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Varvara68 [4.7K]
2 years ago
13

Question 1 and 2 and 3 physics lesson homework

Physics
1 answer:
storchak [24]2 years ago
5 0

Answer:

Jackson 2: Smart 3: Ahmed

Explanation:

square + circle = egg

oops, wrong guy

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1. While driving to Palm Desert you notice that some of the windmills near the freeway are spinning 20 times per minute. The len
igomit [66]

Answer:

a) \alpha = -0.233 rad/s^{2}

b) the rotational acceleration will remain the same,  \alpha = -0.233 rad/s^{2}

c) When r = 36 m, a_{c} = 157.25 m/s^{2}

   When r = 18 m, a_{c} = 78.63 m/s^{2}

d) When r = 36 m, a_{c} = 39.31 m/s^{2}

   When r = 36 m, a_{c} = 19.66 m/s^{2}

Explanation:

The windmills are spinning 20 times per minute, the number of spins in 1 second = 20/60  = 1/3 spins/sec

frequency, f = 1/3 spins/sec

there are 1/3 spins in 1 second,

there will be 50 spins in 50/ (1/3) seconds = 150 seconds

i.e time taken to make 50 spins = 150 seconds

Time interval between 20 and 50 spins, Δt = 150 - 60 = 90

Δt = 90

Angular frequency at 20 spins:

w₁ = 2 π f

w₁ = 2 π * 1/3 = 2π/3 = 2.09 rad/s

Since the blade stops at 50 spins, angular frequency at 50 spins, w₂ = 0

The rotational acceleration,

\alpha = \frac{\triangle w}{\triangle t} \\\alpha = \frac{w_{2} - w_{1} }{\triangle t} \\\alpha = \frac{0 - 2.09 }{90} \\\alpha = -0.233 rad/s^{2}

b) The magnitude of the rotational acceleration does not depend on the radius, r. It depends on the angular frequency, therefore the magnitude of the rotational acceleration doe not change.

c) The centripetal acceleration is given by the formula, a_{c} = w^{2} r

While the windmill spins at 20 times per minute, the angular speed gotten, w = 2.09 rad/s

When r = 36 m

a_{c} = 2.09^{2} * 36\\a_{c} = 157.25 m/s^{2}

At the halfway point, r = 18 m

a_{c} = 2.09^{2} * 18\\a_{c} = 78.63 m/s^{2}

d) If the angular velocity of the blades were cut in halves:

w = 2.09/2

w = 1.045 rad/s

When r = 36 m

a_{c} = 1.045^{2} * 36\\a_{c} = 39.31 m/s^{2}

When r = 18 m

a_{c} = 1.045^{2} * 18\\a_{c} = 19.66 m/s^{2}

8 0
3 years ago
If the weight of the bowling ball acts down with a force of 200 N, what force would the table need to push up with to keep the b
zavuch27 [327]
5858585 8 8 855858 858  585858
3 0
3 years ago
A scientist measures the number of flies in a room over time. The results are
snow_lady [41]
A. The number of files was constant.
8 0
2 years ago
A 10-kg mass slides down a flat hill that makes an angle of 10° with the horizontal. If friction is negligible, what is the resu
slega [8]

Answer:

Resultant force = 17.02 N

Explanation:

As we assume the coefficient of friction is negligible, the normal force won't affect the resultant force.

As a result of this, the sine of the angle times the force of gravity on the 10 kg mass is equal to the resultant force, which is, force due to gravity = m × a = 10 kg × 9.8 m/s² = 98 N

98 sin(10)=?

Sin(10)= 0.1736

98 x 0.1736= 17.02 N.

Therefore, the resultant force = 17.02 N

8 0
3 years ago
Help help help help help
mina [271]
Please mark me brainliest

7 0
3 years ago
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