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DiKsa [7]
2 years ago
7

1. How would the motion of a pendulum change at high altitude like a high mountain top? How would the motion change under weight

less conditions?
Physics
1 answer:
gavmur [86]2 years ago
6 0

Answer:

A pendulum moves with the principle of small angles that show that the frequency and period of the pendulum are not dependent on the initial angular displacement of the mass it carries.

On a high mountain top, a reduced force of gravity is exerted on the mass, therefore, producing long periods of oscillation on the pendulum.

However, under weightless conditions where gravity is zero, the pendulum will not work as what keeps the mass in oscillation is gravity. If therefore gravity is absent there can be no oscillation.

Explanation:

On a high mountain top, a reduced force of gravity is exerted on the mass, therefore, producing long periods of oscillation on the pendulum.

However, under weightless conditions where gravity is zero, the pendulum will not work as what keeps the mass in oscillation is gravity. If therefore gravity is absent there can be no oscillation.

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If temperature and the number of particles remain constant, and increase in pressure will cause volume to
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3 years ago
5. A ball rolls off a 1.5 m tall horizontal table and lands on the floor 0.70 m away.
pav-90 [236]

Take the starting position of the ball 1.5 m above the floor to be the origin. Then at time t, the ball's horizontal and vertical positions from the origin are

x = v₀ t

y = -1/2 gt²

where v₀ is the initial speed with which it rolls off the edge and g = 9.8 m/s².

A. The floor is 1.5 m below the origin, so we solve for t when y = -1.5 m :

-1.5 m = -1/2 gt²

⇒   t² = (3.0 m)/g

⇒   t = √((3.0 m)/g) ≈ 0.55 s

B. It would take the same amount of time.

C. The ball travels a horizontal distance of 0.70 m before reaching the floor, so we solve for v₀ with t = 0.55 s :

0.70 m = v₀ (0.55 s)

⇒   v₀ = (0.70 m) / (0.55 s) ≈ 1.3 m/s

D. At time t, the ball has horizontal and vertical velocity components

v[x] = 1.3 m/s

v[y] = -gt

so the horizontal component of the ball's final velocity vector is the same as the initial one, 1.3 m/s.

E. The vertical component of velocity would be

v[y] = -g (0.55 s) ≈ -5.4 m/s

F. The magnitude of the final velocity would be

√((1.3 m/s)² + (-5.4 m/s)²) ≈ 5.6 m/s

G. The final velocity vector makes an angle θ with the horizontal such that

tan(θ) = (-5.4 m/s) / (1.3 m/s)

⇒   θ = arctan(-5.4/1.3) ≈ -77°

i.e. approximately 77° below the horizontal.

3 0
2 years ago
student conducted an experiment and find the density of an ICEBERGE. A students than recorded the following readings. Mass 425
Fiesta28 [93]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The experimental value of density is   \rho  =  1.05*10^{3} \ kg/m^3  \pm  101 \ kg/m^3

Comparing it with the value of density of water (1.0*10^{3} \ kg/m^3) we can see that the density of ice is greater

Explanation:

From the question we are told

    The mass is M  =  (425 \pm 25) \ g =(0.425  \pm 0.025) \ kg

   The volume is V  =  (405 \pm 15 ) \ mL = (0.000405 \pm 1.5*10^{-5}) \  m^3

The experimental value of density is mathematically evaluated as

        \rho =  \frac{M}{V}

       \rho =  \frac{0.425}{0.000405}

       \rho =  1.05 *10^{3} \ kg/m^3

The possible error in this experimental value of density is mathematically evaluated as

        \frac{\Delta \rho}{\rho} =  \frac{\Delta  M}{M} +\frac{\Delta  V}{V}

substituting value

         \frac{\Delta \rho}{1.05*10^{3}} =  \frac{0.025}{0.425} +\frac{1.5*10^{-5}}{0.000405}

        \Delta \rho  = 101 \ kgm^{-3}

Thus the experimental value of density is

             \rho  =  1.05*10^{3} \ kg/m^3  \pm  101 \ kg/m^3

                     

     

6 0
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