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2 years ago

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A student attaches a block to a vertical spring so that the block-spring system will oscillate if the block-spring system is rel

eased from rest at a vertical position that is not the system’s equilibrium position. The system oscillates near Earth’s surface. The system is then taken to the Moon’s surface, where the gravitational field strength is nearly 16 that of the gravitational field strength near Earth's surface. Which of the following claims is correct about the period of oscillation for the system?

1 answer:

vodka [1.7K]2 years ago

4 0

Answer:

Time period of the motion will remain the same while the amplitude of the motion will change

Explanation:

As we know that time period of oscillation of spring block system is given as

$T= 2\pi\sqrt{\frac{M}{k}}$

now we know that

M = mass of the object

k = spring constant

So here we know that the time period is independent of the gravity

while the maximum displacement of the spring from its mean position will depends on the gravity as

$mg = kx$

$x = \frac{mg}{k}$

so we can say that

Time period of the motion will remain the same while the amplitude of the motion will change

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A pair of glasses is dropped from the top of a 32.0m stadium. A pen is dropped 2.Os later. How high above the ground is the pen

Svetllana [295]

Answer:

$h_p = 30.46\ m$

Explanation:

<u>Free Fall Motion</u>

A free-falling object refers to an object that is falling under the sole influence of gravity. If the object is dropped from a certain height h, it moves downwards until it reaches ground level.

The speed vf of the object when a time t has passed is given by:

$v_f=g\cdot t$

Where $g = 9.8 m/s^2$

Similarly, the distance y the object has traveled is calculated as follows:

$\displaystyle y=\frac{g\cdot t^2}{2}$

If we know the height h from which the object was dropped, we can solve the above equation for t:

$\displaystyle t=\sqrt{\frac{2\cdot y}{g}}$

The stadium is h=32 m high. A pair of glasses is dropped from the top and reaches the ground at a time:

$\displaystyle t_1=\sqrt{\frac{2\cdot 32}{9.8}}=2.56\ sec$

The pen is dropped 2 seconds after the glasses. When the glasses hit the ground, the pen has been falling for:

$t_2=2.56 - 2 = 0.56\ sec$

Therefore, it has traveled down a distance:

$\displaystyle y=\frac{9.8\cdot 0.56^2}{2} = 1.54\ m$

Thus, the height of the pen is:

$h_p = 32 - 1.54\Rightarrow h_p=30.46\ m$

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2 years ago

Calculate the average speed and average velocity of a complete round-trip in which the outgoing 250 km is covered at 95 km/h fol

ser-zykov [4K]

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. From them we will consider speed as the distance traveled per unit of time. Said unit of time will be cleared to find the total time taken to travel the given distance. Later with the calculated average times and distances, we will obtain the average speed.

PART A)

The time taken to travel a distance of 250km with a speed of 95km/h is

$t = \frac{d}{v}$

$t = \frac{250km}{95km/h}$

$t = 2.63h$

Time taken for the lunch is

$t = 1h$

The time taken travel a distance of 250km with a speed of 55km/h

$t = \frac{d}{v}$

$t = \frac{250}{55}$

$t = 4.54h$

The total time taken is

$t = t_{outgoing}+t_{lunch}}t_{return}$

$t = 2.63+1+4.54$

$t = 8.17h$

The average speed is the ratio of total distance and total time

$v = \frac{250+250}{8.17}$

$v = 61.15km/h$

PART B)

As the displacement is zero the average velocity is zero.

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2 years ago

A swimmer does 3,560 J of work in 55 s. What is the swimmer’s power output? Round your answer to two significant figures. The po

Natasha2012 [34]

The value of the swimmer's power output is calculated by dividing the work done by the time it took for the work to be completed. From the given in this item,
P = 3560 J/ 55 s = 64.73 W
Rounding off to two significant figures will give us 65 W.

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2 years ago

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An overnight rainstorm has caused a major roadblock. Three massive rocks of mass m1=584 kg, m2=838 kg, and m3=322 kg have blocke

Elena-2011 [213]

Answer:

Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N

Explanation:

Total force required = Mass x Acceleration,

F = ma

Here we need to consider the system as combine, total mass need to be considered.

Total mass, a = m₁+m₂+m₃ = 584 + 838 + 322 = 1744 kg

We need to accelerate the group of rocks from the road at 0.250 m/s²

That is acceleration, a = 0.250 m/s²

Force required, F = ma = 1744 x 0.25 = 436 N

Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N

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2 years ago

A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.

babymother [125]

Answer

given,

mass of the person, m = 50 Kg

length of scaffold = 6 m

mass of scaffold, M= 70 Kg

distance of person standing from one end = 1.5 m

Tension in the vertical rope = ?

now equating all the vertical forces acting in the system.

T₁ + T₂ = m g + M g

T₁ + T₂ = 50 x 9.8 + 70 x 9.8

T₁ + T₂ = 1176...........(1)

system is equilibrium so, the moment along the system will also be zero.

taking moment about rope with tension T₂.

now,

T₁ x 6 - mg x (6-1.5) - M g x 3 = 0

'3 m' is used because the weight of the scaffold pass through center of gravity.

6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3

6 T₁ = 4263

T₁ = 710.5 N

from equation (1)

T₂ = 1176 - 710.5

T₂ = 465.5 N

hence, T₁ = 710.5 N and T₂ = 465.5 N

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2 years ago