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Murljashka [212]
3 years ago
7

Kinematics Equations

Physics
1 answer:
blagie [28]3 years ago
7 0
Haiaiiiiisbdjdnjssjsa
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netineya [11]

Answer: b is sedimentary. c is metamorphic. and a is igneous.

Explanation:

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A batter hits two baseballs with the same force. One hits the ground near third base. The other is a home run out of the park. W
Basile [38]
I think the answer is "<span>The ball that went out of the park shows more work because the distance was greater."</span>
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3 years ago
A block of mass m = 150 kg rests against a spring with a spring constant of k = 880 N/m on an inclined plane which makes an angl
weeeeeb [17]

Answer:

b)  k Δx - W cos θ - μ mg cos θ = m a ,  c)  θ = 86.6º, d)  Δx = 1.18 m

Explanation:

a) In the attachment we can see a diagram of the forces in this problem and the coordinate axes for its decomposition.

F is the force applied by the spring, while it is compressed, this force disappears when the block leaves the spring

b) Let's apply Newton's second law for when the spring is compressed

let's use trigonometry to break down the weight

            sin θ = Wₓ / W

            cos θ = W_y / W

             Wₓ = W sin θ

             W_y = W cos θ

Y axis  

               N - W_y = 0

               N = W_y

               N = W cos θ

X axis

           F -Wₓ -fr = ma

the force applied by the spring is given by hooke's law

           F = k Δx

friction force has the expression

           fr = μ N

           fr = μ W cos θ

we substitute

            k Δx - W cos θ - μ mg cos θ = m a           ( 1)

c) If the plane has no friction, what is the angle so that Δx = 0.1m

             

We write the equation 1, with fr = 0 and since the system is still a = 0

            k Δx - W cos θ -0 = 0

            cos θ = \frac{k \Delta x}{ m g}

            cos θ = \frac{880 \ 0.1}{ 150 \ 9.8}

            cos θ = 0.0598

            θ = cos⁻¹ 0.0598

            θ = 86.6º

d) In this part they give the angle θ = 45º and there is no friction, they ask the compression

the acceleration is zero, we substitute in 1

            k Δx - W cos θ - 0 = 0

            Δx = \frac{mg \ cos \  \theta}{k}

            Δx = \frac{ 150 \ 9.8 \ cos45}{880}

            Δx = 1.18 m

7 0
3 years ago
A thin film of oil (n = 1.27) is located on smooth, wet pavement. When viewed from a direction perpendicular to the pavement, th
alisha [4.7K]

Answer:

Explanation:

The problem is based on interference in thin films

refractive index of water is more than given oil so there will be phase change of π at upper and lower layer of the film .

a )

for constructive interference , the condition is

2μt = nλ where t is thickness of layer , μ is refractive index , λ is wavelength and n is order of the fringe

Putting the values

2 x 1.27 t = n x 640

2 x 1.27 t =  640 ( for minimum thickness n = 1 )

t = 252 nm .

b )

2 x 1.27 t = m₁  λ₁

for destructive interference

2μt = (2m₂+1)λ₂/2

2 x 1.27 t =(2m₂+1)λ₂/2

m₁ λ₁  = (2m₂+1)λ₂/2

2m₁λ₁  = (2m₂+1)λ₂

2m₁ / (2m₂+1) = λ₂ / λ₁

2m₁ / (2m₂+1) = 548/ 640

2m₁ / (2m₂+1) = .85625

2m₁ = .85625 (2m₂+1)

This is the required relation between m₁ and m₂

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3 years ago
Name the four phases of the lunar cycle​
Serggg [28]

Answer:

Hello there!

This is a picture that helped me learn the phases in an easy way. I hope it helps you too.

Explanation:

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