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MAXImum [283]
3 years ago
14

Identify the forces acting on the object of interest. From the list below, select the forces that act on the piano.

Physics
1 answer:
Nonamiya [84]3 years ago
3 0

Answer:

gravitational force acting on the piano (piano's weight)

force of Chadwick on the piano

force of the floor on the piano (normal force)

Explanation:

Figure is missing: found it in attachment.

In the figure, we notice that the piano is accelerating along the horizontal direction: this means that there is a net force acting along this direction. This force is prodiced by Chadwick, and it acts in the same direction as the acceleration, so one force is:

force of Chadwick on the piano

Also, every object on Earth experencies the force of gravity, which is also called weight. The weight of the piano acts downward, so a second force is:

gravitational force acting on the piano (piano's weight)

Finally, we notice that the piano is in equilibrium along the vertical direction (no acceleration): this is because there is another force acting opposite to the piano's weight (and with equal magnitude), and this force is the normal force exerted by the floor on the piano:

force of the floor on the piano (normal force)

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answer should be 10 because the line goes from (0,0) then to (1,10) and so on

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Two bicycle tires are set rolling with the same initial speed of 3.30 m/s along a long, straight road, and the distance each tra
vredina [299]

Answer:

At low pressure- \mu_{k}=0.02315

At high pressure- \mu_{k}=0.00445

Explanation:

Initial speed, V_{i}=3.3 m/s

Final speed, V_{f}=3.3/2= 1.65 m/s

Net horizontal force due to rolling friction F_{net}=\mu_{k} mg where m is mass, g is acceleration due to gravity, \mu_{k} is coefficient of rolling friction

From kinematic relation, V_{f}^{2}- V_{i}^{2}=2ad

For each tire,

V_{f}^{2}- V_{i}^{2}=2\mu_{k}gd

Making \mu_{k} the subject

\mu_{k}=\frac {V_{f}^{2}- V_{i}^{2}}{2gd}

Under low pressure of 40 Psi, d=18 m

\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*18}=-0.02315

Therefore, \mu_{k}=0.02315

At a pressure of 105 Psi, d=93.7

\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*93.7}=-0.00445

Therefore, \mu_{k}=0.00445

4 0
3 years ago
Two forces act on an object. One force has a magnitude of 10 N directed north, and the other force has a magnitude of 2 N direct
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Answer:

8 N North.

Explanation:

Given that,

One force has a magnitude of 10 N directed north, and the other force has a magnitude of 2 N directed south.

We need to find the magnitude of net force acting on the object.

Let North is positive and South is negative.

Net force,

F = 10 N +(-2 N)

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So, the magnitude of net force on the object is 8 N and it is in North direction (as it is positive). Hence, the correct option is (d) "8N north".

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Answer:

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Explanation:

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a_sh-v [17]

Answer:

The acceleration is 11.25 km/hr

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