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4 years ago

6

Which is an example of a solution?

1 answer:

vladimir1956 [14]4 years ago

6 0

Answer:

C

Explanation:

plato

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Citrus2011 [14]

Answer:

the phone call goes out, transmits through radio waves, then moves to the central hub. then the signals go to the nearest tower to the receiver if the call

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3 years ago

What is there for astronomical evidence of the light spectra?

Romashka-Z-Leto [24]

Answer:

From spectral lines astronomers can determine not only the element, but the temperature and density of that element in the star. The spectral line also can tell us about any magnetic field of the star. The width of the line can tell us how fast the material is moving. We can learn about winds in stars from this.

Explanation:

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3 years ago

Read 2 more answers

How does the kinetic energy of the particles in a hot cup of coffee compare to the kinetic energy of the particles in ice cream?

trasher [3.6K]

Answer:

it has more molecules than the burning match, which equals MORE total kinetic energy.

Explanation:

4 0

4 years ago

Read 2 more answers

Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its

ruslelena [56]

Answer:

a

The orbital speed is $v= 2.6*10^{3} m/s$

b

The escape velocity of the rocket is $v_e= 3.72 *10^3 m/s$

Explanation:

Generally angular velocity is mathematically represented as

$w = \frac{2 \pi}{T}$

Where T is the period which is given as 1.6 days = $1.6 *24 *60*60 = 138240 sec$

Substituting the value

$w = \frac{2 \pi}{138240}$

$= 4.54*10^ {-5} rad /sec$

At the point when the rocket is on a circular orbit

The gravitational force = centripetal force and this can be mathematically represented as

$\frac{GMm}{r^2} = mr w^2$

Where G is the universal gravitational constant with a value $G = 6.67*10^{-11}$

M is the mass of the earth with a constant value of $M = 5.98*10^{24}kg$

r is the distance between earth and circular orbit where the rocke is found

Making r the subject

$r = \sqrt[3]{\frac{GM}{w^2} }$

$= \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }$

$= 5.78 *10^7 m$

The orbital speed is represented mathematically as

$v=wr$

Substituting value

$v= (5.78*10^7)(4.54*10^{-5})$

$v= 2.6*10^{3} m/s$

The escape velocity is mathematically represented as

$v_e = \sqrt{\frac{2GM}{r} }$

Substituting values

$= \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }$

$v_e= 3.72 *10^3 m/s$

7 0

4 years ago

Un bloque de 750 kg es empujado hacia arriba por una pista inclinada 15º respecto de la horizontal. El coeficiente de rozamiento

kotegsom [21]

Answer:

4776.98 N is the minimum force to start the rise.

Explanation:

We can use the first Newton's law to find the minimum force to move the block.

So we will have:

$F-W_{x}-F_{f}=0$

Where:

F is the force
W(x) is the weight of the block in the x direction, W = mg*sin(15)
F(f) is the static friction force (F(f) = μN), μ is the static friction coefficient 0.4.

$F=W_{x}+F_{f}=mgsin(15)+\mu N$

$F=mgsin(15)+\mu mgcos(15)$

$F=mg(sin(15)+\mu cos(15))$

$F=750*9.81(sin(15)+0.4*cos(15))$

$F=4776.98 N$

Therefore 4776.98 N is the minimum force to move the block.

I hope it helps you!

8 0

3 years ago