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3 years ago

Complete.

Physics

1 answer:

Kamila [148]3 years ago

4 0

Answer:

For a positive point charge, the lines radiate outward. While, for a negative point charge, the lines converge inward.

Explanation:

The electric field produced by a single point charge has a radial shape and follows an inverse square law. The magnitude is given by

$E=k\frac{q}{r^2}$

where

k is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

The direction of the field depends on the sign of the charge. In particular, we have:

For a positive charge, the direction of the field is out of the charge
FOr a negative charge, the direction of the field is towards the charge

This is due to the fact that the direction of the field shows the direction of the force that a positive test charge would experience in this field. Therefore, since a positive test charge would feel a repulsive force in the field produced by a positive charge, then the direction of the field is outward; on the other hand, a positive test charge would feel an attractive force in the field produced by a positive charge, therefore the direction of the field is inward.

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I. What is the initial velocity of the car?

qaws [65]

Answer:

I. 0 m/s

II. 20 m/s

III. Part BC

Explanation:

I. Determination of the initial velocity.

From the diagram given above,

The motion of the car starts from the origin. This implies that the car start from rest and as such, the initial velocity of the car is 0 m/s

II. Determination of the maximum velocity attained.

From the diagram given above, we can see clearly that the maximum velocity is 20 m/s.

III. Determination of the part of the graph that represents zero acceleration.

It important that we know the meaning of zero acceleration.

Zero acceleration simply means the car is not accelerating. This can only be true when the car is moving with a constant velocity.

From the graph given above, the car has a constant velocity between B and C.

Therefore, part BC illustrates zero acceleration.

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3 years ago

Which type of muscle cell can have multiple nuclei

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Answer:

Skeletal muscle cells

Explanation:

Skeletal muscle cells are long, cylindrical, and striated. They are multi-nucleated meaning that they have more than one nucleus. This is because they are formed from the fusion of embryonic myoblasts.

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2 years ago

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Sodium hydroxide (NaOH) was added to pure water (H2O). Which most likely were the pH values of the water before and after the so

Anastaziya [24]

Before 7 after 9. A pH smaller than 7 indicates acidity with 0 being completely acidic. A pH greater than 7 shows alkalinity with 14 being completely alkaline. 7 is neutral. Since NaOH is alkaline, adding it to a neutral substance would increase the pH and it would increase from 7 to 9.

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3 years ago

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When the distance between two interacting objects doubles, the gravitational force is

Umnica [9.8K]

The gravitational force will be one quarter.

The gravitational force between two objects is given by the formula

F=GMm/r^2

here, r is the distance between the objects.

Thus the gravitational force is inversely proportional to the square of the distance between the objects, Therefore if the distance between two objects is doubled the force will be one quarter.

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3 years ago

(a) Consider the initial-value problem dA/dt = kA, A(0) = A0 as the model for the decay of a radioactive substance. Show that, i

murzikaleks [220]

Answer:

a) $t = -\frac{ln(2)}{k}$

b) See the proof below

$A(t) = A_o 2^{-\frac{t}{T}}$

c) $t = 3T \frac{ln(2)}{ln(2)}= 3T$

Explanation:

Part a

For this case we have the following differential equation:

$\frac{dA}{dt}= kA$

With the initial condition $A(0) = A_o$

We can rewrite the differential equation like this:

$\frac{dA}{A} =k dt$

And if we integrate both sides we got:

$ln |A|= kt + c_1$

Where $c_1$ is a constant. If we apply exponential for both sides we got:

$A = e^{kt} e^c = C e^{kt}$

Using the initial condition $A(0) = A_o$ we got:

$A_o = C$

So then our solution for the differential equation is given by:

$A(t) = A_o e^{kt}$

For the half life we know that we need to find the value of t for where we have $A(t) = \frac{1}{2} A_o$ if we use this condition we have:

$\frac{1}{2} A_o = A_o e^{kt}$

$\frac{1}{2} = e^{kt}$

Applying natural log we have this:

$ln (\frac{1}{2}) = kt$

And then the value of t would be:

$t = \frac{ln (1/2)}{k}$

And using the fact that $ln(1/2) = -ln(2)$ we have this:

$t = -\frac{ln(2)}{k}$

Part b

For this case we need to show that the solution on part a can be written as:

$A(t) = A_o 2^{-t/T}$

For this case we have the following model:

$A(t) = A_o e^{kt}$

If we replace the value of k obtained from part a we got:

$k = -\frac{ln(2)}{T}$

$A(t) = A_o e^{-\frac{ln(2)}{T} t}$

And we can rewrite this expression like this:

$A(t) = A_o e^{ln(2) (-\frac{t}{T})}$

And we can cancel the exponential with the natural log and we have this:

$A(t) = A_o 2^{-\frac{t}{T}}$

Part c

For this case we want to find the value of t when we have remaining $\frac{A_o}{8}$

So we can use the following equation:

$\frac{A_o}{8}= A_o 2^{-\frac{t}{T}}$

Simplifying we got:

$\frac{1}{8} = 2^{-\frac{t}{T}}$

We can apply natural log on both sides and we got:

$ln(\frac{1}{8}) = -\frac{t}{T} ln(2)$

And if we solve for t we got:

$t = T \frac{ln(8)}{ln(2)}$

We can rewrite this expression like this:

$t = T \frac{ln(2^3)}{ln(2)}$

Using properties of natural logs we got:

$t = 3T \frac{ln(2)}{ln(2)}= 3T$

8 0

3 years ago