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iren [92.7K]
2 years ago
7

A 550N object has a coefficient of .012 against a smooth surface. What

Physics
1 answer:
Brrunno [24]2 years ago
4 0

Vertically, the object is in equilibrium, so that the net force in this direction is

∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0

where <em>n</em> is the magnitude of the normal force due to the contact between the object and surface. You're given that the object's weight is <em>mg</em> = 550 N, so <em>n</em> = 550 N as well.

Horizontally, the net force would be

∑ <em>F</em> (horizontal) = <em>p</em> - <em>f</em> = 0

where <em>p</em> is the magnitude of the applied force and <em>f</em> is the magnitude of (kinetic) friction opposing <em>p</em>. Now,

<em>f</em> = 0.012<em>n</em> = 0.012 (550 N) = 6.6 N

so that you need to apply a force of <em>p</em> = 6.6 N to keep the object sliding at a steady pace.

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A box of mass 3.1kg slides down a rough vertical wall. The gravitational force on the box is 30N . When the box reaches a speed
Tems11 [23]

Answer:

The box will be moving at 0.45m/s. The solution to this problem requires the knowledge and application of newtons second law of motion and the knowledge of linear motion. The vertical component of the force Fp acts vertically upwards against the directio of motion. This causes a constant upward force of 23sin45° to act on the box. Fhe frictional force of 13N also acts vertically upwards and so two forces act upwards against rhe force of gravity resulting un a net force of 0.7N acting kn the box. This corresponds to an acceleration of 0.225m/s². So in w.0s after i start to push v = 0.45m/s.

Explanation:

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3 years ago
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What is formula for time and velocity
Juliette [100K]
Divide distance by the time it takes to travel that distance
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3 years ago
Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
alexandr1967 [171]

Answer:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

m_e is the mass of the electron

Solving for v, we find

v=\sqrt{k\frac{e^2}{m_e r}}

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

r=5.3\cdot 10^{-11}m

while the electron mass is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

Substituting into the formula, we find

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

7 0
3 years ago
Can I answer my own question
Andre45 [30]

Answer:

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4 0
2 years ago
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Ford used to make a 7.0 liter engine. Calculate how many many cubic inches would be the same size engine.
Anestetic [448]

Answer : The volume of engine in cubic inches is, 427.166\text{ inch}^3

Explanation :

As we are given that the volume of engine is, 7.0 liters. Now we have to calculate the volume of engine in cubic inches.

Conversion used :

1\text{ liter}=1000cm^3

So, 7\text{ liter}=7000cm^3

1\text{ inch}=2.54cm

or,

1cm=\frac{1}{2.54}\text{ inch}

1cm^3=(\frac{1}{2.54}\text{ inch})^3

As,

1cm^3=(\frac{1}{2.54}\text{ inch})^3

So,

7000cm^3=\frac{7000cm^3}{1cm^3}\times (\frac{1}{2.54}\text{ inch})^3=427.166\text{ inch}^3

Thu, the volume of engine in cubic inches is, 427.166\text{ inch}^3

3 0
3 years ago
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