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3 years ago

A water wave has a frequency of 2 Hertz and a wavelength of 5 cm. Calculate it speed

Physics

1 answer:

Llana [10]3 years ago

8 0

Answer: 0.1 m/s

Explanation:

Use formula,

v = f * w where, v is speed, f is frequency and w is wavelength.

Now,

v = 2 * 5 * 10 ^ -2 ( Remember to convert all the units to SI units. Here 5 cm becomes 5 * 10 ^ -2 m. )

v = 0.1 m/s.

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I don’t know how to answer this question? Can anyone help?

VMariaS [17]

Answer:

F=ma

here F is force, m is mass and a is accelaration,

According to the question,

F=3*F= 3F

m= 1/3 of m= m/3

a= ?

so the equation becomes,

3F= m/3*a

3F*3= ma

9F=ma

F= ma/9

Therefore accelaration reduces by 1/9.

I am not very sure.

7 0

3 years ago

A car is traveling at 70 km/h. It then uniformly decelerates to a complete stop in 12 s. Find its acceleration ( in m/s2 ).

Yakvenalex [24]

Answer:

Acceleration will be $-1.620m/sec^2$

Explanation:

We have given initial speed of the car is 70 km/hr

We know that 1 km = 1000 m

And 1 hour = 3600 sec

So $70km/hr=70\times \frac{1000m}{3600sec}=19.444m/sec$

It is given that car stops in 12 sec

So final speed of the car v = 0 m/sec

Time t = 12 sec

From first equation of motion v = u+at

So $0=19.444+a\times 12$

$a=\frac{-19.444}{12}=-1.620m/sec^2$ ( negative sign indicates that speed of the car will constantly decrease )

5 0

3 years ago

Read 2 more answers

The rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between

erma4kov [3.2K]

Answer:

Considering first question

Generally the coefficient of performance of the air condition is mathematically represented as

$COP = \frac{T_i}{T_o - T_i}$

Here $T_i$ is the inside temperature

while $T_o$ is the outside temperature

What this coefficient of performance represent is the amount of heat the air condition can remove with 1 unit of electricity

So it implies that the air condition removes $\frac{T_i}{T_o - T_i}$ heat with 1 unit of electricity

Now from the question we are told that the rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between inside and outside. This can be mathematically represented as

$Q \ \alpha \ (T_o - T_i)$

=> $Q= k (T_o - T_i)$

Here k is the constant of proportionality

since 1 unit of electricity removes $\frac{T_i}{T_o - T_i}$ amount of heat

E unit of electricity will remove $Q= k (T_o - T_i)$

$E = \frac{k(T_o - T_i)}{\frac{T_i}{ T_h - T_i} }$

=> $E = \frac{k}{T_i} (T_o - T_i)^2$

given that $\frac{k}{T_i}$ is constant

=> $E \ \alpha \ (T_o - T_i)^2$

From this above equation we see that the electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square of the temperature difference.

Considering the second question

Assuming that $T_i = 30 ^oC$

and $T_o = 40 ^oC$

Hence

$E = K (T_o - T_i)^2$

Here K stand for a constant

$E = K (40 - 30)^2$

=> $E = 100K$

Now if the $T_i = 20 ^oC$

Then

$E = K (40 - 20)^2$

=> $E = 400 \ K$

So from this see that the electricity require (cost of powering and operating the air conditioner)when the inside temperature is low is much higher than the electricity required when the inside temperature is higher

Considering the third question

Now in the case where the heat that enters the building is at a rate proportional to the square-root of the temperature difference between inside and outside

We have that

$Q = k (T_o - T_i )^{\frac{1}{2} }$

$E = \frac{k (T_o - T_i )^{\frac{1}{2} }}{\frac{T_i}{T_o - T_i} }$

=> $E = \frac{k}{T_i} * (T_o - T_i) ^{\frac{3}{2} }$

Assuming $\frac{k}{T_i}$ is a constant

Then

$E \ \alpha \ (T_o - T_i)^{\frac{3}{2} }$

From this above equation we see that the electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square root of the cube of the temperature difference.