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erik [133]
3 years ago
11

Provide an explanation tok please 80 points!!

Physics
1 answer:
nalin [4]3 years ago
8 0

Answer:

4th answer

Explanation:

have a good rest of ur day

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Gold has a density of 19300 kg/m³. Calculate the mass of 0.02 m³ of gold in kilograms.
hjlf

Answer:

Mass = 386 kg

Explanation:

<u><em>Density = Mass / Volume</em></u>

Mass = Density × Volume

Where D = 19300 kg/m³ , V = 0.02 m³

<em>Putting the given in the above formula</em>

Mass = 19300 × 0.02

Mass = 386 kg

7 0
3 years ago
Read 2 more answers
The beginning of the Phanerozoic is marked by what occurrence
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The beginning of the Phanerozoic is marked by the development of hard body parts, such as shells and bones.
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4 years ago
The net force acting on an object equals the applied force plus the force of friction.
Georgia [21]

Answer:

False

Explanation:

The net force is equal to the applied force minus the force of friction. It is possible for friction to act in the same direction as an applied force, but that would mean there would have to be more than two forces acting on the object.

3 0
3 years ago
What causes the balloon to move along the string?
ExtremeBDS [4]
Gravity? Im almost sure thats it
3 0
3 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
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