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3 years ago

Anybody wanna play among us?

Physics

2 answers:

kherson [118]3 years ago

8 0

Answer:

yess

Explanation:

Grace [21]3 years ago

3 0

Answer:

yes me

<h3>mark me brainliest</h3>

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Ocean waves of wavelength 26 m are moving directly toward a concrete barrier wall at 4.0 m/s . The waves reflect from the wall,

Genrish500 [490]

Answer:

a) the distance between her and the wall is 13 m

b) the period of her up-and-down motion is 6.5 s

Explanation:

Given the data in the question;

wavelength λ = 26 m

velocity v = 4.0 m/s

a) How far from the wall is she?

Now, The first antinode is formed at a distance λ/2 from the wall, since the separation distance between the person and wall is;

x = λ/2

we substitute

x = 26 m / 2

x = 13 m

Therefore, the distance between her and the wall is 13 m

b) What is the period of her up-and-down motion?

we know that the relationship between frequency, wavelength and wave speed is;

v = fλ

hence, f = v/λ

we also know that frequency is expressed as the reciprocal of the time period;

f = 1/T

Hence

1/T = v/λ

solve for T

Tv = λ

T = λ/v

we substitute

T = 26 m / 4 m/s

T = 6.5 s

Therefore, the period of her up-and-down motion is 6.5 s

6 0

3 years ago

Pooping in my room and my room is upstairs and upstairs bathroom upstairs

diamong [38]

Answer:

huh? do you need help on math?

Explanation:

what do you mean?

7 0

2 years ago

On a Saturday afternoon, you decide to pay a neighborhood kid to mow your lawn. The kid usesa manual push lawn mower with a mass

Mars2501 [29]

Answer with Explanation:

We are given that

A.Mass,m=12 kg

$\theta=53^{\circ}$

$\mu_k=0.16$

Speed,v=1.5m/s

Net force in x direction must be zero

$F_{net}=0$

$Fsin\theta-f=0$

$Fsin\theta=f$

Net force in y direction

$N-mg-Fcos\theta=0$

$N=mg+Fcos\theta$

$f=\mu_kN=\mu_k(mg+Fcos\theta)$

$Fsin\theta=\mu_k(mg+Fcos\theta)$

$Fsin\theta=\mu_kmg+\mu_kFcos\theta$

$Fsin\theta-\mu_kFcos\theta=\mu_kmg$

$F(sin\theta-\mu_kcos\theta)=\mu_kmg$

$F=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}$

Power,P=Fv

$P=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}v$

Where $g=9.8m/s^2$

B.Substitute the values

$P=\frac{0.16\times 12\times 9.8}{sin53-0.16cos53}\times 1.5$

$P=40.17W$

6 0

3 years ago

A horizontal clothesline is tied between 2 poles, 12 meters apart. When a mass of 4 kilograms is tied to the middle of the cloth

Olegator [25]

Answer:

T=26.03 N

Explanation:

Given that

Distance between poles = 12 m

Mass of block m= 4 kg

Sag distance = 5 m

Lets take tension in the clothesline is T.

The component of tension in vertical direction will be T cosθ.

By force balancing

2 T cosθ = 40

here $tan\theta =\dfrac{5}{6}$

θ=39.80°

2 T cos39.8 = 40

T=26.03 N

6 0

3 years ago

Why is it wrong to leave our light on

Dennis_Churaev [7]

Answer:

you will get huge electricity bills ............

8 0

2 years ago

Anybody wanna play among us?​

Anybody wanna play among us?