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grin007 [14]
3 years ago
7

When a board with a box on it is slowly tilted to larger and larger angle, common experience shows that the box will at some poi

nt "break loose" and start to accelerate down the board.
The box begins to slide once the component of gravity acting parallel to the boardequals the force of static friction. Which of the following is the most general explanation for why the box accelerates down the board?

The force of kinetic friction is smaller than that of static friction, butF_gremains the same.
Once the box is moving,F_gis smaller than the force of static friction but larger than the force of kinetic friction.
Once the box is moving,F_gis larger than the force of static friction.
When the box is stationary,F_gequals the force of static friction, but once the box starts moving, the sliding reduces the normal force, which in turn reduces the friction.
Physics
1 answer:
eimsori [14]3 years ago
6 0

Answer: The force of kinetic friction is smaller than that of static friction, but F_g  remains the same.

Explanation:

The situation is same as when a book is pushed with an increasing force on a table; When the force is low, book doesn't move, until that under a given force starts moving, and then it goes on movement even if the force decreases a bit.

The physical explanation for this, that friction force adopts any value needed to avoid to move the object, till a limit value is achieved, called static friction force, equal to the normal force times the static friction coefficient.

Once in movement, the kinetic friction coefficient replaces the static one , and  in general is lower than the static one, so the force diminishes.

In the case of the box sliding down the board, the force that tries to move the object down the board, is the component of the weight parallel to the board, that can be showed that being equal to the weight times the sinus of the angle of the board with the horizontal, as follows:

F_g = m g sin θ

When θ increases, F_g does the same, so friction force always has the same magnitude than F_g (but opposite direction) so the box doesn't move, till that θ takes a value that produces a F_g equal to static friction force.

Beyond this limit angle, F_g (remaining the same for a given angle) is greater than the kinetic friction force, and the box slides.

In the limit, when θ=90º, sin θ =1⇒ F_g = mg, so the object is in free fall.

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Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh
frosja888 [35]

Answer:

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

Explanation:

Physically speaking, the resulting velocity of the helicopter (\vec v_{H}), measured in meters per second, is equal to the absolute velocity of the wind (\vec v_{W}), measured in meters per second, plus the velocity of the helicopter relative to wind (\vec v_{H/W}), also call velocity at still air, measured in meters per second. That is:

\vec v_{H} = \vec v_{W}+\vec v_{H/W} (1)

In addition, vectors in rectangular form are defined by the following expression:

\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha) (2)

Where:

\|\vec v\| - Magnitude, measured in meters per second.

\alpha - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W}) (3)

\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)

If we know that \|\vec v_{W}\| = 25\,\frac{m}{s}, \|\vec v_{H/W}\| = 125\,\frac{m}{s}, \alpha_{W} = 240^{\circ} and \alpha_{H/W} = 325^{\circ}, then the resulting velocity of the helicopter is:

\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

8 0
3 years ago
calculate the amount of work done by a person while taking a bag of mass 100kg to the top of the building hight 10m. The mass of
vredina [299]

Explanation:

Total mass=100+10=110

Total weight=mass×gravitational field strength

=110×10

=1100N

Work done=force×distance

=1100×10

=11000J

<em>Please mark me as brainliest if this helped you!</em>

6 0
2 years ago
The electromagnetic waves that have the lowest frequencies are called
o-na [289]
A. Radio waves
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3 years ago
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mestny [16]

Answer:bsnz

Explanation:nslsosols

7 0
2 years ago
The energy required to ionize magnesium is 738 kj/mol. What minimum frequency of light is required to ionize magnesium
Cerrena [4.2K]

Answer:

The minimum frequency required to ionize the photon is 111.31 × 10^{37} Hertz

Given:

Energy = 378 \frac{kJ}{mol}

To find:

Minimum frequency of light required to ionize magnesium = ?

Formula used:

The energy of photon of light is given by,

E = h v

Where E = Energy of magnesium

h = planks constant

v = minimum frequency of photon

Solution:

The energy of photon of light is given by,

E = h v

Where E = Energy of magnesium

h = planks constant

v = minimum frequency of photon

738 × 10^{3} = 6.63 × 10^{-34} × v

v = 111.31 × 10^{37} Hertz

The minimum frequency required to ionize the photon is 111.31 × 10^{37} Hertz


7 0
3 years ago
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