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3 years ago

So... if bedbugs live in beds, where do cockroaches live?

Physics

2 answers:

Deffense [45]3 years ago

6 0

Answer:

oo.p i wish I could answer that

Explanation:

mel-nik [20]3 years ago

6 0

Answer:

Sewers and possibly your basement

Explanation:

Anywhere warm, dark and wet

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Rufina [12.5K]

Manganese has 2 (two) electron that would free floating and able to form a metallic bond.
The electronic configuration of manganese is (Ar) 3d5 4s2. The two electron in 4s orbital are the valence electron which can freely move from one place to another.

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3 years ago

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Use your knowledge of waves and the graph shown to determine the frequency of wave C. What does it tell you about speed of the w

stealth61 [152]

Answer

Explanation:

goes up by 10 each time 10 to 20 to 30

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2 years ago

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$

and re-arranging the equation, we find

$v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s$

7 0

2 years ago

HELP PLEASE!!!!!

MakcuM [25]

Answer:C

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3 years ago

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Help ASAP thanks !!!

Ann [662]

Answer:

true

Explanation:

3 0

3 years ago