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2 years ago

Use relative dating in a sentence.

1 answer:

3 0

Answer:Relative dating is the science of determining the relative order of past events (i.e., the age of an object in comparison to another), without necessarily determining their absolute age (i.e. estimated age).

Explanation:

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What is the change in velocity if the final velocity is 80 mph and the initial velocity is 20 mph? Is the object speeding up or

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Answer: Δv=vf-vi=80mph-20mph=60mph

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3 years ago

Which statements describe the energy in a food web? Select three options.

hichkok12 [17]

Answer:

it loses engry it follows difernt paths

Explanation:

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3 years ago

If distance between two charges increased 5 times then force between them

Marrrta [24]

decreased 5 times

Explanation: if the force increases 5 times between them would decrease 5 times

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3 years ago

Which of the following is an effect of the electric force?

stich3 [128]

As we know that two charges exert force on each other when they are placed near to each other

The force between two charges is given as

$F = \frac{kq_1q_2}{r^2}$

here we know that

$q_1, q_2$ = two different point charges

r = distance between two point charges

also we know that two similar charges always repel each other while two opposite charges always attract each other

so here correct answer would be

<em>A. A positive and negative charge attract each other.</em>

6 0

3 years ago

Read 2 more answers

At what position or positions on the x-axis is the electric field zero?

ElenaW [278]

Answer:

The electric field will be zero at x = ± ∞.

Explanation:

Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.

We know that,

The electric field is

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

The electric field vector due to charge one

$\vec{E_{1}}=\dfrac{kq_{1}}{r_{1}^2}(\hat{x})$

The electric field vector due to charge second

$\vec{E_{2}}=\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})$

We need to calculate the electric field

Using formula of net electric field

$\vec{E}=\vec{E_{1}}+\vec{E_{2}}$

$\vec{E_{1}}+\vec{E_{2}}=0$

Put the value into the formula

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})+\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})=0$

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})=\dfrac{kq_{2}}{r_{2}^2}(\hat{x})$

$(\dfrac{r_{2}}{r_{1}})^2=\dfrac{q_{2}}{q_{1}}$

$\dfrac{r_{2}}{r_{1}}=\sqrt{\dfrac{q_{2}}{q_{1}}}$

Put the value into the formula

$\dfrac{2.0+x}{x}=\pm\sqrt{\dfrac{2.0}{2.0}}$

$2.0+x=x$

If x = ∞, then the equation is be satisfied.

Hence, The electric field will be zero at x = ± ∞.

4 0

3 years ago