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3 years ago

Can anyone help?6th question.free brainliest answer...

Physics

1 answer:

DIA [1.3K]3 years ago

6 0

There's a crest and a trough in each complete wave. So the question is describing 10 complete waves.

After that, the question becomes somewhat murky. It goes on to say "its time period is 0.2 seconds".

-- The "time period" of a wave is usually defined as the time for one complete wave. If that's what the phrase means, then ...

Frequency = ( 1/0.2sec )

Frequency = 5 Hz.

= = = = = = = = = =

BUT ... Is the question awkwardly trying to tell us that the 10 waves take 0.2 seconds ? If that's what it's saying, then ...

Frequency = (10) / (0.2 sec)

Frequency = 50 Hz .

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A chain 72 meters long whose mass is 29 kilograms is hanging over the edge of a tall building and does not touch the ground. How

dangina [55]

Answer:

Work done required is 3567.2 J

Explanation:

Given :

Length of chain, l = 72 m

Mass of chain, M = 29 kg

Linear mass density of chain, μ = $\frac{Mass\ of\ chain }{Length\ of\ chain}$ = $\frac{29}{72}$ = 0.40 kg/m

Let x be the length of the chain which lift to the top of the building.

Work done required to lift the chain is equal to the potential energy of the chain.

W = ∫μg (72 - x ) dx

Here g is acceleration due to gravity.

The limit of integration is from 0 to 14.

W = μg ( 72x - x²/2)

Substitute 0.40 kg/m for μ, 9.8 m/s² for g and 14 m for x in the above equation.

W = $0.40\times9.8\times(72\times14\ - \frac{14^{2} }{2})$

W = 3567.2 J

5 0

3 years ago

Find the scaler product of this two vector

OLga [1]

Answer:

The scalar product of a and b is: a · b = |a||b| cosθ

5 0

3 years ago

Explain why nuclear fission and nuclear fusion release large amounts of energy

levacccp [35]

Answer:

Because of the formula $E=mc^2$

Explanation:

In this problem we are describing two different processes:

Nuclear fission occurs when a heavy, unstable nucleus breaks apart into two or more lighter nuclei

Nuclear fusion occurs when two (or more) light nuclei fuse together producing a heavier nucleus

In both cases, the total mass of the final products is smaller than the total mass of the initial nuclei.

According to Einsten's formula, this mass difference has been converted into energy, as follows:

$E=\Delta mc^2$

where:

E is the energy released in the reaction

$\Delta m$ is the mass defect, the difference between the final total mass and the initial total mass

$c=3.0 \cdot 10^8 m/s$ is the speed of light

From the formula, we see that the factor $c^2$ is a very large number, therefore even if the mass defect $\Delta m$ is very small, nuclear fusion and nuclear fission release huge amounts of energy.

8 0

3 years ago

Help please..................

Shtirlitz [24]

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7 0

3 years ago

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400

liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, $f_{ra} = \mu mg\\$ ............(1)

Since frictional force is a type of force, we are safe to say $f_{ra} = ma$ .......(2)

Equating (1) and (2)

$ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}$

a) Speed of A at the start of the sliding

$d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s$

b) Speed of B at the start of the sliding

$d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s$