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mylen [45]
2 years ago
6

Anne has a sample of a substance. Its volume is 20 cm3, and its mass is 100 grams. What is the sample’s density? The sample’

s density is g/cm3.
Physics
1 answer:
Tanzania [10]2 years ago
6 0

Answer:

<h2>5 g/cm³</h2>

Explanation:

The density of a substance can be found by using the formula

d =  \frac{m}{v}  \\

m is the mass

v is the volume

From the question

m = 100 g

v = 20 cm³

We have

d =  \frac{100}{20}  = 5 \\

We have the final answer as

<h3>5 g/cm³</h3>

Hope this helps you

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To determine the height of a tall building such as Sears Tower in Chicago, Illinois a ball was dropped from the top of the build
Darya [45]

Answer:

The height of Sears Tower is 1448.5 feet.

Explanation:

<h3>We apply the free fall formula to the ball: </h3><h3>y=v_{o} *t+\frac{1}{2} *g*t^{2}</h3><h3>y: The vertical distance the ball moves at time t  </h3><h3>v_{o}i: Initial speed </h3><h3>g=Gravity acceleration=9.8*(\frac{\frac{1ft}{0.305m} }{s^{2} } )</h3>

Known information

We know that the vertical distance (y) that the ball moves in 9,5s  is equal to height of Sears Tower (h).  

Too we know that the ball is released from rest, then,v_{0}=0

Height of Sears Tower calculation:

We replace  in the equation 1 the data following;

y=h

v_{o} =0

g=32,1\frac{ft}{s^{2} }

t= 9,5s

h=0*9.5+\frac{1}{2} *32.1*9.5^{2}

h=1448.5 ft

Answer: The height of Sears Tower is 1448.5 ft

6 0
3 years ago
An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r
bixtya [17]

1) Linear charge density of the shell:  -2.6\mu C/m

2)  x-component of the electric field at r = 8.7 cm: 1.16\cdot 10^6 N/C outward

3)  y-component of the electric field at r =8.7 cm: 0

4)  x-component of the electric field at r = 1.15 cm: 1.28\cdot 10^7 N/C outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

1)

The linear charge density of the cylindrical insulating shell can be found  by using

\lambda_2 = \rho A

where

\rho = -567\mu C/m^3 is charge volumetric density

A is the area of the cylindrical shell, which can be written as

A=\pi(b^2-a^2)

where

b=4.7 cm=0.047 m is the outer radius

a=2.7 cm=0.027 m is the inner radius

Therefore, we have :

\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m

 

2)

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

E=E_1+E_2

where:

E_1=\frac{\lambda_1}{2\pi r \epsilon_0}

where

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

E_2=\frac{\lambda_2}{2\pi r \epsilon_0}

where

\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m

And this field points radially inward, because the charge is negative.

Therefore, the net field is

E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C

in the outward direction.

3)

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

E_y=0

4)

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

E=\frac{\lambda_1}{2\pi \epsilon_0 r}

where:

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C

5)

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0
3 years ago
What is an Atwood Machine?
Lady_Fox [76]
The best answer is letter (A) a double pulley system. Atwood Machine is normally used as a measurement in balancing to object to verify the mechanical law of motion with constant acceleration.
6 0
3 years ago
Describe the story of the constellation Capricornus. WILL MARK BRAINLEST IF NOT STOLEN FROM A WEBSITE OR ANYTHING WITH A PROPER
ICE Princess25 [194]

Your answer:

In Greek mythology, this constellation is related with the time the Olympian gods sought refuge in Egypt. Unfortunately, following their epic fighting with the Titans, peace did not closing for long, as the monster Typhon, son of the Titan Tartarus and Earth, sought revenge. Typhon was once a fearsome fire-breathing creature, taller than mountains and with palms which possessed dragons' heads in region of fingers. The Olympian gods sought to break out by way of adopting a number disguises: Zeus, a ram - Hera, a white cow, Bacchus (another model of the fable suggests Pan) a goat. As Typhon approached, Bacchus/Pan threw himself into the Nile but, in a panic, solely succeeded in altering part of his body, ending up with a goat's physique and the tail of a fish. Meanwhile, Zeus had been dismembered via Typhon, however was saved when Bacchus/Pan let out an ear-splitting yell, distracting the monster lengthy ample for an agile Hermes to gather the supreme god's limbs and cautiously fix him. In gratitude, Zeus transferred Bacchus/Pan to the heavens.

5 0
3 years ago
Read 2 more answers
A simple experiment to measure the speed of sound doesn't involve a stopwatch. You can fill up along tube with water and put a t
Serhud [2]

Answer:

Explanation:

In order to answer this problem you have to know the depth of the column, we say R, this information is important because allows you to compute some harmonic of the tube. With this information you can compute the depth of the colum of air, by taking tino account that the new depth is R-L.

To find the fundamental mode you use:

f_n=\frac{nv_s}{4L}

n: mode of the sound

vs: sound speed

L: length of the column of air in the tube.

A) The fundamental mode id obtained for n=1:

f_1=\frac{v_s}{4L}

B) For the 3rd harmonic you have:

f_3=\frac{3v_s}{4L}

C) For the 2nd harmonic:

f_2=\frac{2v_s}{4L}

7 0
2 years ago
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