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Brrunno [24]
2 years ago
7

A simple pendulum consists of a brass sphere of mass m = 1 kg suspended on a string of length L ≈ 1 meter. The pendulum oscillat

es with amplitude A = 1 cm and period T = 2 seconds. Which of the following alterations will change the period of the pendulum’s oscillations to T ′ = 1 second? 1. Using a lighter brass sphere of mass m ′ ≈ 0.25 kg. 2. Using a shorter string of length L ′ ≈ 0.25 meters. 3. Using a longer string of length L ′ ≈ 2 meters. 4. Using a longer string of length L ′ ≈ 4 meters. 5. Using a shorter string of length L ′ ≈ 0.5 meters. 6. Increasing the oscillation amplitude to A ′ = 4 cm. 7. Using a heavier brass sphere of mass m ′ ≈ 4 kg. 8. Reducing the oscillation amplitude to A ′ = 0.25 cm.
Physics
1 answer:
hram777 [196]2 years ago
6 0

Answer:

2. Using a shorter string of length L ′ ≈ 0.25 meters

5. Using a shorter string of length L ′ ≈ 0.5 meters

Explanation:

The period of a pendulum is given by

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration due to gravity

We see from the formula that the period of the pendulum depends only on its length, not on its mass or its amplitude of ocillation. Therefore, the only alterations that can change the period of the pendulum are the ones where its length is changed.

Moreover, we notice that the period is proportional to the square of the length: this means that in order to decrease the period of the pendulum (the problem asks us which alterations will reduce the period of the pendulum from 2 s to 1 s), the length of the pendulum should also be reduced.

Therefore, the only alterations that will reduce the period of the pendulum are:

2. Using a shorter string of length L ′ ≈ 0.25 meters

5. Using a shorter string of length L ′ ≈ 0.5 meters

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Studentka2010 [4]

(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration (g=9.8 m/s^2) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

v=18.9 m/s

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

v_y^2 - u_y^2 = 2ah (1)

where

u_y = 16.4 m/s is the initial vertical velocity

v_y is the vertical velocity at height h

a=g=-9.8 m/s^2 is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, v_y = 0, so we can use the equation to find the maximum height

h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m

So, at half of the maximum height,

h = \frac{13.7}{2}=6.9 m

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s

And so, the speed of the stone at half of the maximum height is

v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

v_1 = 18.9 m/s

while the speed at half of the maximum height (part b) is

v_2 = 22.2 m/s

So the difference is

\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s

And so, in percentage,

\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%

So, the stone in part (b) is moving 17.4% faster than in part (a).

4 0
2 years ago
A nonuniform, 80.0-g, meterstick balances when the support is placed at the 51.0-cm mark. At what location on the meterstick sho
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Answer:34 cm

Explanation:

Given

mass of meter stick m=80 gm

stick is balanced when support is placed at 51 cm mark

Let us take 5 gm tack is placed at x cm on meter stick so that balancing occurs at x=50 cm mark

balancing torque

80\times 10^{-3}(51-50)=5\times 10^{-3}(50-x)

80=5(50-x)

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x=\frac{170}{5}

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tamaranim1 [39]

Answer:

Explanation:

Given

Weight of car W=14,000\ N

mass of car m=\frac{14,000}{9.8}=1428.57\ N

velocity of car v=25\ m/s

radius r=200\ m

(a)Centripetal acceleration is given by

a_c=\frac{v^2}{r}

a_c=\frac{25^2}{200}

a_c=3.125\ m

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F=F_c=\frac{mv^2}{r}

F=\frac{1428.57\times 25^2}{200}

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(c)Friction force between car and tires is given by

=\mu N

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F_c=F_r

4464.285=\mu \times 1428.57\times 9.8

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