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3 years ago

PLEASE help me with this science question.

Physics

2 answers:

frez [133]3 years ago

7 0

What’s the question

MissTica3 years ago

4 0

Happy to help what is the question

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A thin layer of liquid methylene iodide (n = 1.756) is sandwiched between two flat, parallel plates of glass (n = 1.50). What mu

Harrizon [31]

Answer:

t = 96.1 nm

Explanation:

For strong reflection through liquid layer we know that the path difference between two reflected light rays must be integral multiple of wavelength

now we know that the path difference of two reflected light from thin liquid layer is given as

$2\mu t - \frac{\lambda}{2} = N\lambda$

here we know that

$\mu = 1.756$

t = thickness of layer

N = 0 (for minimum thickness of layer)

$\lambda = 675 nm$

now we have

$2(1.756) t = \frac{675 nm}{2}$

$t = 96.1 nm$

5 0

3 years ago

A merry-go-round is shaped like a uniform disk and has moment of inertia of 50,000 kg m 2 . It is rotating so that it has an ang

lapo4ka [179]

Answer:

r = 20 m

Explanation:

The formula for the angular momentum of a rotating body is given as:

L = mvr

where,

L = Angular Momentum = 10000 kgm²/s

m = mass

v = speed = 2 m/s

r = radius of merry-go-round

Therefore,

10000 kg.m²/s = mr(2 m/s)

m r = (10000 kg.m²/s)/(2 m/s)

m r = 5000 kg.m ------------- equation 1

Now, the moment of inertia of a solid uniform disc about its axis through its center is given as:

I = (1/2) m r²

where,

I = moment of inertia = 50000 kg.m²

Therefore,

50000 kg.m² = (1/2)(m r)(r)

using equation 1, we get:

50000 kg.m² = (1/2)(5000 kg.m)(r)

(50000 kg.m²)/(2500 kg.m) = r

5 0

3 years ago

Following a collision between a large spacecraft and an asteroid, a copper disk of radius 28.0 m and thickness 1.20 m, at a temp

coldgirl [10]

Answer:

A. 9.31 x10^10J

B. -8.47x10 ^ 12J

C. 8.38x 10^12J

Explanation:

See attached file pls

3 0

3 years ago

Mercury has a radial acceleration of 3.96 × 10−2 m/s2 and its orbital period is T = 88 days. What is the radius of Mercury’s orb

Maslowich

Answer: 58,045,522,878.8 meters

Explanation:

Ok, the data we have is

Period = T = 88 days

Radial acceleration = ar = 3.96x10^-2 m/s^2

And we know that the equation for the radial acceleration is:

ar = v^2/r = r*w^2

Where v is the velocity. r is the radius and w is the angular velocity.

And we know that:

w = 2*pi*f

where f is the frequency, and:

T = 1/f.

Then we can write:

w = 2*pi/T

and our equation becomes:

ar = r*(2*pi/T)^2

Now we solve this for r.

First we need to use the same units in both equations, so we want to write T in seconds.

T = 88 days,

A day has 24 hours, and one hour has 3600 seconds:

T = 88*24*3600 s =7,603,200s

Then:

3.96x10^-2 m/s^2 = r*(2*3.14/7,603,200s)^2

r = (3.96x10^-2 m/s^2) /(2*3.14/7,603,200s)^2 = 58,045,522,878.8 meters

5 0

3 years ago

A hailstone traveling with a velocity of 43 meters/second comes to a virtual stop 0.28 seconds after hitting water. What is the

shutvik [7]

Acceleration=(change in speed)/(time for the change). 43/0.28 = 153.6 m/s^2.

6 0

3 years ago

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