturally occurring minerals such as lodestone. Although ferromagnetic (and ferrimagnetic) materials are the only ones that are strongly attracted to a magnet and are widely thought to be magnetic, all other substances respond weakly to a magnetic field via one of many different forms of magnetism.
Hence option D is correct.
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A diver comes off a board with arms straight up and legs straight down, giving her a moment of inertia about her rotation axis of 18kg⋅m2. She then tucks into a small ball, decreasing this moment of inertia to 3.6kg⋅m2. While tucked, she makes two complete revolutions in 1.2s.
Required:
If she hadn't tucked at all, how many revolutions would she have made in the 1.5 s from board to water?
Answer:
θ₁ = 0.5 revolution
Explanation:
We will use the conservation of angular momentum as follows:
[tex]L_1=L_2\\I_1\omega_1=I_2\omega_2[/tex]
where,
I₁ = initial moment of inertia = 18 kg.m²
I₂ = Final moment of inertia = 3.6 kg.m²
ω₁ = initial angular velocity = ?
ω₂ = Final Angular velocity = [tex]\frac{\theta_2}{t_2} = \frac{2\ rev}{1.2\ s}[/tex] = 1.67 rev/s
Therefore,
[tex](18\ kg.m^2)\omega_1 = (3.6\ kg.m^2)(1.67\ rev/s)\\\\\omega_1 = \frac{(3.6\ kg.m^2)(1.67\ rev/s)}{(18\ kg.m^2)}\\\\\omega_1 = \frac{\theta_1}{t_1} = 0.333\ rev/s\\\\\theta_1 = (0.333\ rev/s)t_1[/tex]
where,
θ₁ = revolutions if she had not tucked at all = ?
t₁ = time = 1.5 s
Therefore,
[tex]\theta_1 = (0.333\ rev/s)(1.5\ s)\\[/tex]
θ₁ = 0.5 revolution