All categories

3 years ago

What period can you find the element thallium in??

1 answer:

4 0

Other families might be more private, so a girl might not know exactly what to say or whom to tell. But this news needs to be shared, at least with one grown-up in the family. You might just pull the person aside and say it clearly, like this: "I got my first period today. Do you think you can help me get some supplies?"

You might be interested in

Mercury is characterized by

cupoosta [38]

<span>It's close to the sun without much atmosphere, so it's characterized by </span><span>very extreme temperatures.

Happy studying ^_^</span>

8 0

2 years ago

A baseline for an experimental investigation is provided by the ??

ki77a [65]

Hypothesis

I think

I hope this helps tho

4 0

3 years ago

Using evidence from the article, defend the concept that Earth’s magnetic poles have swapped places over time.

Mamont248 [21]

Answer: Scientists found evidence of Earths magnetic field reversal in rocks on the ocean floor at plate boundaries. These rocks have alternating polarity due to magnetization that occurred during their cooling period. Using radio metric dating, scientist estimate that reversals occur approximately every several hundred thousand years.

Explanation:

6 0

3 years ago

4. Which of the following is the term used to describe a body's resistance to a change in motion?

Alborosie

Inertia ..................

5 0

2 years ago

A test charge of -3x10^-7 C is located 7 cm to the right of a charge of -9x10^-6 C and 20 cm to the left of a charge of +10x10^-

musickatia [10]

Answer:

5.634 N rightwards

Explanation:

qo = - 3 x 10^-7 C

q1 = - 9 x 10^-6 C

q2 = 10 x 10^-6 C

r1 = 7 cm = 0.07 m

r2 = 20 cm = 0.2 m

The force on test charge due to q1 is F1 which is acting towards right

According to the Coulomb's law

$F_{1}=\frac{Kq_{1}q_{0}}{r_{1}^{2}}$

F1 = (9 x 10^9 x 9 x 10^-6 x 3 x 10^-7) / (0.07 x 0.07)

F1 = 4.959 N rightwards

The force on test charge due to q2 is F1 which is acting towards right

According to the Coulomb's law

$F_{2}=\frac{Kq_{2}q_{0}}{r_{2}^{2}}$

F2 = (9 x 10^9 x 10 x 10^-6 x 3 x 10^-7) / (0.2 x 0.2)

F2 = 0.675 N rightwards

Net force on the test charge

F = F1 + F2 = 4.959 + 0.675 = 5.634 N rightwards

3 0

3 years ago