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3 years ago

A negatively charged ion formed from one atom of and one atom of

Physics

1 answer:

pogonyaev3 years ago

6 0

Answer:

An atom that loses one or more valence electrons to become a positively charged ion is known as a cation, while an atom that gains electrons and becomes negatively charged is known as an anion.

Explanation:

i hope this helps some

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A squirrel falls from this tree after being shocked by the falling apples. If the

Alchen [17]

Answer:

0.85m

Explanation:

Given parameters:

Height of fall = 3.5m

Unknown:

Duration of fall = ?

Solution:

To solve this problem, we apply the right motion equation.

Since we know the height, we can use the equation below;

S = ut + $\frac{1}{2} gt^{2}$

S is the height

u is the initial velocity = 0m/s

t is the time

g is the acceleration due to gravity

3.5 = 0 + $\frac{1}{2}$ x 9.8 x t²

3.5 = 4.9t²

t² = $\frac{3.5}{4.9}$

t² = 0.71

t = √0.71 = 0.85m

3 0

3 years ago

How to find net horizontal force with friction and applied force?

wlad13 [49]

Just subtract the both of them.

7 0

4 years ago

A bug flying horizontally at 0.65 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, th

irina [24]

The angular momentum is defined as,

$L=I\omega$

Acording to this text we know for conservation of angular momentum that

$L_i=L_f$

Where $L_i$ is initial momentum

$L_f$ is the final momentum

How there is a difference between the stick mass and the bug mass, we define that

Mass of the bug= m

Mass of the stick=10m

At the point 0 we have that,

$L_i=mvl$

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

vector from the point of reference (O), and ve is the velocity

At the end with the collition we have

$L_f=(I_b+I_s)\omega$

Substituting

$L_f=(ml^2+\frac{10ml^2}{3})\omega$

$L_f=\frac{13}{10}ml^2w$

$m(0.65)l=\frac{13}{10}ml^2 \omega$

$\omega=\frac{1}{2l}$

Applying conservative energy equation we have

$\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'$

$\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)$

Replacing the values and solving

$l=\frac{13}{0.54g}$

Substituting

l=\frac{13}{0.54(9.8)}

$l=2.45cm$

7 0

4 years ago

A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total

Harlamova29_29 [7]

Answer:

a)0.48 m/s

b) 0.583 m/s

Explanation:

As the wagon rolls,

momentum'p'= m x v => 95.8 x 0.530 = 50.774 Kgm/s

(a)Rock is thrown forward,

momentum of rock = 0.325 x 15.1 = 4.9075 Kgm/s

Conservation of momentum says momentum of wagon is given by

50.774 - 4.9075 = 45.8665

Therefore, Speed of wagon = 45.8665 / (95.8-0.325) = 0.48 m/s

(b) Rock is thrown backward,

momentum of wagon = 50.774 + 4.9075 = 55.68 Kgm/s

Therefore, speed of wagon = 55.68 / (95.8-0.325) = 0.583 m/s

4 0

4 years ago

An incompressible fluid flows steadily through a pipe that has a change in diameter. The fluid speed at a location where the pip

Tema [17]

Answer:

5.10 m/s

Explanation:

The volumetric flow rate for an incompressible fluid through a pipe is constant, so we can write:

$A_1 v_1 = A_2 v_2$ (1)

where

$A_1$ is the cross-sectional area of the first part of the pipe

$A_2$ is the cross-sectional area of the second part of the pipe

$v_1$ is the speed of the fluid in the first part of the pipe

$v_2$ is the speed of the fluid in the second part of the pipe

Here we have:

$v_1 = 1.28 m/s$

$r_1 = \frac{8.0 cm}{2}=4.0 cm = 0.04 m$ is the radius in the first part of the pipe, so the area is

$A_1 = \pi r_1^2 = \pi (0.04 m)^2 =5.02\cdot 10^{-3}m^2$

$r_2 = \frac{4.0 cm}{2}=2.0 cm = 0.02 m$ is the radius in the first part of the pipe, so the area is

$A_2 = \pi r_2^2 = \pi (0.02 m)^2 =1.26\cdot 10^{-3}m^2$

Using eq.(1), we find the fluid speed at the second location:

$v_2 = \frac{A_1 v_1}{A_2}=\frac{(5.02\cdot 10^{-3} m^2)(1.28 m/s)}{1.26\cdot 10^{-3} m^2}=5.10 m/s$

8 0

3 years ago

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