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3 years ago

You push a ball to start rolling it across a frictionless surface how far will the ball roll ?

1 answer:

6 0

If the surface is really truly indeed frictionless,
then the ball will never stop.

It'll keep going, in a straight line, at whatever speed
it had when you stopped pushing it.

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A 18.0-kg rock is sliding on a rough, horizontal surface at 7.10 m/s and eventually stops due to friction. the coefficient of ki

Bond [772]

A = .3*g = 2.94 m/s²

<span>t = v/a = 9/2.94 = 3.061 sec </span>

<span>W = E/t = ½mv²/t = ½*40*9²/3.061 = 529.2 watts</span>

4 0

3 years ago

This is a problem about a child pushing a stack of two blocks along a horizontal floor. The masses of the blocks, and the coeffi

Rufina [12.5K]

Answer:

N = 23.4 N

Explanation:

After reading that long sentence, let's solve the question

The contact force is the so-called normal in this case we can find it by writing the translational equilibrium equation for the y axis

N - w₁ -w₂ =

N = m₁ g + m₂ g

N = g (m₁ + m₂)

let's calculate

N = 9.8 (0.760 + 1.630)

N = 23.4 N

This is the force of the support of the two blocks on the surface.

7 0

3 years ago

A 2.93 kg particle has a velocity of (2.98 i hat - 3.98 j) m/s.

cupoosta [38]

Answer:

a) The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

b) The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

Explanation:

a) The vectorial equation of momentum is represented by the following expression:

$\vec p = m\cdot \vec v$ (1)

Where:

$\vec p$ - Vector momentum, measured in kilogram-meters per second.

$m$ - Mass of the particle, measured in kilograms.

$\vec v$ - Vector velocity, measured in meters per second.

If we know that $m = 2.93\,kg$ and $\vec v = 2.98\,\hat{i}-3.98\,\hat{j}\,\,\,\left[\frac{m}{s} \right]$ , then the momentum is:

$\vec p = (2.93)\cdot (2.98\,\hat{i}-3.98\,\hat{j})\,\,\,\left[\frac{kg\cdot m}{s} \right]$

$\vec p = 8.731\,\hat{i}-11.661\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]$

The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

b) The magnitude and direction of momentum are represented by the following expressions:

$\|\vec p \| = \sqrt{p_{x}^{2}+p_{y}^{2}}$ (2)

$\theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}} \right)$ (3)

Where:

$\|\vec p\|$ - Magnitude of momentum, measured in kilogram-meters per second.

$\theta$ - Direction of momentum, measured in sexagesimal degrees.

If we know that $p_{x} = 8.731\,\frac{kg\cdot m}{s}$ and $p_{y} = -11.661\,\frac{kg\cdot m}{s}$ , then the magnitude and direction of momentum are, respectively:

$\|\vec p\| = \sqrt{\left(8.731\,\frac{kg\cdot m}{s} \right)^{2}+\left(-11.661\,\frac{kg\cdot m}{s} \right)^{2}}$

$\|\vec p\| \approx 14.567\,\frac{kg\cdot m}{s}$

$\theta =\tan^{-1}\left(\frac{-11.661\,\frac{kg\cdot m}{s} }{8.731\,\frac{kg\cdot m}{s} } \right)$

$\theta \approx 306.823^{\circ}$

The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

6 0

3 years ago

Motion is always compared to something called

HACTEHA [7]

Movement can be without change in positions. As only a part of it can be made to move. If you move there is a motion and if there is a motion there is a movement so they're same thing. In motion whole of body moves, but in movements only the parts of body moves.

7 0

4 years ago

Work is being done in which of these situations? all motions are at a constant velocity

fredd [130]

Where the force is not perpendicular to the path of motion

are you missing the the situations ?

7 0

3 years ago