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2 years ago

8. A 30-kg box is sliding down a frictionless plane that is sloped at 24º. Assuming the object starts at rest,

Physics

1 answer:

Katena32 [7]2 years ago

6 0

The net force on the box parallel to the plane is

∑ F[para] = mg sin(24°) = ma

where mg is the weight of the box, so mg sin(24°) is the magnitude of the component of its weight acting parallel to the surface, and a is the box's acceleration.

Solve for a :

g sin(24°) = a ≈ 3.99 m/s²

The box starts at rest, so after 7.0 s it attains a speed of

(3.99 m/s²) (7.0 s) ≈ 28 m/s

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Alekssandra [29.7K]

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I think it's a chemical change.

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3 years ago

Jason and Guy are throwing watermelons straight up from the ground. (They’re making fruit salad.) Jason throws his watermelon at

NISA [10]

Answer:2t

Explanation:

Given

Jason throwing watermelon at speed v and it spend t time in air

time to reach max height

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3 years ago

What happend to the egg in the fresh water glass?

Molodets [167]

Answer:

Nothing in the fresh water

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2 years ago

Read 2 more answers

Two narrow, parallel slits separated by 0.85 mm are illuminated by 600 nm light, and the viewing screen is 2.8 m away from the s

AURORKA [14]

Answer:

Phase difference = pi/4 radians

Explanation:

Given:

- The wavelength of incident light λ = 600 nm

- The split separation d = 0.85 mm

- Distance of screen from split plane L = 2.8 m

Find:

What is the phase difference between the two interfering waves on a screen, at a point 2.5 mm from the central bright fringe?

Solution:

- The phase difference can be evaluated by determining the type of interference that occurs at point y = 2.5 mm above central order. We will use the derived results from Young's double slit experiment.

sin ( Q ) = m*λ /d

m = d*sin(Q) / λ

- Where, m is the order number and angle Q is the angle for mth order of fringe from central bright fringe.

r = sqrt ( L^2 + 0.0025^ )

Where, r is the distance from split to the interference bright fringe.

r = sqrt(2.8^ + 0.0025^) = 2.8

sin(Q) = 0.0025 / 2.8

Hence. m = 0.00085*0.0025 / 2.8*(600*10^-9)

m = 1.26

- We know that constructive interference would occurred at m = 1 and destructive interference @ m = 1.5. They have a phase difference of pi/2 radians.

- The order number lies in between constructive and destructive interference i.e m ≈ 1.25 then the corresponding phase difference = 0.5*(pi/2).

Answer: Phase difference = pi/4 radians

6 0

3 years ago

If you walk 1.2 km north and then 1.6 km east, what are the magnitude and direction of your resultant displacement?

vodomira [7]

<u>Answer:</u>

Option A is the correct answer.

<u>Explanation:</u>

Let the east point towards positive X-axis and north point towards positive Y-axis.

First walking 1.2 km north, displacement = 1.2 j km

Secondly 1.6 km east, displacement = 1.6 i km

Total displacement = (1.6 i + 1.2 j) km

Magnitude = $\sqrt{1.2^2+1.6^2} = 2 km$

Angle of resultant with positive X - axis = $tan^{-1}(1.2/1.6)=36.87^0$ = 36.87⁰ east of north.

5 0

3 years ago