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2 years ago

Which of the following are heterogeneous mixtures (not homogeneous mixtures)?

1 answer:

5 0

The following are heterogeneous mixture: MILK THAT IS SOLD AT THE STORE, A CAKE MIX AND MIXED NUT.An heterogeneous mixture is one which contained many parts which can be separated out.A smoothie is an homogeneous mixture because the juice of different fruits that are use can not be separated out.
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What is the force of gravity between you and a standard tennis ball if you are 1 meter away

faltersainse [42]

Answer:

WHY MUST I CRYYYYYYYY

Explanation:

3 0

2 years ago

A mass on a spring A oscillates at twice the frequency of the same mass on spring B. Which statement is correct?A.The spring con

Nataliya [291]

Answer:

A.The spring constant for B is one quarter of the spring constant for A.

Explanation:

If spring A oscillates at twice the frequency of spring B, and period is frequency inverted. It means spring B has a period twice of spring A's.

$T_B = 2T_A$

As $T = 2\pi\sqrt{\frac{m}{k}}$ , and the 2 springs have the same mass

$2\pi\sqrt{\frac{m}{k_B}} = 2\pi\sqrt{\frac{m}{k_A}}$

$\sqrt{k_A} = 2\sqrt{B}$

$k_A = 4k_B$

$k_B = k_A/4$

So A.The spring constant for B is one quarter of the spring constant for A. is the correct answer.

3 0

3 years ago

Can someone please help me

Mars2501 [29]

The correct answer is letter b.

To find the answer follow the following steps.
1. 6524.96 x .25 = X

2. 1631.24 = X

This works for all of the given answers to find the correct answer.

5 0

2 years ago

Given a force of 100 N and an acceleration of 10 m/s2, what is the mass?

Drupady [299]

F=ma, so 100=m×10. Solve for m by dividing by 10. The mass is 10 kg.

5 0

2 years ago

Suppose that a comet that was seen in 563 A.D. by Chinese astronomers was spotted again in year 1951. Assume the time between ob

Mars2501 [29]

Answer:

$a=2.77*10^{13}m$

$R_a=5.49*10^{13}m$

Explanation:

The period of the comet is the time it takes to do a complete orbit:

T=1951-(-563)=2514 years

writen in seconds:

$2514years*\frac{3,154*10^7s}{1year}=7.93 *10^{10}s$

Since the eccentricity is greater than 0 but lower than 1 you can know that the trajectory is an ellipse.

Therefore, if the mass of the sun is aprox. 1.99e30 kg, and you assume it to be much larger than the mass of the comet, you can use Kepler's law of periods to calculate the semimajor axis:

$T^2=\frac{4\pi^2}{Gm_{sun}}a^3\\ a=\sqrt[3]{\frac{Gm_{sun}T^2}{4\pi^2} } \\a=1.50*10^{6}m$

Then, using the law of orbits, you can calculate the greatest distance from the sun, which is called aphelion:

$R_a=a(1+e)\\R_a=2.77*10^{13}(1.986)\\R_a=5.49*10^{13}m$

8 0

2 years ago