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Ksju [112]
3 years ago
6

A teacher stands well back from an outside doorway 0.88 m wide, and blows a whistle of frequency 780 Hz. Ignoring reflections, e

stimate at what angle(s) it is not possible to hear the whistle clearly on the playground outside the doorway.
Physics
1 answer:
AlladinOne [14]3 years ago
5 0

the wavelength is to be 56.67 cm and I know that i should be applying the double slit equation for destructive interference.

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For this problem, imagine that you are on a ship that is oscillating up and down on a rough sea. Assume for simplicity that this
ikadub [295]

Answer:

no idea

Explanation:

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3 years ago
A LETTER FROM THE LORAX
KengaRu [80]

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Explanation:

You could try to say how helpful they are what they are and what they do

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Read 2 more answers
Calculate the mass of an object with a density of 102.5 g/mL and volume of 375 mL.
AveGali [126]

Answer:

38,437.5

Explanation:

Density(d)= 102.5g/ml

Volume (v)=375ml

Mass(m) = ?

D =m/v

102.5= m/375

102.5*375=m

38,437.5=m

therefore Mass = 38,437.5g/ml.

6 0
3 years ago
The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti
Colt1911 [192]

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

5 0
2 years ago
A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder weigh
DiKsa [7]

Answer:(a)360N,(b)171N,(c)2.702m

Explanation:

(a)Maximum Friction Force =\mu \left ( N\right )=0.4\times \left ( 740+160\right )

=360 N

cos\theta =\frac{3}{5}

sin\theta =\frac{4}{5}

(b)Moment about Ground Point

740\times 1\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta

N_1tan\theta =1140

N_1=171 N

N_1=f=171 N

(c)

740\times x\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta

Here maximum friction force can be 360 N

Therefore N_1=360 N

Where x is the maximum distance moved by man along the ladder

360\times 5\times \frac{4}{3}=740x+160\times 2.5

740x=2000

x=2.702m

5 0
3 years ago
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