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marin [14]
4 years ago
7

Un libro del peso di 12 N è in equilibrio su un tavolo. Sapendo che il coefficiente di attrito statico vale 0,5, la forza di att

rito quanto vale?
Physics
1 answer:
Tanzania [10]4 years ago
4 0

Answer:

60

Explanation:

Translation -

A book weighing 12 N is balanced on a table. Knowing that the static friction coefficient is 0.5, how much is the friction force worth?

Friction force is

f = u * n

f = 0.5 * 12N

f = 60

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Answer:

152642.2

Explanation:

the pressure at the bottom of the cylinder is

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=101300+806(9.8)(6.5)

=152642.2 Pa

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3 years ago
There are several forms of energy. Which of the following is not a form of energy?
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A. Is the right answer.
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If a platinum atom has 78 protons. How many electrons does it contain?​
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I think the answer would be 78 electrons, maybe?
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4 years ago
In class we calculated the range of a projectile launched on flat ground. Consider instead, a projectile is launched down-slope
zysi [14]

Answer:

With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.

Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude g.

If we have the initial velocity v_o and its angle \theta, we can obtain the vertical component of the velocity v_{oy} using trigonometry:

v_{oy}=v_osin\theta

Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:

v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }

Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}

Finally, from the equation of horizontal motion with constant speed, we have that:

x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s}  }{9.8\frac{m}{s^{2} } } =7.49m

In words, the projectile travels 7.49m horizontally before it lands.

8 0
4 years ago
1. What net force is required to accelerate a car at a rate of 2 m/s2 if the car has a mass of 3,000 kg?
nataly862011 [7]

Answer:

Net force required to accelerate the car is 6000 N

Explanation:

Force is calculated by the equation, F = Mass × Acceleration

This is based on Newton's Second Law of Motion which states that the force acting on an object is its mass times the acceleration of the object.

Here, mass = 3000 kg and acceleration = 2 m/s²

⇒ Force = Mass × Acceleration

             = 3000 × 2 = 6000 N

⇒ F = 6000 N

⇒ M = 3000 kg

⇒ a = 2 m/s²

7 0
3 years ago
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