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Irina-Kira [14]

3 years ago

15

two-point charges are 10.0 cm apart and have charges of 2.0 uc and -2.0uc respectively What is the magnitude of the electrical f

ield at the midpoint between the two charges?

1 answer:

Scorpion4ik [409]3 years ago

6 0

Answer:

Electric field due to two charges is given as

$E = 1.44 \times 10^7 N/C$

Explanation:

As we know that two charges are opposite in nature

So the electric field at the mid point of two charges will add together

so the net field is given as

$E = 2\frac{kq}{r^2}$

now we have

$q = 2\times 10^{-6} C$

$r = 5 cm = 0.05 m$

now we have

$E = 2\frac{(9\times 10^9)(2\times 10^{-6})}{0.05^2}$

$E = 1.44 \times 10^7 N/C$

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A boat race runs along a triangular course marked by buoys A, B, and C. The race starts with the boats headed west for 3700 mete

ale4655 [162]

Answer:

The last two bearings are

49.50° and 104.02°

Explanation:

Applying the Law of cosine (refer to the figure attached):

we have

x² = y² + z² - 2yz × cosX

here,

x, y and z represents the lengths of sides opposite to the angels X,Y and Z.

Thus we have,

$cos X=\frac{x^2-y^2-z^2}{-2yz}$

or

$cos X=\frac{y^2 + z^2-x^2}{2yz}$

substituting the values in the equation we get,

$cos X=\frac{2900^2 + 3700^2-1700^2}{2\times 2900\times 3700}$

or

$cos X=0.8951$

or

X = 26.47°

similarly,

$cos Y=\frac{1700^2 + 3700^2-2900^2}{2\times 1700\times 3700}$

or

$cos Y=0.649$

or

Y = 49.50°

Consequently, the angel Z = 180° - 49.50 - 26.47 = 104.02°

The bearing of 2 last legs of race are angels Y and Z.

7 0

3 years ago

Please help !!!!! I’ll give brainliest !

goblinko [34]

I think A but I dont really know

3 0

3 years ago

I need help with this. -3x-7=18

miskamm [114]

the answer is

-8.33333333

4 0

3 years ago

Read 2 more answers

The force of magnitude F acts along the edge of the triangular plate. Determine the moment of F about point O. Find the general

Sever21 [200]

Answer:

The moment is 81.102 k N-m in clockwise.

Explanation:

Given that,

Force = 260 N

Side = 580 mm

Distance h = 370 mm

According to figure,

Position of each point

$O=(0,0)$

$A=(0,-b)$

$B=(h,0)$

We need to calculate the position vector of AB

$\bar{AB}=(h-0)i+(0-(-b))j$

$\bar{AB}=hi+bj$

We need to calculate the unit vector along AB

$u_{AB}=\dfrac{\bar{AB}}{|\bar{AB}|}$

$u_{AB}=\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}}$

We need to calculate the force acting along the edge

$\hat{F}=F(u_{AB})$

$\hat{F}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})$

We need to calculate the net moment

$\hat{M}=\hat{F}\times OA$

Put the value into the formula

$\hat{M}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})\times(-b\hat{j})$

$\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}((h\hat{i}+b\hat{j})\times(-b\hat{j}))$

$\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}(-bh\hat{k})$

$\hat{M}=-\dfrac{bhF}{\sqrt{h^2+b^2}}$

Put the value into the formula

$\hat{M}=-\dfrac{580\times10^{-3}\times370\times10^{-3}\times260}{\sqrt{(370\times10^{-3})^2+(580\times10^{-3})^2}}$

$\hat{M}=-81.102\ \hat{k}\ N-m$

Negative sign shows the moment is in clockwise.

Hence, The moment is 81.102 k N-m in clockwise.

5 0

4 years ago

Two blocks on a frictionless horizontal surface are on a collision course. one block with mass 0.6 kg moves at 0.8 m/s to the ri

Elanso [62]

First we can say that since there is no external force on this system so momentum is always conserved.

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

$0.6*0.8 + 1.2*0 = 0.6*v_{1f} + 1.2*v_{2f}$

$0.48= 0.6*v_{1f} + 1.2*v_{2f}$

$0.8 = v_{1f} + 2v_{2f}$

now by the condition of elastic collision

$v_{2f} - v_{1f} = 0.8 - 0[\tex]now add two equations[tex]3*v_{2f} = 1.6$

$v_{2f} = 0.533 m/s$

also from above equation we have

$v_{1f} = -0.267 m/s$

So ball of mass 0.6 kg will rebound back with speed 0.267 m/s and ball of mass 1.2 kg will go forwards with speed 0.533 m/s.

8 0

4 years ago