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3 years ago

When the wavelength is tripled at a constant frequency, what happens to the velocity of the wave?

Physics

1 answer:

Afina-wow [57]3 years ago

5 0

B is the correct answer

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Given that the velocity of blood pumping through the aorta is about 30 cm/s, what is the total current of the blood passing thro

TiliK225 [7]

Answer:

94.248 g/sec

Explanation:

For solving the total current of the blood passing first we have to solve the cross sectional area which is given below:

$A_1 = \pi R^2\\\\A_1 = \pi (1)^2\\\\A_1 = 3.1416 cm^2$

And, the velocity of blood pumping is 30 cm^2

Now apply the following formula to solve the total current

$Q = \rho A_1V_1\\\\Q = (1)(3.1416)(30)\\\\$

Q = 94.248 g/sec

Basically we applied the above formula So, that the total current could come

3 0

4 years ago

How to find the magnitude of a centripetal force?<br>(formula)

frozen [14]

Answer:

F=mv^2÷r

Explanation:

i know every thing

the magnitude f of the centripetal force is equal to the mass m of the body times it veloctiy squared v^2 divided by the radius r of its path

4 0

3 years ago

Which of the following requires freshwater supplies?

Nitella [24]

Agriculture would use fresh water

7 0

3 years ago

What is the speed of light?

Sergeu [11.5K]

The speed of light : 299 792 458 m / s

5 0

3 years ago

Read 2 more answers

A 0.537-kg basketball is dropped out of a window that is 5.88 m above the ground. The ball is caught by a person whose hands are

Monica [59]

Answer:

a) 20.54 J

b) 30.97 J

c) 10.43 J

d) -20.54 J

Explanation:

m = Mass

g = Acceleration due to gravity = 9.81 m/s²

h = Height

$W=mgh\\\Rightarrow W=0.537\times 9.81\times (5.88-1.98)\\\Rightarrow W=0.537\times 9.81\times 3.9\\\Rightarrow W=20.54\ J$

Work done by the ball's weight is 20.54 J

$W=mgh\\\Rightarrow W=0.537\times 9.81\times (5.88)\\\Rightarrow W=30.97\ J$

Gravitational potential energy of the basketball, relative to the ground, when it is released is 30.97 J

$W=mgh\\\Rightarrow W=0.537\times 9.81\times (1.98)\\\Rightarrow W=10.43\ J$

Gravitational potential energy of the basketball, relative to the ground, when it is released is 10.43 J

Change in gravitational potential energy

$\Delta U=10.43-30.97=-20.54\ J$

Change in gravitational potential energy is given by -20.54 J

4 0

4 years ago