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3 years ago

5

How would you describe the behavior of particles in a solid?

2 answers:

Minchanka [31]3 years ago

4 0

How would you describe the behavior of particles in a solid?

vichka [17]3 years ago

4 0

You could say they are not very active but still have very little movement rather then a gas where the particles are moving fast

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Which of these diagrams best represents the steps in the formation of planets?

sineoko [7]

The right answer for the question that is being asked and shown above is that: "<span>C) The clouds of dust and gases rotate at high speed > The clouds condense > The sun is born > The planets are born " This is the </span><span>diagram that best represents the steps in the formation of planets</span>

6 0

2 years ago

Read 2 more answers

To convert centimeters to kilometers, which conversion factors would you need?

Lelechka [254]

Answer:

The last option is the only correct one if you like to multiply

The second last option is good if you like to divide.

Explanation:

Each fraction in the last two options has a value of 1

example

dividing by 1

15 cm /(100 cm/ 1 m) = 0.15 m 0.15 m / (1000 m/ 1km) = 0.00015 km

and

multiplying by 1

15 cm(1 m / 100cm) = 0.15 m 0.15m(1 km/1000m) = 0.00015 km

only one of the two fractions in each of the top two options has a value of 1.

3 0

2 years ago

he membrane that surrounds a certain type of living cell has a surface area of 6.0 x 10-9 m2 and a thickness of 1.6 x 10-8 m. As

k0ka [10]

Answer:

$1.54481175\times 10^{-12}\ C$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

A = Area = $6\times 10^{-9}\ m^2$

d = Thickness = $1.6\times 10^{-8}\ m$

k = Dielectric constant = 5.4

V = Voltage = 86.2 mV

Charge is given by

$Q=CV\\\Rightarrow Q=k\epsilon\dfrac{A}{d}V\\\Rightarrow Q=5.4\times 8.85\times 10^{-12}\times \dfrac{6\times 10^{-9}}{1.6\times 10^{-8}}\times 86.2\times 10^{-3}\\\Rightarrow Q=1.54481175\times 10^{-12}\ C$

The charge on the outer surface is $1.54481175\times 10^{-12}\ C$

4 0

2 years ago

A particle moving along the x-axis has a position given by m, where t is measured in s. What is the magnitude of the acceleratio

8090 [49]

Question:

A particle moving along the x-axis has a position given by x=(24t - 2.0t³)m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero

Answer:

24 m/s

Explanation:

Given:

x=(24t - 2.0t³)m

First find velocity function v(t):

v(t) = ẋ(t) = 24 - 2*3t²

v(t) = ẋ(t) = 24 - 6t²

Find the acceleration function a(t):

a(t) = Ẍ(t) = V(t) = -6*2t

a(t) = Ẍ(t) = V(t) = -12t

At acceleration = 0, take time as T in velocity function.

0 =v(T) = 24 - 6T²

Solve for T

$T = \sqrt{\frac{-24}{6}} = \sqrt{-4} = -2$

Substitute -2 for t in acceleration function:

a(t) = a(T) = a(-2) = -12(-2) = 24 m/s

Acceleration = 24m/s

4 0

3 years ago

A child whirls a ball in a vertical circle. Assuming the speed of the ball is constant (an approximation), when would the tensio

BARSIC [14]

Answer:

C. At the bottom of the circle.

Explanation:

Lets take

Radius of the circle = r

Mass = m

Tension = T

Angular speed = ω

The radial acceleration towards = a

a= ω² r

Weight due to gravity = mg

<h3>At the bottom condition</h3>

T - m g = m a

T = m ω² r + m g

<h3>At the top condition</h3>

T + m g = m a

T= m ω² r -m g

From above equation we can say that tension is grater when ball at bottom of the vertical circle.

Therefore the answer is C.

C. At the bottom of the circle.

8 0

3 years ago