How far out from the sun is the rock line?

Pls help

belka [17]

Answer:

2156 J

Explanation:

From the question,

Work done = Combined mass of the bucket and water×height×gravity.

W = (M+m)hg............................. Equation 1

Where M = mass of water, m = mass of the bucket, h = height, g = acceleration due to gravity.

Given: M = 20 kg, m = 2 kg, h = 10 m

Constant: g = 9.8 m/s²

Substitute these value into equation 1

W = (20+2)×10×9.8

W = 22×98

W = 2156 J

4 0

2 years ago

if vector u has lenght 70 and direction 40 degrees, and vector v has length 85 and direction 335 degrees what is the length and

Anastaziya [24]

Answer:

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

Explanation:

||u|| = 70

θ = 40°

$\vec{u}_x=||u||cos\theta \\\Rightarrow \vec{u}_x=70cos40=53.62$

$\vec{u}_y=||u||sin\theta \\\Rightarrow \vec{u}_y=70sin40=44.99$

||v|| = 85

θ = 335°

$\vec{v}_x=||v||cos\theta \\\Rightarrow \vec{v}_x=85cos335=77.03$

$\vec{v}_y=||v||sin\theta \\\Rightarrow \vec{v}_y=85sin335=-35.92$

Resultant

$R=\sqrt{R_x^2+R_y^2}\\\Rightarrow R=\sqrt{(\vec{u}_x+\vec{v}_x)^2+(\vec{u}_y+\vec{v}_y)^2}\\\Rightarrow R =\sqrt{(70cos40+85cos335)^2+(70sin40+85sin335)^2}\\\Rightarrow R =131.15$

$\theta=tan^{-1}\frac{R_y}{R_x}\\\Rightarrow \theta=tan^{-1}\frac{70sin40+85sin335}{70cos40+85cos335}\\\Rightarrow \theta=tan^{-1}0.069=3.97^{\circ}$

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

4 0

2 years ago

The total momentum of two marbles before a collision is 0.06 kg-m / s. no outside forces act on the marbles. what is the total m

SpyIntel [72]

Hi my friend, since momentum is always conserved without external forces, the momentum after the collosion will still be 0.06 kg*m/s. Hope it helps☺

4 0

3 years ago

Volcanic eruption affects water drainage but what may that result in

dmitriy555 [2]

Explosive eruptions can severely disturb landscapes downwind or downstream of volcanoes by damaging vegetation and depositing large volumes of erodible fragmental material. As a result, fluxes of water and sediment in affected drainage basins can increase dramatically.

6 0

3 years ago

Consider the interference pattern produced by two parallel slits of width a and separation d, in which d = 3a. The slits are ill

laila [671]

Answer:

a) m =1 θ = sin⁻¹ λ / d, m = 2 θ = sin⁻¹ ( λ / 2d) , c) m = 3

Explanation:

a) In the interference phenomenon the maxima are given by the expression

d sin θ = m λ

the maximum for m = 1 is at the angle

θ = sin⁻¹ λ / d

the second maximum m = 2

θ = sin⁻¹ ( λ / 2d)

the third maximum m = 3

θ = sin⁻¹ ( λ / 3d)

the fourth maximum m = 4

θ = sin⁻¹ ( λ / 4d)

b) If we take into account the effect of diffraction, the intensity of the maximums is modulated by the envelope of the diffraction of each slit.

I = I₀ cos² (Ф) (sin x / x)²

Ф = π d sin θ /λ

x = pi a sin θ /λ

where a is the width of the slits

with the values of part a are introduced in the expression and we can calculate intensity of each maximum

c) The interference phenomenon gives us maximums of equal intensity and is modulated by the diffraction phenomenon that presents a minimum, when the interference reaches this minimum and is no longer present

maximum interference d sin θ = m λ

first diffraction minimum a sin θ = λ

we divide the two expressions

d / a = m

In our case

3a / a = m

m = 3

order three is no longer visible

7 0

2 years ago