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3 years ago

9

Hi, I am having some difficulties in solving this question. Could someone please explain this question to me in detail. The thin

g is my physics textbook is horrible and I don't understand a thing in there. Help would be appreciated!

1 answer:

aniked [119]3 years ago

8 0

Answer:

a) 3.0×10⁸ m

b) 0 m

Explanation:

Displacement is the distance from the starting position to the final position.

a) In half a year, the Earth travels from one point on the circle to the point on the exact opposite side of the circle (from 0° to 180°). The distance between the points is the diameter of the circle.

x = 2r

x = 2 (1.5×10⁸ m)

x = 3.0×10⁸ m

b) In a full year, the Earth travels one full revolution, so it ends up back where it started. The displacement is therefore 0 m.

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The cooled lava pools on the surface of a crater are called what?

zimovet [89]

The answer i believe is craters

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4 years ago

Read 2 more answers

Walk 42 miles due north, deviate 78 degrees to east, and walk 65miles. What is the displacement? ( magnitude and direction with

pochemuha

Answer:84.405m, $\theta =48.876^{\circ}$

Explanation:

Given

Person walk 42 miles due to north so its position vector is

$r_1=42\hat{j}$

Now he deviates $78^{\circ}$ to east and walk 65 miles

so its new position vector

$r_2=42\hat{j}+65cos78\hat{j}+65sin78\hat{i}$

$r_2=65sin78\hat{i}+\left ( 42+65cos78\right )\hat{j}$

So magnitude of acceleration is

$|r_2|=\sqrt{\left ( 65sin78\right )^2+\left ( 42+65cos78\right )^2}$

$|r_2|=\sqrt{63.58^2+55.514^2}$

$|r_2|=84.405 m$

for direction

$tan\theta =\frac{42+65cos78}{65sin78}$

$tan\theta =0.8731$

$\theta =41.124^{\circ}with\ respect\ to\ east$

$\theta =48.876^{\circ}with\ respect\ to\ North$

7 0

4 years ago

A 75kg Tibetan is trekking along flat, but icy ledge with his 450kg yak when he slips over the edge. Luckily, he is holding the

Tcecarenko [31]

Answer:

minimal coefficient of static friction: $\mu_s=0.1667$

Explanation:

Once the Tibetan is hanging from the strap, he is exerting a horizontal force on the yak equal to his weight which is the product of his mass times the acceleration of gravity (g) as written below:

$w = m\,*\,g= 75\,kg\,*\,g$

The other forces acting on the yak are (see attached diagram):

* the force of gravity on the yak (identified in blue color in the image as $F_g$ ,

* the normal force (indicated in green in the image and identified by the letter "n") of the ledge on the yak as reaction to the yak's weight

* the force of static friction between the yak's hooves and the ledge (pictured in red in the image and identified with $f_s$ )

Since the normal force and the force of gravity on the yak cancel each other (balance - the yak is not moving vertically), the only forces we need to analyse are the force of the Tibetan's weight via the strap, and the force of static friction which should at least be equal in magnitude so the Tibetan doesn't fall. We assume these two forces are acting horizontally (one to the right: the Tibetan's weight, and one to the left: the static friction).

As we said, we want them to be at least equal so thy are in balance.

We recall that the force of static friction is the product of the normal force (n) times the coefficient of static friction ( $\mu_s$ ), such that: $f_s=\mu_s\,*\,n$

In our case these are the forces at play:

$F_g= M\,*\,g=450\, kg \,*\,g\\n=F_g=450\,kg\,*\,g\\f_s=\mu_s\,*\,n=\mu_s\,*450\,kg\,*\,g\\w=m\,*\,g=75\,kg\,*\,g$

So we need to find what is the minimum coefficient of static friction that precludes the Tibetan from falling. We therefore proceed to make an equality between the force of static friction on the yak and the weight of the Tibetan:

$f_s=w\\\mu_s\,*450\,kg\,*g=75\,kg\,*\,g$

and proceed to solve for the coefficient of friction by dividing both sides by "g" (which by the way cancels out), and by the yak's mass:

$\mu_s\,*450\,kg\,*g=75\,kg\,*\,g\\\mu_s=\frac{75}{450} \\\mu_s=0.1667$

where we have rounded to four decimal places the periodic number that the quotient generates. Notice that as expected, the coefficient of friction has no units (they all cancelled out in the division).

5 0

3 years ago

500mL of He at 98 kPa expands to 750 mL. Find P2

GarryVolchara [31]

The value of P₂ is 65.33 kPa

<u>Explanation:</u>

Given:

Volume, V₁ = 500mL

Pressure, P₁ = 98 kPa

Volume, V₂ = 750 mL

Pressure, P₂ = ?

According to Boyle's law:

P₁V₁ = P₂V₂

On substituting the value we get:

$98 X 500 = 750 X P_2\\\\P_2 = \frac{98 X 500}{750} \\\\P_2 = 65.33 kPa$

The value of P₂ is 65.33 kPa

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3 years ago

What evidence is there that work is being done?

suter [353]

Answer:

The boy uses force to move the rock a distance.

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2 years ago