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tatyana61 [14]
3 years ago
13

Please answer my question!,

Physics
1 answer:
Karolina [17]3 years ago
3 0

Answer:

Hope this helps! Mark as brainliest if liked thanks!

Explanation:

Your reasoning that the shadow is the shortest at mid-day is spot-on!

The wording of the question is the key to the answer. It says that the measurements were made in Summer. So this means that British Summer Time (BST) is being applied. BST is one hour ahead of Greenwich Mean Time and so what looks like 1pm is really 12 noon.

The safest sort of answer is to say that the shadow is shortest when the sun is at its highest point, and in this particular question that is at 1 pm because it is BST.

You might be interested in
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
2 Jim keeps a can on the ramp. The can
IceJOKER [234]

Answer:

The description of the statement is summarized throughout the explanation section below.

Explanation:

  • The layer of the ramp must've been harsh, which exacerbated the toy movement to occur gradually. This same friction force was indeed starting to function on the remote control car as well as trying to stop that one to react quickly.
  • Jim does have to simplify the exterior including its ramp by scrubbing or greasing to overcome impact as well as make driving travel quicker.

6 0
3 years ago
A circular loop of wire is in the plane of the paper. The south pole of a bar magnet is being moved from a position in front of
drek231 [11]

Answer:

Counterclockwise

explanation in attachment

5 0
3 years ago
L. What does it mean to "convert" from one unit to another? Give an example.
TiliK225 [7]

These are four questions and four answers


Question 1. What does it mean to "convert" from one unit to another? Give an example.


Convert from one unit to another is transforming or changing the magnitude to an equivalent measure according to a different pattern of comparison or unit of measure.


For example, converting from meter to centimeters, means to transform the unit of measure (the pattern of comparison) from meter to centimeter.


When your measure is 3 meters, means the length is 3 times the standard pattern named meter.


Therefore, converting it to centimeters means that you want to know how many times is 3 meters relative to the a new pattern named centimeter.


Since, 1 meter equals 100 cm, you build a conversion factor which you can use to make the conversion:


1 m = 100 cm ⇒ 1 = 100 cm / 1 m


∴ 3 m = 3 m × 100 cm / 1 m = 300 cm.


Question 2. What are the general mathematical operations you use when converting from one SI unit to another?


The most common mathematical operations are the basic ones: addition, substraction, multiplication and division.


See one example of each one:


a) Addition:


Transform 20°C to K: 20 + 273.15 K = 293.15 K


b) Subtraction:


Convert 300 K to celsius degrees: 300 - 273.15°C = 26.85°C


c) Multiplication


Convert 1000 Km to meter


1000 Km × 1000 m/Km = 1,000,000 m


d) Division


Convert 1540 mm Hg to atm


1 atm = 760 mmHg ⇒ 1 = 760 mmHg / atm

1540 mmHg / (760 mmHg / atm) = 2.026 atm


Question 3. How many meters do you cover in a 10 km (10-K) race?


This requires to convert 10 km to the equivalent number of meters.


Conversion factor: 1 km = 1000 m ⇒ 1 = 1000 m/Km


∴ 10 Km = 10 Km × 1000 m/km = 10,000 m ← answer.


Quesiton 4. An Olympic swimming pool is 50 meters long. You swim from one end to the other four times.


a. How many meters do you swim?

4 laps × 50 m/lap = 200 m ← answer


b. How many kilometers do you swim?


1 = 1 km / 1000m

200 m = 200 × 1 km / 1000 m = 0.200 km ← answer


c. How many centimeters do you swim?


1 m = 100 cm ⇒ 1 = 100 cm / m

200 m = 200 m × 100 cm/m = 20,000 cm ← answer

5 0
4 years ago
Read 2 more answers
An element that is unable to create a ion and therefore always has a charge of 0 is most likely a member of a family called
Allisa [31]

Explanation:

'noble gases' which are also known as group VIII

7 0
3 years ago
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