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mr Goodwill [35]
3 years ago
6

Your 64-cm-diameter car tire is rotating at 3.5 rev/s when suddenly you press down hard on the accelerator. After traveling 200

m, the tire’s rotation has increased to 6.0 rev/s. What was the tire’s angular acceleration? Give your answer in rad/s2.
Physics
1 answer:
andreyandreev [35.5K]3 years ago
4 0

Answer: angular acceleration = 0.748rad/s²

Explanation: according to the question, our answer needs to be in rad/s², thus all units in rev/s will be converted to rad/s

Assuming the motion of the object is of a constant angular acceleration, then newton's laws of motion is applicable.

The formulae below is used

v² = u² + 2αθ

v = final angular speed =6rev/s = 6*2π = 12π rad/s

u =initial angular speed =3.5rev/s = 3.5 *2π = 7π rad/s

Note 1 rev = 2π rad.

α = angular acceleration.

θ = angular displacement.

Diameter = 64cm = 0.64m, radius = 64/2 = 32cm = 0.32m

The angular displacement can be gotten using the formulae below

S = rθ, where s= linear distance covered = 200m, r = radius = 0.32m

θ = S/r = 200/0.32=625 rad.

By substituting the parameter we have that

(12π)² = (7π)² + 2α(625)

1421.22 = 486.31 + 1250α

1421.22 - 486.31 = 1250α

934.91 = 1250α

α = 934.91/1250

α= 0.748 rad/s²

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According to Bernoulli's equation, the pressure in a fluid will tend to decrease if its velocity increases. Assuming that a wind
Pie

Answer:

The pressure drop predicted by Bernoulli's equation for a wind speed of 5 m/s

= 16.125 Pa

Explanation:

The Bernoulli's equation is essentially a law of conservation of energy.

It describes the change in pressure in relation to the changes in kinetic (velocity changes) and potential (elevation changes) energies.

For this question, we assume that the elevation changes are negligible; so, the Bernoulli's equation is reduced to a pressure change term and a change in kinetic energy term.

We also assume that the initial velocity of wind is 0 m/s.

This calculation is presented in the attached images to this solution.

Using the initial conditions of 0.645 Pa pressure drop and a wind speed of 1 m/s, we first calculate the density of our fluid; air.

The density is obtained to be 1.29 kg/m³.

Then, the second part of the question requires us to calculate the pressure drop for a wind speed of 5 m/s.

We then use the same formula, plugging in all the parameters, to calculate the pressure drop to be 16.125 Pa.

Hope this Helps!!!

7 0
3 years ago
How might an intense solar storm affect people on Earth?
kow [346]

An intense solar storm COULD disrupt communications and damage the power grid. <em> (A)</em>

8 0
3 years ago
What is the electric field strength 7.50 x 10E-1 meters from a 2.8 x 10 E-3 C charged object?
Nadusha1986 [10]

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5 0
3 years ago
Which statement best describes beaches? Beaches change over time as waves move further up the shore. Beaches are always changing
NemiM [27]

Answer:

I think the 1st statement is right.

Explanation:

Wind patterns doesn't stay the same.

Waves don't follow the same patterns.

Waves move further up the shore.

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3 0
3 years ago
The speed of sound in air is 10 times faster than the speed of a wave on a certain string. The density of the string is 0.002kg/
nata0808 [166]

Answer:

The tension on the string is 2.353 N.

Explanation:

Given;

the speed of sound in air, v₀ = 343 m/s

then, the speed of sound on the string, v = 343 / 10 = 34.3 m/s

mass per unit length, m/l = μ = 0.002 kg/m

The speed of sound on the string is given as;

v = \sqrt{\frac{T}{\mu} } \\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu

where;

T is the tension on the string

T = (34.3)²(0.002)

T = 2.353 N

Therefore, the tension on the string is 2.353 N.

3 0
3 years ago
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