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I need help with my physics homework agh! Please help it's due tomorrow. <br>

storchak [24]

Rubbing both pieces cause each piece to have a negative charge.

When two parts have the same they repel each other, so holding one piece up tot he end of the other piece would push it away.

Because one piece is held in the middle by a string, it would rotate the piece in a circle.

If they held the piece to the other end of the one held by a string it would start to rotate in the opposite direction.

4 0

3 years ago

Andre, whose mass is 77.1 kg, sits on a diving board above a dunk tank. If he is sitting 0.90 m above the water, what is his tot

Ganezh [65]

Answer:

680 J

Explanation:

Mechanical energy = potential energy + kinetic energy

ME = PE + KE

ME = mgh + ½ mv²

ME = (77.1 kg) (9.8 m/s²) (0.90 m) + ½ (77.1 kg) (0 m/s)²

ME = 680 J

8 0

4 years ago

Read 2 more answers

"relate the fertile phase of the menstrual cycle to the process of fertilisation"

DerKrebs [107]

Explanation:

During your menstrual cycle , Harmones make the eggs in your Ovaries mature -

• when an egg is mature , That means it's ready to be fertilized by a sperm cell .

• These hormones also make the lining of your uterus thick and spongy . So if your egg does get Fertilised , It has a nice cushy place to land and start a pregnancy .

<h3>Hope this helps </h3>

8 0

2 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

galina1969 [7]

Answer:

44.64 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.2\times 1180+80.6^2}\\\Rightarrow v=128.01\ m/s$

$v=u+at\\\Rightarrow 128.01=80.6+4.2t\\\Rightarrow t=\frac{128.01-80.6}{4.2}=11.29\ s$

<u>Time taken to reach 1180 m is 11.29 seconds</u>

$v=u+at\\\Rightarrow 0=128.01-9.8t\\\Rightarrow t=\frac{128.01}{9.8}=13.06\ s$

<u>Time the rocket will keep going up after the engines shut off is 13.06 seconds.</u>

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128.01^2}{2\times -9.8}\\\Rightarrow s=836.05\ m$

The distance the rocket will keep going up after the engines shut off is 836.05 m

Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m

The rocket will fall from this height

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2016.05=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2016.05\times 2}{9.8}}\\\Rightarrow t=20.29\ s$

<u>Time taken by the rocket to fall from maximum height is 20.29 seconds</u>

Time the rocket will stay in the air is 11.29+13.06+20.29 = 44.64 seconds

5 0

3 years ago

From an h = 53 feet observation tower on the coast, a Coast Guard officer sights a boat in difficulty. The angle of depression o

maksim [4K]

Answer:

757,93 feets

Explanation:

We can make a right triangle between the boat (A), the Coast Guard officer (B) and the base of the observation tower (C), like in the graph attached. Now, you could also made a rectangle, adding the horizontal at the height of the Coast Guard, starting in B and ending in D, the vertex opossing C.

The angle of depression, its O in the graph.

Now, as we got an rectangle, of course, the segment AD its the same length as CB, and CA, the distance from the boat to shoreline, its the same length as DB.

ADB its an right triangle, with AB, the hypothenuse, and BD and DA, the catheti (or <em>legs</em>).

Now, we know the lenght BC, the height of the tower, 53 feets, so we also know the lenght of DA. DA its the opposite cathetus to the angle O. We wish to know the length AC, equal to the lenght DB, the adjacent cathetus of the angle O.

Know, the trigonometric function that connects the adjacent cathetus with the opossite cathetus its the tangent.

$tangent( O ) = \frac{opposite}{adjacent}$

We can take that the angle O = 4 °, and knowing that the opossite cathetus its 53 feets, we got:

$tangent( 4) = \frac{53 feets}{DB}$

$DB= \frac{53 feets}{tangent( 4)}$

$DB= 757,93 feets$

This its equal to the distance from the boat to the shoreline.

4 0

3 years ago