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OLEGan [10]
3 years ago
13

Suppose we wish to use a 8.0 m iron bar to lift a heavy object by using it as a lever. If we place the pivot point at a distance

of 1.0 m from the end of the bar that is in contact with the load and we can exert a downward force of 562 N on the other end of the bar, find the maximum load that this person is able to lift (pry) using this arrangement (neglect the mass of the bar in this problem).
Physics
1 answer:
lesya692 [45]3 years ago
5 0

Answer:

W = 3934 N ,     m = 401.43 kg

Explanation:

This problem can be solved using the rotational equilibrium relation, where we place the reference system at the pivot point and assume that the counterclockwise turns are positive.

     ∑τ = 0

     F 7 - W 1 = 0

    W = F 7/1

     W = 562 7

     W = 3934 N

   

W = mg

     m = W / g

     m = 3934 / 9.8

     m = 401.43 kg

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marysya [2.9K]

Answer:

a.) high amplitude, high frequency

Explanation:

Frequency and amplitude are properties of sound. Varying these properties changes how people perceive sound.

While hearing sound of a particular frequency we call it pitch i.e., the perception of a frequency of sound.

High pitch means high frequency and high frequency is perceived to have a shrill sound.

The loudness of a sound is measured by the intensity of sound i.e., the energy the sound possesses per unit area. As the amplitude increases the intensity increases. So, a loud sound will have higher density.

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4 0
3 years ago
A 15 kg mass is lifted to a height of 2m. What is gravitational potential energy at this position
7nadin3 [17]

Answer: 294J

Explanation:

U_g=mg(y_f-y_i)

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7 0
3 years ago
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A hot-air balloon plus cargo has a mass of 308 kg and a volume of 2910 m3 on a day when the outside air density is 1.22 kg/m3. T
lesya [120]

9514 1404 393

Answer:

  1.114 kg/m³

Explanation:

The total mass of the air in the balloon and the balloon + cargo will be the mass of the displaced air. If d is the density of the air in the balloon, then we have ...

  2910d +308 = 2910×1.22

Solving for d, we find ...

  2910d = 2919(1.22) -308

  d = 1.22 -308/2910

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The density of the hot air is about 1.114 kg/m³.

6 0
3 years ago
Consider the following four objects: a hoop, a solid sphere, a flat disk, a hollow sphere. Each of the objects has mass M and ra
Svetlanka [38]

Answer:

Hoop.

Explanation:

The angular acceleration performed at a given torque:

\alpha = \frac{\tau}{I}

The moments of inertia of each element are described below:

Hoop

I = M\cdot R^{2}

Solid sphere

I = \frac{2}{5}\cdot M \cdot R^{2}

Flat disk

I = \frac{1}{2}\cdot M \cdot R^{2}

Hollow sphere

I = \frac{2}{3}\cdot M \cdot R^{2}

The greater the moment of inertia, the greater the torque to obtain the same angular acceleration. Therefore, the hoop requires the largest torque to receive the same angular acceleration.

8 0
3 years ago
On a certain planet, which is perfectly spherically symmetric, the free-fall acceleration has magnitude g = go at the north pole
ohaa [14]
The reason why there is a difference between free-fall acceleration is a centrifugal force.
I attached a diagram that shows how this force aligns with the force of gravity.
From the diagram we can see that:
F=F_g-F_{cf}=mg'-mw^2r'cos(\alpha)\\ ma=mg'-mw^2r'cos(\alpha)\\ a=g'-w^2rcos^2(\alpha)\\
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We can calculate g' and wr^2 from the given conditions in the problem.
g(90)=g_0;\ g_0= g'-w^2rcos^2(90)\\
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w^2r=g_0(a-1)

Our final equation is:
g=g_0-g_0(a-1)cos^2(\alpha)
Colatitude is:
\alpha_c=90^\circ-\alpha
The answer is:
g=g_0-g_0(a-1)cos^2(90-9)=g_0-g_0(a-1)sin^2(9)

5 0
3 years ago
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