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3 years ago

8

A small, uncharged metal sphere is placed near a larger, negatively charged sphere. Which diagram best represents the charge dis

tribution on the smaller sphere

1 answer:

Rudik [331]3 years ago

3 0

Answer:

I'm sorry I don't know what is the answer

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A heavy piece of hanging sculpture is suspended by a90-cm-long, 5.0 g steel wire. When the wind

Yuliya22 [10]

As we know that fundamental frequency is given as

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

here we know that

$\mu = \frac{m}{l}$

here we have

m = mass of wire = 5 g

l = length of wire = 90 cm

$\mu = \frac{0.005}{0.90} kg/m$

$\mu = 5.56 \times 10^{-3} kg/m$

from above formula now

$80 = \frac{1}{2(0.90)}\sqrt{\frac{T}{5.56\times 10^{-3}}}$

$144 = \sqrt{180 T}$

$T = 115.2 N$

now we know that tension is due to weight of the sculpture so we will have

$Mg = 115.2 N$

$M = 11.76 kg$

so its mass will be 11.76 kg

6 0

4 years ago

A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t

vladimir1956 [14]

a) $\omega=\frac{-mvR}{I+MR^2}$

b) $v=\frac{-mvR^2}{I+MR^2}$

Explanation:

a)

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

$L_1=0$

Later, after the girl throws the rock, the angular momentum will be:

$L_2=(I_M+I_g)\omega +L_r$

where:

$I$ is the moment of inertia of the merry-go-round

$I_g=MR^2$ is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

$\omega$ is the angular speed of the merry-go-round and the girl

$L_r=mvR$ is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

$L_1=L_2$

So we find:

$0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}$

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

b)

The linear speed of a body in rotational motion is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

$\omega=\frac{-mvR}{I+MR^2}$ is the angular speed

$r=R$ is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

$v=\omega R=\frac{-mvR^2}{I+MR^2}$

5 0

3 years ago

Use the worked example above to help you solve this problem. A car traveling at a constant speed of 27.9 m/s passes a trooper hi

Sidana [21]

H h h huh h jvjvhvuvuvuvyv

7 0

4 years ago

The initial concentration of acid ha in solution is 0.39 m. if the ph of the solution at equilibrium is 0.76, what is the percen

Alexus [3.1K]

The percent ionization of the acid is 44.56%

<h3>How can we calculate the percent ionization of the acid?</h3>

To calculate the percent ionization of the acid we are using the formula,

The H⁺ ion concentration [H⁺] = C x,

where, we are given,

C= concentration of the acid.

=0.39 M

x= degree of dissociation of the acid.

And one more thing we are given that, the pH of the acid=0.76.

So from the above statement we can say that,

pH = - log [H⁺]

Or,0.76 = -log [H⁺]

Or, log [H⁺] = -0.76

Or, [H⁺] = antilog -0.76

Or,[H⁺]= 10^-0.76

Or,[H⁺]=0.1738.

Now from the above calculation we know, the H⁺ ion concentration= 0.1738 M.

Now we put the known values in the above equation,

[H+]= Cx

Or,0.1739= 0.39 x

Or, x= 0.4459

From the above calculation we can conclude that the percent Ionization of the acid= 0.4459 X 100= 44.59%≈45%

Learn more about Ionization:

brainly.com/question/1445179

#SPJ4

4 0

2 years ago

Based on the chemical equation, use the drop-down menu to choose the coefficients that will balance the chemical

Vadim26 [7]

Answer:

2 1 1

Explanation:

8 0

3 years ago