All categories

3 years ago

Find the net force on q3. Include the direction ( +or-).

Physics

1 answer:

daser333 [38]3 years ago

8 0

Answer:

72.16 N

Explanation:

Given:

q₁ = -53.0 μC

q₂ = 105 μC

q₃ = -88.0 μC

q₁ to q₂ = 0.50 m

q₂ to q₃ = 0.95 m

To find:

Net force on q₃

Solution:

First compute net electric field on q₃

E = F/q = k.Q/d²

The formula of electric field at q₃:

E = k.Q / r²

Where

r is distance

Q is magnitude of charge

k is a constant with a value of 8.99 x 10⁹ N m²/C²

When

q₂ to q₃ = 0.95m and

q₂ = 105 μC then

Find electric field due to charge q₂

E = ( (8.99 x 10⁹)x( 105 x 10⁻⁶) ) / 0.95²

= (8990000000 x 0.000105) / 0.9025

= 943950 / 0.9025

= 1045927.977839

= 1.046 x 10⁶ N/C

This interprets that it will act or point away from q₂

q₁ to q₂= 0.50m

q₂ to q₃ = 0.95m and

q₁ = -53 μC then

Find electric field due to charge q₁

E = (8.99 x 10⁹) x (53 x 10⁻⁶) / (0 .50 + 0.95)²

= (8990000000 x 0.000053) / (1.45)²

= 476470 /2.1025

= 226620.689655

= 0.227 x 10⁶ N/C

This interprets that it will act or point towards q₁

Since these fields are opposite in direction.

Compute Net Field

Net Field = 1.046 x 10⁶ - 0.227 x 10⁶ N/C

= 1046000 - 227000

= 819000

= 0.819 x 10⁶

≈ 0.82 x 10⁶

This interprets that it will act or point away from q₂

Compute force on q3

q₃ E = 88 x 10⁻⁶ x 0.82 x 10⁶

= 88000000 x 820000

= 72160000000000

= 72.16 N

Force on -ive charge in a field is always in a direction opposite to direction of field

So this interprets that direction of this field will be towards q₂.

You might be interested in

Read the scenario and solve these two problems.

Burka [1]

Answers:

a) 5400000 J

b) 45.92 m

Explanation:

a) The kinetic energy $K$ of an object is given by:

$K=\frac{1}{2}mV^{2}$

Where:

$m=12000 kg$ is the mass of the train

$V=30 m/s$ is the speed of the train

Solving the equation:

$K=\frac{1}{2}(12000 kg)(30 m/s)^{2}$

$K=5400000 J$ This is the train's kinetic energy at its top speed

b) Now, according to the Conservation of Energy Law, the total initial energy is equal to the total final energy:

$E_{i}=E_{f}$

$K_{i}+P_{i}=K_{f}+P_{f}$

Where:

$K_{i}=5400000 J$ is the train's initial kinetic energy

$P_{i}=0 J$ is the train's initial potential energy

$K_{f}=0 J$ is the train's final kinetic energy

$P_{f}=mgh$ is the train's final potential energy, where $g=9.8 m/s^{2}$ is the acceleration due gravity and $h$ is the height.

Rewriting the equation with the given values:

$5400000 J=(12000 kg)(9.8 m/s^{2})h$

Finding $h$ :

$h=45.918 m \approx 45.92 m$

7 0

2 years ago

Read 2 more answers

An object that completes 20 vibrations in 10 seconds has a frequency of

nika2105 [10]

Answer:

The object has frequency of 2 Hz and time period of 0.5 s.

Explanation:

Frequency is defined as number of oscillation per second ie back and forth swings done in single second.

Here it is given that the object oscillates 20 times in 10 seconds.

So f = $\frac{20}{10}$ = 2Hz

The time period is defined as time taken by the object to complete one full oscillation.

T = $\frac{1}{f}$

T= $\frac{1}{2}$ =0.5 s

Thus the object has frequency of 2 Hz and time period of 0.5 s.

7 0

3 years ago

What is the difference between m/s and m.s.

vfiekz [6]

Answer

m/s rate of change of dispalcement per sec. ie velocity

m/s^2 is (m/s)/s ie rate of change of velocity per sec. ie accelerationplanation:

5 0

3 years ago

Read 2 more answers

I need help with questions 6-8. Thank you!! Image is attached

Digiron [165]

6) b) 2.7 m/s

7) b) DCA

8) b) B

Explanation:

In a displacement-time plot, the slope of the line is given by

$m=\frac{\Delta y}{\Delta x}$

where

$\Delta y$ is the change in the y-variable, so it is the displacement

$\Delta x$ is the change in the x-variable, so it is the time elapsed

So, the slope of the line in a displacement-time plot corresponds to the velocity:

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

Therefore, to find the velocity of the object, we have to estimate the slope of its curve.

To estimate the velocity of object B, we have to estimate the slope of the line tangent to curve B at 10 seconds.

By doing an estimate by eye, we see that the displacement of object B changes from -10 m to 0 m when time increases from about 8 s to 12 s, so the velocity is about:

$v=\frac{0-(-10)}{12-8}\sim 2.5 m/s$

So the closest option is b) 2.7 m/s.

As we said in part A, the velocity of each object is given by the slope of each curve.

Therefore:

- The steeper the curve, the higher the velocity

- The less steep the curve, the lower the velocity

From the graph, we observe that, among A, C and D:

- Curve D has the largest slope (in absolute value), so object D has the largest magnitude of the velocity

- Curve C is less steep than curve C, so object C has the second largest magnitude of velocity

- Curve A is flat, so the slope is zero, so its velocity is zero

So, from greatest magnitude to lowest magnitude of velocity, we have:

b) DCA

In the graph, the overall displacement of each object is given by the change in the y-variable, $\Delta y$ .

This means that the object with largest displacement is the object whose curve has the largest variation in y.

From the graph, we see that:

- Object b has the largest variation in y, from -15 m to 30 m, so

$\Delta y=30-(-15)=45 m$

- Then, object D has the second largest displacement (in magnitude), from -15 m to 25 m,

$|\Delta y| = 25 -(-15)=40 m$

Finally, object C has displacement

$\Delta y = 20-(-5)=25 m$

While object A has displacement zero. Therefore, the correct option is

b) B

3 0

3 years ago

The spectra of most galaxies show redshifts. this means that their spectral lines _________.

spin [16.1K]

Have wavelengths that are longer than normal.

3 0

2 years ago