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Levart [38]
3 years ago
6

When a point charge of q is placed on one corner of a square, an electric field strength of 2 N/C is observed at the center of t

he square. Suppose three identical charges of q are placed on the remaining three corners of the square. What is the magnitude of the net electric field at the center of the square
Physics
1 answer:
bearhunter [10]3 years ago
3 0

Answer:

Explanation:

Net electric field at the centre will be zero .

Since all the charges are equal and they all are symmetrically situated around the centre . So the electric field produced by each will cancel  out each other and hence the resultant electric field will be zero . It happens because electric field is a vector quantity and therefore it adds up vectorially . All the four electric field will form two pairs , in each pair electric fields are acting in opposite direction . So they all cancel out to zero .

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Galileo _____.
AnnZ [28]

Answer:

friction help to slow motion in other word it oppose motion, but in a frictionless environment object would move with difficult stopping point.

5 0
3 years ago
Read 2 more answers
A cylinder is fitted with a piston, beneath which is a spring, as in the drawing. The cylinder is open to the air at the top. Fr
astraxan [27]

Answer:

x =  0.0734 m = 7.34 cm

Explanation:

First we shall calculate the area of the piston:

Area = \pi radius^2\\Area = \pi (0.028\ m)^2\\Area = 0.00246\ m^2

Now, we will calculate the force on the piston due to atmospheric pressure:

Atmospheric\ Pressure = \frac{Force}{Area}\\\\Force = (Atmospheric\ Pressure)(Area)\\Force = (101325\ N/m^2)(0.00246\ m^2) \\Force = F = 249.56\ N

Now, for the compression of the spring we will use Hooke's Law as follows:

F = kx\\

where,

k = spring constant = 3400 N/m

x = compression = ?

Therefore,

<u>x =  0.0734 m = 7.34 cm</u>

8 0
3 years ago
A 150 kg line backer sacks the 120 kg quarterback. With what force is the quarterback sacked if the line backer has an accelerat
Gekata [30.6K]

Answer:

The force required to move the quarterback with linebacker is <u>1215 N</u>

Explanation:

\text { Mass of linebacker } \mathrm{m}_{2}=150 \mathrm{kg}

\text { Mass of quarterback } \mathrm{m}_{2}=120 \mathrm{kg}

\text { Moved at an acceleration }(a)=4.5 \mathrm{m} / \mathrm{s}^{2}

Using Newton's second law, it is established that  F = Ma

Where F is net force acting on the system, a is the acceleration and M is mass of the two object \left(m_{1}+m_{2}\right)

Now consider both \mathrm{m}_{1} \text { and } \mathrm{m}_{2}as a system, so net force acting on the system is \text { Force }=\left(m_{1}+m_{2}\right) a

Substitute the given values in the above formula,

\text { Force }=(150+120) \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}

\text { Force }=270 \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}

Force = 1215 N

<u>1215 N </u>is the force required to move the quarterback with linebacker.

5 0
3 years ago
You need to focus a 10 mW, 632.8 nm Gaussian laser beam that is 5.0 mm in diameter into a sample. You have access to a lens with
Anna [14]

Answer:

ee that the lens with the shortest focal length has a smaller object

           

Explanation:

For this exercise we use the constructor equation or Gaussian equation

        \frac{1}{f}  = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distance to the object and the image respectively.

Magnification a lens system is

          m = \frac{h'}{h} = - \frac{q}{p}

             h ’= -\frac{h q}{p}

In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞

Let's calculate the distance to the image for each lens

f = 6.0 cm

           \frac{1}{q} = \frac{1}{f }  - \frac{1}{p}

as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf

          q = f = 6.0 cm

for the lens of f = 12.0 cm q = 12.0 cn

to find the size of the image we use

           h ’= h q / p

where p has a high value and is the same for all systems

           h ’= h / p q

Thus

f = 6 cm h ’= fo 6 cm

 

f = 12 cm h ’= fo 12  cm

therefore we see that the lens with the shortest focal length has a smaller object

8 0
3 years ago
A continental polar air mass forms in ____
Sladkaya [172]
I believe it is A bit not completely sure
7 0
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