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mylen [45]
3 years ago
12

A 6.0-kg block initially at rest is pulled to the right along a frictionless, horizontal surface by a constant horizontal force

of magnitude 12 N. Find the block’s speed after it is has moved through a horizontal distance of 3.0 m
Physics
1 answer:
Citrus2011 [14]3 years ago
3 0

Answer:

Explanation:

Given that,

Block of mass M= 6kg

Initially at rest u =0

A force F = 12N is used in pulling it along the horizontal

Distance moved s = 3m

Speed.

Using Newton second law

ΣFx = ma

12 = 6a

Then, a = 12/6

a = 2m/s²

Then, using equation of motion

v² = u² + 2as

v² = 0² + 2×2×3

v² = 0 + 12

v = √ 12

v = 3.46 m/s

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max2010maxim [7]

Answer:

E. all of these

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all are necessary to designate a point in space. Hence option E is correct.

For example in simple harmonic motion we need to specify all the above factors of the object in order to designate the position of the object.  

8 0
3 years ago
Use this free body diagram to help you find the magnitude of the force F2 needed to keep this block in static equilibrium. WILL
oksian1 [2.3K]
Static equilibrium means that all forces are equal, so make this easiest you want to break F1 into it's horizontal and vertical components. As there are no other forces acting in the horizontal, we know the horizontal component of F1 is 40N. This allows the vertical component to be found using pythagorus theorem. After finding the vertical and horizontal components, you just have to add the vertical components to find the difference between the up and down.

4 0
3 years ago
If the mass of the sun is 1x, at least one planet will fall into the habitable zone if I place a planet in orbits___, ____, ____
nasty-shy [4]

If the mass of the sun is 1x, at least one planet will fall into the habitable zone. if I place a planet in orbits 2, 6, and 75, and all planets will orbit the sun successfully.

If the mass of the sun is 2x, at least one planet will fall into the habitable zone. if I place a planet in orbits 84, 1, and 5, and all planets will orbit the sun successfully.

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8 0
2 years ago
A cubical block of wood, 10.0 cm on a side, floats at the interface between oil and water with its lower surface 1.50 cm below t
bezimeni [28]

Answer:

the gauge pressure at the upper face of the block is 116 Pa

Explanation:

Given the data in the question;

A cubical block of wood, 10.0 cm on a side.

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density ρ = 790 kg/m³

Using expression for the gauged pressure;

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we know that, acceleration due to gravity g = 9.8 m/s²

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= 116.13 ≈ 116 Pa

Therefore, the gauge pressure at the upper face of the block is 116 Pa

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