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2 years ago

7

Imagine two people standing at placemark A and placemark E, looking at each other across the fault. Which of the following state

ments correctly describes the relative motion of the land opposite each observer?
a. To both observers, the land opposite them is moving to the right.
b. To both observers, the land opposite them is moving to the left.
c. To the observer at placemark A, the land at placemark Eis moving to the left
d. To the observer at placemark E, the land at placemark A is moving to the right.

1 answer:

lord [1]2 years ago

5 0

Answer:

To both observers, the land opposite them is moving to the right.

Explanation:

I have this class too. and there is also a quizlet with all the answers to the rest of the other questions. Trust me its right .

https://quizlet.com/261219090/oce-1001-chapter-2-flash-cards/

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The advantage of the SI unit over CGS unit are:

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2 years ago

A 6.60-kg block slides with an initial speed of 1.56 m/s up a ramp inclined at an angle of 28.4° with the horizontal. The coeffi

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Answer:

The distance travel by block before coming to rest is 0.122 m

Explanation:

Given:

Mass of block $m = 6.60$ kg

Initial speed of block $v _{i} = 1.56$ $\frac{m}{s}$

Final speed of block $v_{f} = 0$ $\frac{m}{s}$

Coefficient of kinetic friction $\mu _{k} = 0.62$

Ramp inclined at angle $\theta =$ 28.4°

Using conservation of energy,

Work done by frictional force is equal to change in energy,

$\mu _{k} mgd \cos 28.4 = \Delta K - \Delta U$

Where $\Delta U = mg d\sin 28.4$

$\mu _{k} mgd \cos 28.4 = \frac{1}{2}mv_{i} ^{2} - mgd\sin 28.4$

$\mu _{k} mgd \cos 28.4 +mgd\sin 28.4 = \frac{1}{2}mv_{i} ^{2}$

$d(6.60 \times 9.8 \times 0.62 \times 0.879 + 6.60 \times 9.8 \times 0.475) = \frac{1}{2} \times 6.60 \times (1.56)^{2}$

$d = 0.122$ m

Therefore, the distance travel by block before coming to rest is 0.122 m

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2 years ago

A lens of focal length 15.0 cm is held 10.0 cm from a page (the object ). Find the magnification .

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Answer:

Magnification, m = 3

Explanation:

It is given that,

Focal length of the lens, f = 15 cm

Object distance, u = -10 cm

Lens formula :

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

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2 years ago

Which changes would result in a decrease in the gravitational force between two objects? Check all that apply.

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2 years ago

A particle with a charge of +4.20nC is in a uniform electric field E⃗ directed to the left. It is released from rest and moves t

Morgarella [4.7K]

Answer:

(A). The work done is $1.50\times10^{-6}\ J$ .

(B). The potential of the starting point with respect to the endpoint is 357.14 V.

(C). The magnitude of E is 5952.38 N/C.

Explanation:

Given that,

Charge = 4.20 nC

Distance = 6.00 cm

Kinetic energy $K.E=1.50\times10^{-6}\ J$

The particle start from rest.

So, the initial kinetic energy i zero.

(A). We need to calculate the work by the electric force

Using formula of work done

$W = \Delta K.E$

$W=K.E_{f}-K.E_{i}$

Put the value into the formula

$W= 1.50\times10^{-6}-0$

$W=1.50\times10^{-6}\ J$

The work done is $1.50\times10^{-6}\ J$ .

(B). We need to calculate the potential of the starting point with respect to the endpoint

We know that.

Change in potential energy = change in kinetic energy

$\Delta P.E=\Delta K.E$

So, $U = 1.50\times10^{-6}$

Using formula of potential

$V=\dfrac{U}{q}$

Put the value into the formula

$V=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}}$

$V=357.14\ V$

The potential of the starting point with respect to the endpoint is 357.14 V.

(C). We need to calculate the magnitude of E

Using formula of work done

$W=F\times r$ ....(I)

Using formula of force

$F=qE$

Put the value in the equation (I)

$W=qE\times r$

$E=\dfrac{W}{q\times r}$

Put the value into the formula

$E=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}\times6.00\times10^{-2}}$

$E=5952.38\ N/C$

The magnitude of E is 5952.38 N/C.

Hence, This is the required solution.

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2 years ago