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2 years ago

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Physics

1 answer:

joja [24]2 years ago

3 0

Answer:

Electrical force is directly proportional to the product of the charges on the two objects.

Explanation:

The electric force between two charges is given by :

$F=\dfrac{kq_1q_2}{d^2}$

Where

k is electrostatic constant

q₁ and q₂ are charges

d is the distance between charges

It can be very clear that the electric force is directly proportional to the product of charges and inversely proportional to the square of the distance between them.

Hence, the correct option is (a) "Electrical force is directly proportional to the product of the charges on the two objects".

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A man starts walking from home and walks 2 miles at 20° north of west, then 4 miles at 10° west of south, then 3 miles at 15° no

Rzqust [24]

Answer:

a) R = 2.5 mi b) To return to your case you must walk in the opposite direction or θ = 98º

This is 8º north west

Explanation:

This is a distance exercise with vectors the best way to work these is to decompose the vectors and perform the sum on each axis separately

To use the Cartesian system all angles must be measured from the positive side of the x-axis or the signs of the components must be assigned manually depending on the quadrant where they are.

First vector A = 2 to 20º north west

Measured from the positive x axis is θ = 180 -20 = 160º

We use trigonometry to find the components

Cos 20 = Aₓ / A

sin 20 = $A_{y}$ / A

Aₓ = A cos 160 = 2 cos 160

$A_{y}$ = A sin160 = 2 sin160

Aₓ = -1,879 mi

$A_{y}$ = 0.684 mi

Second vector B = 4 mi 10º west of the south

Angle θ = 270 - 10 = 260º

cos 2600 = Bₓ / B

sin 260 = $B_{y}$ / B

Bₓ = B cos 260

$B_{y}$ = B sin 260

Bₓ = 4 cos 260

$B_{y}$ = 4 sin 260

Bₓ = -0.6946mi

$B_{y}$ = - 3,939 mi

Third vector C = 3 mi to 15 north east

cos 15 = Cₓ / C

sin15 = $C_{y}$ / C

Cₓ = C cos 15

$C_{y}$ = C sin15

Cₓ = 3 cos 15

$C_{y}$ = 3 sin 15

Cₓ = 2,898 mi

$C_{y}$ = 0.7765 mi

Now we can find the final position of the person

X = Aₓ + Bₓ + Cₓ

X = -1.879 -0.6949 + 2.898

X = 0.3241 mi

Y = $A_{y}$ + $B_{y}$ + $C_{y}$

Y = 0.684 - 3.939 +0.7765

Y = -2.4785 mi

a) We use Pythagoras' theorem

R = √ (x2 + y2)

R = √ (0.3241 2 + (-2.4785) 2)

R = 2.4996 mi

R = 2.5 mi

b) let's use trigonometry

Tan θ = y / x

Tanθ = -2.4785 / 0.3241

θ = tan⁻¹ (-7,647)

θ = -82

Measured from the positive side of the x axis is Te = 360 - 82 = 278º

(90-82) south east

To return to your case you must walk in the opposite direction or Te = 98º

This is 8º north west

3 0

3 years ago

xxTIMURxx [149]

Idk what to say to this

3 0

2 years ago

How are the land area and the oceans affected during an ice age

amm1812

The land was frozen everywhere and a part of the ocean froze over to create a bridge between two countries. Some people may disagree with me over the ice bridge, though.

3 0

3 years ago

Read 2 more answers

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from re

spayn [35]

The magnitude of the electric field can be calculated using the equation

$E = \frac{F}{q}$ , where F is the Force acting on the proton and q is the charge of the proton.

We know that the charge of the proton is $q = 1.6 x 10^{-19} C$

We have to calculate the Force first.

We know that F = ma from Newton's 2nd law, where m is the mass of the proton and a is the acceleration.

We know that the mass of the proton $m = (1.67) X 10^{-27} kg$

So it turns out that we have to calculate acceleration before anything else.

In order to calculate the acceleration, we make use of the following data from the question:

Initial Velocity of the proton $V_{i} = 0$

Distance traveled $D = 1.70 cm$ = 0.017 m

Time taken for the travel between the plates $t = (1.48) X 10^{-6} s$

Acceleration a = ?

Using the equation, $D = V_{i}t + \frac{1}{2} at^{2}$ , we get

Knowing that initial velocity is 0, the equation reduces to $D = \frac{1}{2}at^{2}$

Rearranging the equation so as to make a the subject of the formula, we have

$a = \frac{2D}{t^{2} }$

Plugging in the numbers and simplifying gives us a = 1.5 x $10^{10} m/s^{2}$

We can now calculate the Force using F = ma

Plugging in the known values, we get F = 2.5 x $10^{-17} N$

Using this, we can calculate E through the equation $E = \frac{F}{q}$

Plugging the numbers and simplifying gets us E = 156.25 N/C

Thus, the magnitude of the electric field between the plates of the capacitor is 156.25 N/C

B) To calculate the Final Velocity of the proton, we can make use of the equation

$V_{f} = V_{i} + at$

Plugging the numbers in and simplifying gets us $V_{f} = (2.22) * 10^{4} m/s$

5 0

3 years ago

Read 2 more answers

A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected

gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta$

Put the value into the fomrula

$5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45$

$15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}$

$15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}$ ....(I)

Along y -axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta$

Put the value into the formula

$0+0=5\times v_{1}\sin30-3\times v_{2}\sin45$

$\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0$ ...(II)

From equation (I) and (II)

$v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}$

$v_{1}=2.19\ m/s$

Put the value of v₁ in equation (I)

$\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0$

$v_{2}=\dfrac{5.475\times\sqrt{2}}{3}$

$v_{2}=2.58\ m/s$

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

3 0

3 years ago