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ValentinkaMS [17]

3 years ago

14

sabendo que uma onda sonora tem um periodo de 0,002s e um comprimento de onda de 0,5m determine sua velocidade nesse meio

1 answer:

Artemon [7]3 years ago

3 0

Answer: Sorry i need more Details. I want to help but I dont speak spanish.

Explanation: Maybe come back later with a english question. So sorry. I hope ypu have a good Day.

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Identify and explain one way in which we try to deal with infiltration? What are the benefits and cost of that particular soluti

tigry1 [53]

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i vigilate

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5 0

3 years ago

A 10.00 kg block is placed at the top of a long frictionless inclined plane angled at 37.9 degrees relative to the horizontal. T

NeX [460]

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4 0

2 years ago

What heat transfer do sun heats earth go through

Eddi Din [679]

Yes because the heat is to hot and would melt the earth layers so it would go through

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7 0

3 years ago

Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2

Bess [88]

Answer:

The acceleration of $M_2$ is $a = 0.7156 m/s^2$

Explanation:

From the question we are told that

The mass of first block is $M_1 = 2.25 \ kg$

The angle of inclination of first block is $\theta _1 = 43.5^o$

The coefficient of kinetic friction of the first block is $\mu_1 = 0.205$

The mass of the second block is $M_2 = 5.45 \ kg$

The angle of inclination of the second block is $\theta _2 = 32.5^o$

The coefficient of kinetic friction of the second block is $\mu _2 = 0.105$

The acceleration of $M_1 \ and\ M_2$ are same

The force acting on the mass $M_1$ is mathematically represented as

$F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

=> $M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

Where T is the tension on the rope

The force acting on the mass $M_2$ is mathematically represented as

$F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

$M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

At equilibrium

$F_1 = F_2$

So

$T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

making a the subject of the formula

$a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}$

substituting values $a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}$

=> $a = 0.7156 m/s^2$

3 0

4 years ago

A small spinning asteroid is in a circular orbit around a star, much like the earth's motion around our sun. The asteroid has a

Fudgin [204]

Answer:

Temperature will be 305 K

Explanation:

We have given The asteroid has a surface area $A=7.70m^2$

Power absorbed P = 3800 watt

Boltzmann constant $\sigma =5.67\times 10^{-8}Wm/K^4$

According to Boltzmann rule power radiated is given by

$P=\sigma AT^4$

$3800=5.67\times 10^{-8}\times 7.70\times T^4$

$T^4=87.0381\times 10^8$

$T=305K$

So temperature will be 305 K

8 0

3 years ago