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ipn [44]
3 years ago
11

If the wavelength of a wave increases does its frequency also increase?

Physics
1 answer:
irinina [24]3 years ago
4 0
Frequency decreases whilst wavelength increases and the opposite also occurs
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A 3m plank AB of mass 20kg has a center of gravity 2m from end A and rests on supports R1 and R2 both 0,5m from either end. It a
postnew [5]

Answer:

130.7 N

Explanation:

5 0
2 years ago
A car is moving with a constant acceleration. At t = 5.0 s its velocity is 8.0 m/s and at t = 8.0 s its
spayn [35]
Distance=average velocity x time =((12+8)/2) x 3 =30m
4 0
3 years ago
Use the information below for the next five questions:
sergey [27]

Answer:  Please see answer in explanation column.

Explanation:

Given that

v≈(331 + 0.60T)m/s

where Temperature, T =  14°C

v≈(331 + 0.60 x 14)m/s

v =331+ 8.4 = 339.4m/s

In our solvings, note that

f= frequency

 λ=wavelength

L = length

v= speed of sound

a) Length of the pipe is calculated using the fundamental frequency formulae that

f=v/2L

Length = v/ 2f

= 339.4m/s/ 2 x 494Hz ( s^-1)= 0.3435m

b) wavelength of the fundamental standing wave in the pipe

L = nλ/2,

λ = 2L/ n

λ( wavelength )= 2 x 0.3435/ 1

= 0.687m

c) frequency of the fundamental standing wave in the pipe

F = v/  λ

= 339.4m/s/0.687m=

494.03s^-1 = 494 Hz

d) the frequency in the traveling sound wave produced in the outside air.

This is the same as the frequency in the open organ pipe = 494Hz

e)The wavelength of the travelling sound wave produced in the outside air is the same as the wavelength calculated in b above = 0.687m

f) To play D above middle c . the distance is given by

L =v/ 2 f

= 343/ 2 x 294

=0.583m

7 0
3 years ago
How do I go about this?
Anna71 [15]

Hi there!

(a)

Recall that:
W = F \cdot d = Fdcos\theta

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

W =248(56)cos(30) = 12027.36 J

To the nearest multiple of ten:
W_A = \boxed{12030 J}

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
\boxed{W_g = 0 J}

(c)
Similarly, the normal force is perpendicular to the displacement, so:
\boxed{W_N = 0 J}

(d)

Recall that the force of kinetic friction is given by:
F_{f} =\mu_k mg

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
W_f = -\mu_k mgd\\W_f = - (0.1)(56)(9.8)(56) = -3073.28 J

In multiples of ten:
\boxed{W_f = -3070 J}

(e)
Simply add up the above values of work to find the net work.

W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}}

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
F_{net} = F_{Ax} - F_f

W = F_{net} \cdot d = (F_{Ax} - F_f)

W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}

5 0
2 years ago
Can I improve the design of my simple machine? How?
blsea [12.9K]

Answer:

12345

Explanation:

yan na po answer ko hehehe

5 0
3 years ago
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