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hammer [34]
2 years ago
11

Okay, I have three more questions.

Physics
2 answers:
Blababa [14]2 years ago
5 0
When a liquid or a gas is heated, it expands and becomes less dense, so it rises, while the cooler, denser liquid or gas sinks. (Not sure if this helps but this is what I hit)
Solnce55 [7]2 years ago
4 0
1. Less
2. More
3. Convection current
I am unsure if 3 is correct
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Two children are riding on a merry-go-round that is rotating with a constant angular speed. Abbie is one meter from the center o
Ainat [17]

Answer:

  • <em>Abbie’s acceleration is (1/2) Zak’s acceleration.</em>

Explanation

1. <u>Data</u>:

a) ω = constant

b) Abbie: r₁ = 1 m

c) Zak: r₂ = 2 m

d) Ac₁ = ? Ac₂

2. <u>Formulae</u>

  • Ac = ω² r

3. <u>Solution</u>:

a) Abbie:

  • Ac₁ = ω² r₁  =  ω² (1m)

b) Zack:

  • Ac₂ = ω² r₂  = ω² (2m)

c) Divide Ac₁ / Ac₂

  • Ac₁ / Ac₂ =  ω² (1m) / [ω² (2m) ] = 1/2

⇒      Ac₁ = (1/2) Ac₂ = Ac₂ / 2 = 0.5 Ac₂

5 0
3 years ago
A boat floats south on the Amazon River at a speed of 6 m/s. The boat and
EleoNora [17]

Explanation:

Take south to be negative.

a. Momentum is mass times velocity.

p = mv

p = (540 kg) (-6 m/s)

p = -3240 kg m/s

p = 3240 kg m/s south

b. Impulse = change in momentum

J = Δp

Since the mass is constant:

J = mΔv

J = (540 kg) (-4 m/s − (-6 m/s))

J = 1080 kg m/s

J = 1080 kg m/s north

7 0
3 years ago
3. What are the challenges of looking for Dyson spheres?
S_A_V [24]

1. it is difficult to search for it . Because infrared rays will never penetrate through earth atmosphere.

2. we are unaware of how it looks like and we only know it is red and will glow . A damaged star also looks like this.

3. Dust also makes is hard to detect Dyson spheres . So we will get confused between Dyson sphere and a star surrounded by dust.

5 0
3 years ago
Read 2 more answers
Submarines need to be extremely strong to withstand the extremely high pressure of water pushing down on them. An experimental r
Brrunno [24]
1500 I think so but I not sure
8 0
2 years ago
A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is
aniked [119]

1) At the moment of being at the top, the piston will not only tend to push the penny up but will also descend at a faster rate at which the penny can reach in 'free fall', in that short distance. Therefore, at the highest point, the penny will lose contact with the piston. Therefore the correct answer is C.

2) To solve this problem we will apply the equations related to the simple harmonic movement, hence we have that the acceleration can be defined as

a = -\omega^2 A

Where,

a = Acceleration

A = Amplitude

\omega= Angular velocity

From a reference system in which the downward acceleration is negative due to the force of gravity we will have to

a = -g

-\omega^2 A = -g

\omega = \sqrt{\frac{g}{A}}

From the definition of frequency and angular velocity we have to

\omega = 2\pi f

f = \frac{1}{2\pi} \sqrt{\frac{g}{A}}

f = \frac{1}{2\pi} \sqrt{\frac{9.8}{4*10^{-2}}}

f = 2.5Hz

Therefore the maximum frequency for which the penny just barely remains in place for the full cycle is 2.5Hz

6 0
3 years ago
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