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3 years ago

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The surface area of the earth's crust is 5.10*10^8km^2. The average thickness of the earth's crust is 35km. The mean density of

the earth's crust is 2.8g/cm^3. The most abundant element in the earth's crust is oxygen with an abundance of 4.55*10^3g/t. Calculate the total volume of the earth's crust.

1 answer:

Mars2501 [29]3 years ago

4 0

Answer:

The answer is $V = 1.785 \times 10^{10}\ Km^3$ .

Explanation:

You are asking only for the total volume, so the important data here is the surface area of the earth's crust,

$A = 5.10 \times 10^8\ Km^2$

and the average thickness of the earth's crust,

$h = 35\ Km$ .

Imagine that you can stretch the surface area as if it were a blanket, now if you want to calculate its volume, you just need to multiply its area by its thickness,

$V = A*h = 1.785 \times 10^{10}\ Km^3$ .

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Which statement about dwarf planets is true?

sineoko [7]

"Pluto was the first dwarf planet to be discovered" is the one statement among the following choices given in the question that is true <span>about dwarf planets. The correct option among all the options that are given in the question is the first option or option "a". Pluto was classified as a planet at first but in the year 1930 it was classified as a dwarf planet.</span>

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3 years ago

Read 2 more answers

A: How far did she travel? <br><br><br><br> B: How long did she take?

allochka39001 [22]

A. 60 miles
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Unless you are looking for slope, in which case the answer is different

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3 years ago

Your companion on a train ride through Illinois notices that telephone poles near the tracks appear to be passing by very quickl

Nikitich [7]

Answer: Relative motion

Explanation: If two objects are moving either towards or away from each other with both having their velocities in a reference frame and someone is outside this reference frame seeing the motion of the two objects.

The observer ( in his own frame of reference) will measure a different velocity as opposed to the velocities of the two object in their own reference frame. p

Both the velocity measured by the observer in his own reference frame and the velocity of both object in their reference is correct.

Velocities of this nature that have varying values based on motion referenced to another body is known as relative velocity.

Motion of this nature is known as relative motion.

<em>Note that the word reference frame is simply any where the motion is occurring and the specified laws of motion is valid</em>

<em />

For this example of ours, the reference frame of the companion is the train and the telephone poles has their reference frame as the earth.

The companion will measure the velocity of the telephone poles relative to him and the velocity of the telephone pole relative to an observer outside the train will be of a different value.

6 0

3 years ago

A car and a train move together along straight, parallel paths with the same constant cruising speed v(initial). At t=0 the car

kogti [31]

Answer:

$t_1 = \frac{v_i}{a_i}$

$t_2 = \frac{v_i}{a_i}$

Δd = $v_it_1 = v_i^2/a_i$

Explanation:

As $v(t) = v_i + at$ , when the car is making full stop, $v(t_1) = 0$ . $a = -a_i$ . Therefore,

$0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}$

Apply the same formula above, with $v(t_2) = v_i$ and $a = a_i$ , and the car is starting from 0 speed, we have

$v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}$

As $s(t) = vt + \frac{at^2}{2}$ . After $t = t_1 + t_2$ , the car would have traveled a distance of

$s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}$

Hence $s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}$

As $t_1 = t_2$ we can simplify $s(t) = v_it_1$

After t time, the train would have traveled a distance of $s(t) = v_i(t_1 + t_2) = 2v_it_1$

Therefore, Δd would be $2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i$

8 0

3 years ago

A ball rolls down an incline with an acceleration of 10 cm/s^2. If it starts with an initial velocity of 0 cm/s and has a velocy

11Alexandr11 [23.1K]

Given a = 10 cm/s²
u = 0 cm/s
v = 50 cm/s
we know that
v²=u²+2aS
2500=2×10×S
2500÷20 = S
S= 125 cm
The ramp is 125 cm

8 0

3 years ago