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timama [110]
3 years ago
11

Yes or no

Physics
1 answer:
Tju [1.3M]3 years ago
8 0

Answer:

1. No, it is a colloid

2. Yes

Explanation:

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A 10.0 kg box is dragged 5.00 meters across a rough horizontal floor by a rope with a tension
Lemur [1.5K]

Answer:

Explanation:

Net work done on the box = increase in kinetic energy

= 1 / 2 x 10 ( 5² - 3² )

= 80 J .

5 0
3 years ago
10 kg mass sliding along a horizontal rough surface at speed 30 m/s and rests in 6 seconds. what is the coefficient of friction
Aleks [24]

The coefficient of friction is 0.5.

<h3>What is Coefficient of friction?</h3>

The ratio of friction force to normal force is known as the coefficient of friction (COF), which has no dimensions. Those materials are said to as lubricous if their COF is less than 0.1. Surface roughness and COF are dependent on the composition of the materials.

Initial velocity, u=30m/s

time taken = 6s

Final velocity, v=0

Acceleration of friction, a=μg=10μms⁻²

Apply first kinematic equation of motion,

v - u = at

0 - 30 = (-10μms⁻²) × 6

-30 = -60μ

μ = 1/2

μ = 0..5  

Hence, coefficient of friction is 0.5

to learn more about Coefficient of friction go to - brainly.com/question/14121363

#SPJ4

7 0
2 years ago
A box is moved 20 m across a smooth floor by a force making a downward angle with the floor, so that there is effectively a 10 N
Usimov [2.4K]

Answer:

100 J

Explanation:

From the question, The work done by the forces in moving the box is given as

W = FxdcosФ+Fydcosα................... Equation 1

Where W = Work done, Fx = force acting parallel to the floor, d = distance moved by the box, Ф = angle the parallel force makes with the floor, Fy = force acting perpendicular to the floor, α = angle the perpendicular force make with the floor.

Give: Fx = 10 N, d = 20 m, Fy = 5 N, Ф = 0°, α = 90°

Substitute into equation 1

W = 10×10×cos0°+5×20×cos90°

W = 10×10×1+0

W = 100 J.

Note: The work done by the perpendicular force is zero

Hence the work done = 100 J

5 0
3 years ago
Calculate the force required to accelerate<br> a 600 g ball from rest to 14 m/s in 0.1 s.
Evgen [1.6K]

<u>Statement</u><u>:</u>

A force is required to accelerate a 600 g ball from rest to 14 m/s in 0.1 s.

<u>To </u><u>find </u><u>out</u><u>:</u>

The force required to accelerate the ball.

<u>Solution</u><u>:</u>

  • Mass of the ball (m) = 600 g = 0.6 Kg
  • Initial velocity (u) = 0 m/s [it was at rest]
  • Final velocity (v) = 14 m/s
  • Time (t) = 0.1 s

  • Let the acceleration be a.
  • We know the equation of motion,
  • v = u + at

  • Therefore, putting the values in the above formula, we get
  • 14 m/s = 0 m/s + a × 0.1 s
  • or, 14 m/s ÷ 0.1 s = a
  • or, a = 140 m/s²

  • Let the force be F.
  • We know, the formula : F = ma

  • Putting the values in the above formula, we get
  • F = 0.6 Kg × 140 m/s²
  • or, F = 84 N

<u>Answer</u><u>:</u>

The force required to accelerate the ball is 84 N and this force acts along the direction of motion.

Hope you could understand.

If you have any query, feel free to ask.

5 0
3 years ago
The Bohr model of the atom addressed the problem of _____.
Ghella [55]

The Bohr model of atom addressed the problem of why electrons did not fall into the positive nucleus.

As per  Rutherford's atomic model, electrons are revolving around the nucleus just like planets are revolving around the sun. But his description about electronic motion was against the Clark Maxwell's electromagnetic theory .

As per Maxwell any charged particle under accelerated motion will release energy. The electron is a negatively charged particle which under acceleration will come closer and closer to the nucleus and finally it will fall on the nucleus for which the whole atom will be destroyed. The path of electron will be a spiral one instead of circular. Hence Rutherford's theory could not explain the stability of nucleus.

Bohr addressed that electron in particular orbit has a fixed amount of energy .There will  be no emission or absorption of energy as long as electron revolves around the nucleus.The electron will absorb energy when it will jump from ground state to excited state and will emit energy when it will jump from excited state to ground state.

Hence the option A is right,.


7 0
3 years ago
Read 2 more answers
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