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Contact [7]
2 years ago
7

Is it possible to have sex then after a week you get your periods and get pregnant?

Physics
1 answer:
mrs_skeptik [129]2 years ago
7 0

Answer:

Yes it stilll possible

Explanation:

You might be interested in
25 points and i will give brainliest +5 star rating to a correct a correct answer!!!!
Gnesinka [82]

Answer:

1.97 * 10^8 m/s

Explanation:

Given that:

n = 1.52

Recall : speed of light (c) = 3 * 10^8 m/s

Speed (v) of light in glass:

v = speed of light / n

v = (3 * 10^8) / 1.52

v = 1.9736 * 10^8

Hence, speed of light in glass :

v = 1.97 * 10^8 m/s

4 0
3 years ago
Kalea throws a baseball directly upward at time t = 0 at an initial speed of 13.7 m/s. How high h does the ball rise above its r
Trava [24]

Answer:

h = 9.57 seconds

Explanation:

It is given that,

Initial speed of Kalea, u = 13.7 m/s

At maximum height, v = 0

Let t is the time taken by the ball to reach its maximum point. It cane be calculated as :

v=u-gt

u=gt

t=\dfrac{u}{g}

t=\dfrac{13.7}{9.8}

t = 1.39 s

Let h is the height reached by the ball above its release point. It can be calculated using second equation of motion as :

h=ut+\dfrac{1}{2}at^2

Here, a = -g

h=ut-\dfrac{1}{2}gt^2

h=13.7\times 1.39-\dfrac{1}{2}\times 9.8\times (1.39)^2

h = 9.57 meters

So, the height attained by the ball above its release point is 9.57 meters. Hence, this is the required solution.

7 0
3 years ago
The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, th
Inga [223]

According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.

To solve this problem we must apply the concept related to the conservation of energy theorem.

PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so

E_i = E_f

0 = \frac{1}{2} (m_1+m_2)v_f^2-m_2gh+m_1gh

v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}

PART B) Replacing the values given as,

h= 1.7m\\m_1 = 3.5kg\\m_2 = 4.3kg \\g = 9.8m/s^2 \\

v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}

v_f = \sqrt{2(9.8)(1.7)(\frac{4.3-3.5}{3.5+4.3})}

v_f = 1.8486m/s

Therefore the speed of the masses would be 1.8486m/s

6 0
3 years ago
is the following sentence true or false? the faster the particles of a substance are moving, the more energy they have.
svetlana [45]
I think true. I'm pretty sure, but check w/ others too.
8 0
3 years ago
Read 2 more answers
How much time will it take to do 33j of work with 11 w of power
Yuri [45]

Time required : 3 s

<h3>Further explanation </h3>

Power is the work done/second.

\tt P=\dfrac{W}{t}\\\\P=power,j/s,watt\\\\W=work, J\\\\t=times=s

 

To do 33 J of work with 11 W of power

P = 11 W

W = 33 J

\tt t=\dfrac{W}{P}\\\\t=\dfrac{33}{11}\\\\t=\boxed{\bold{3~s}}

8 0
3 years ago
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