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3 years ago

A wrecking ball of mass 500 kg hangs form a crane by a cable of length 25 mit

Physics

1 answer:

Ne4ueva [31]3 years ago

5 0

Given :

A wrecking ball of mass 500 kg hangs form a crane by a cable of length 25 m

this wrecking ball is released from an angle of 40 degrees.

To Find :

The height from the lowest point of the arc.

Solution :

From the attached image we have to find the value of h.

So,

$cos \ 40^o = \dfrac{L-h}{L}\\\\0.76=\dfrac{25-h}{25}\\\\h = 25( 1- 0.76)\\\\h = 6 \ m$

Hence, this is the required solution.

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One of the primary goals of the Kepler space telescope is to search for Earth-like planets. Data gathered by the telescope indic

amid [387]

Answer:

85.62 m

168.75 years

101.04 years

Explanation:

$L_0$ = Length of ship = 143 m

v = Velocity of ship = 0.8c

c = Speed of light

s = Distance to Boralis orbit = 135 ly

Gamma value

$\gamma=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\\\Rightarrow \gamma=\dfrac{1}{\sqrt{1-\dfrac{0.8^2c^2}{c^2}}}\\\Rightarrow \gamma=1.67$

Length contraction is given by

$L=\dfrac{L_0}{\gamma}\\\Rightarrow L=\dfrac{143}{1.67}\\\Rightarrow L=85.62\ m$

The length is 85.62 m

Time taken

$t=\dfrac{s}{v}\\\Rightarrow t=\dfrac{135}{0.8}\\\Rightarrow t=168.75\ years$

Time taken from the perspective one Earth is 168.75 years

Time dilation is given by

$t'=\dfrac{t}{\gamma}\\\Rightarrow t'=\dfrac{168.75}{1.67}\\\Rightarrow t'=101.04\ years$

The time taken from the perspective of the ship is 101.04 years

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4 years ago

Describe at least two ways nuclear fission was used in the twentieth century

allsm [11]

Is there any options or just a type up answer?

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3 years ago

In which of the following cases would sound reach each ear out of phase

Sonja [21]

When standing behind the sound source

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4 years ago

What type of eclipse makes the sun appear in the form of a ring

Vitek1552 [10]

Annular solar eclipse.

8 0

4 years ago

Read 2 more answers

fomenos

Answer:

The correct option is;

(A) 1 m/s

Explanation:

Given that Ramu and Somu are running towards North with 3 m/s and 4 m/s respectively.

Their friend Srinu is running towards South with a speed of 2 m/s

The magnitude of the relative velocity of Ramu with respect to Ramu is found as follows;

Velocity of Ramu = 3 m/s North ↑

Velocity of Srinu = 2 m/s South↓ = -2 m/s North ↑

Relative velocity of Ramu to Srinu = Velocity of Ramu North + Velocity of Srinu North

Relative velocity of Ramu to Srinu = 3 m/s + (-2) m/s = (3 - 2) m/s = 1 m/s

Relative velocity of Ramu to Srinu = 1 m/s.

4 0

4 years ago