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Alika [10]
3 years ago
11

A stationary block of mass 35 Kg is suspended on a string.What is the tension in the string ?( Neglect the mass of the string)

Physics
1 answer:
SVEN [57.7K]3 years ago
8 0

Answer:

Tstring = 343.35 [N].

Explanation:

Since the block is in equilibrium, we can say that the sum of the forces on the Y-axis or vertical is equal to zero.

∑Fy = 0

T_{string}-W=0

where:

Tstring = tension of the string [N]

W = weight of the block [N]

And the weight can be calculated multiplying the mass of the block by the gravity acceleration.

Now replacing:

T_{string}=m*g\\T_{string}=35*9.81\\T_{string}=343.35[N]

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An ultra intense laser is one with which intensities greater than 1015 W cm-2 can be achieved.

Explanation:

This intensity, which was the upper limit of lasers until the invention of the Chirped Pulse Amplification, CPA technique, is the value around which nonlinear effects on the transport of radiation in materials begin to appear.

Currently, the most powerful lasers reach intensities of the order of 1021W cm-2 and powers of Petawatts, PW, in each pulse. This range of intensities has opened the door for lasers to a multitude of disciplines and scientific areas traditionally reserved for accelerators and nuclear reactors, applying as generators of high-energy electron, ion, neutron and photon beams, without the need for expensive infrastructure.

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Where coastal rock is softer, waves erode the land faster ​
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3 years ago
Gold has a density of 19300 kg/m³. Calculate the mass of 0.02 m³ of gold in kilograms.
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Answer:

Mass = 386 kg

Explanation:

<u><em>Density = Mass / Volume</em></u>

Mass = Density × Volume

Where D = 19300 kg/m³ , V = 0.02 m³

<em>Putting the given in the above formula</em>

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What is the displacement of the object from 2 seconds to 6 seconds?
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3 years ago
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Allen and Jason are chucking a speaker around. On one particular throw, Allen throws the speaker, which is playing a pure tone o
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Answer:

Explanation:

We shall apply Doppler's effect of sound .

speaker is the source , Jason is the observer . Source is moving at 10 m /s , observer is moving at  6 m/s .

apparent frequency = f_o\times\frac{V+v_o}{ V-v_s}

V is velocity of sound , v₀ is velocity of observer and v_s is velocity of source and f_o is real frequency of source .

Here V = 340 m/s , v₀ is 6 m/s , v_s is 10 m/s . f_o = f

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So m = 346 , n = 330 .

8 0
3 years ago
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