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3 years ago

6. What is the wavelength of a wave that is traveling at 30 m/s and has a frequency of 3.2 Hz?

Physics

1 answer:

lidiya [134]3 years ago

6 0

Answer:

6. 9.4 m

7. 1050 m/s.

Explanation:

6. Determination of the wavelength

Velocity (v) = 30 m/s

Frequency (f) = 3.2 Hz

Wavelength (λ) =?

v = λf

30 = λ × 3.2

Divide both side by 3.2

λ = 30 / 3.2

λ = 9.4 m

Thus, the wavelength of the wave is 9.4 m.

7. Determination of the speed of the wave.

Wavelength (λ) = 350 m

Frequency (f) = 3 Hz

Velocity (v) =.?

v = λf

v = 350 × 3

v = 1050 m/s

Thus, the speed of the wave is 1050 m/s

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A lunar module weighs 12 metric tons on the surface of the Earth. How much work is done in propelling the module from the surfac

zhuklara [117]

Answer:

W=76.55 miles.metric tons

Explanation:

Given that

Weight on the earth = 12 tons

So weight on the moon =12/6 = 2 tons

( because at moon g will become g/6)

As we know that

$F=\dfrac{K}{x^2}$

Here x= 1100 miles

F 2 tons

$2=\dfrac{K}{1100^2}$

$K=2.4\times 10^6$

We know that

Work = F. dx

$W=\int_{x_1}^{x_2}F.dx$

$W=\int_{1100}^{1140}\dfrac{2.4\times 10^6}{x^2}.dx$

$W=-2.4\times 10^6\left[\dfrac{1}{x}\right]_{1100}^{1140}$

$W=-2.4\times 10^6\left[\dfrac{1}{1140}-\dfrac{1}{1100}\right]$

W=76.55 miles.metric tons

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3 years ago

A cardio kickboxing class challenges 3 things. List them.

bagirrra123 [75]

Answer:

In a cardio kickboxing class, you will learn proper form for the famous jab, hooks,crosses, uppercuts, back kicks, front kicks.

Explanation:

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3 years ago

A farmer lifts his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 1 ft/s. Dete

Lesechka [4]

Answer:

The velocity of the hay bale is - 0.5 ft/s and the acceleration is $6.25\times 10^{- 3} ft/s^{2}$

Solution:

As per the question:

Constant velocity of the horse in the horizontal, $v_{x} = 1 ft/s$

Distance of the horse on the horizontal axis, x = 10 ft

Vertical distance, y = 20 ft

Now,

Apply Pythagoras theorem to find the length:

$20^{2} + 10^{2} = l^{2}$

$l^{2}= 500$

Now,

$x^{2} + y^{2} = 500$ (1)

Differentiating equation (1) w.r.t 't':

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

$x\frac{dx}{dt} = - y\frac{dy}{dt}$

where

$\frac{dx}{dt}$ = Rate of change of displacement along the horizontal

$\frac{dy}{dt}$ = Rate of change of displacement along the vertical

$v_{x}$ = velocity along the x-axis.

$v_{y}$ = velocity along the y-axis

$xv_{x} = -yv_{y}$

$v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s$

$|v_{y}| = 0.5\ ft/s$

Acceleration of the hay bale is given by the kinematic equation:

$v_{y}^{2} = u_{y} + 2ay$

$(-0.5)^{2} =0 + 2ay$

$0.25 = 2ay$

$\frac{0.25}{2y} = a$

$a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}$

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monitta

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