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Wewaii [24]
2 years ago
8

A person is trying to ride a bike all the way round the inside of a pipe for a stunt in a film. The filmmaker wants to know what

speeds are involved. The pipe has a diameter of 8 m. The mass of the bike and rider is 400 kg. The rider goes at a constant speed of 5 m/s.
a) What is its acceleration at the bottom?
b) What is the force on the bike at an angle of 30° up from the bottom?
c) What is the minimum velocity at the top for the bike and rider to stay moving in a circle?
d) Do the bike and rider have sufficient velocity to stay movins a on a circle at the top?
​
Physics
1 answer:
nirvana33 [79]2 years ago
4 0

Answer:

1) The acceleration will be g = 9.8 m/s²

2) F = Wsinα = mgsinα = 400kg·9.8m/s²·1/2 = 2000N

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Therefor, the right choice is:
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2 years ago
If the opening to the harbor acts just like a single-slit which diffracts the ocean waves entering it, what is the largest angle
soldi70 [24.7K]

Answer:

The angle that the wave would be \theta = sin ^{-1}\frac{2 \lambda}{D}

Explanation:

From the question we are told that  the  opening to  the  harbor acts just like a single-slit so a boat in the harbor that at angle equal to the second diffraction minimum would be safe and the  on at angle greater than the diffraction first minimum would be slightly affected

  The minimum is as a result of destructive interference

       And for single-slit this is mathematically represented as

               D sin \ \theta =m \lambda

where D is the slit with

          \theta is the angle relative to the original direction of the wave

         m is the order of the minimum j

        \lambda is the wavelength

Now since in the question we are told to obtain the largest angle at which the boat would be safe

      And the both is safe at the angle equal to the second minimum then

    The the angle is evaluated as

           \theta = sin ^{-1}[\frac{m\lambda}{D} ]

Since for second minimum m= 2

The  equation becomes

               \theta = \frac{2 \lambda}{D}

3 0
3 years ago
Electromagnetic waves differ from other types of waves because they are (2 points)
saveliy_v [14]

Answer:

b. able to travel through a vacuum.

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7 0
3 years ago
A hockey puck slides off the edge of a platform with an initial velocity of 20 m/s horizontally. The height of the platform abov
Rina8888 [55]

Answer:

20.96 m/s

Explanation:

Using the equations of motion

y = uᵧt + gt²/2

Since the puck slides off horizontally,

uᵧ = vertical component of the initial velocity of the puck = 0 m/s

y = vertical height of the platform = 2 m

g = 9.8 m/s²

t = time of flight of the puck = ?

2 = (0)(t) + 9.8 t²/2

4.9t² = 2

t = 0.639 s

For the horizontal component of the motion

x = uₓt + gt²/2

x = horizontal distance covered by the puck

uₓ = horizontal component of the initial velocity = 20 m/s

g = 0 m/s² as there's no acceleration component in the x-direction

t = 0.639 s

x = (20 × 0.639) + (0 × 0.639²/2) = 12.78 m

For the final velocity, we'll calculate the horizontal and vertical components

vₓ² = uₓ² + 2gx

g = 0 m/s²

vₓ = uₓ = 20 m/s

Vertical component

vᵧ² = uᵧ² + 2gy

vᵧ² = 0 + 2×9.8×2

vᵧ = 6.26 m/s

vₓ = 20 m/s, vᵧ = 6.26 m/s

Magnitude of the velocity = √(20² + 6.26²) = 20.96 m/s

4 0
3 years ago
Read 2 more answers
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