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4 years ago

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A pilot can withstand an acceleration of up to 9g, which is about 88 m/s2, before blacking out. What is the acceleration experie

nced by a pilot flying in a circle of constant radius at a constant speed of 495 m/s if the radius of the circle is 3340 m?

1 answer:

Harman [31]4 years ago

4 0

Answer:

Explanation:

Given that, the pilot can withstand 9g acceleration which is approximately 88m/s².

Now, the pilot is traveling in a circle of radius

r = 3340 m

And the speed is

v = 495 m/s

Then, acceleration?

The acceleration of a circular motion can be determine using centripetal acceleration

a = v² / r

a = 495² / 3340

a = 73.36 m/s².

Since the acceleration is less that the acceleration the pilot can withstand, then, I think the pilot makes the turn without blacking out and successfully

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A thin, uniform stick of length 1.9 m and mass 3.1 kg is pinned through one end and is free to rotate. The stick is initially ha

nirvana33 [79]

Answer:

The acceleration is $\alpha = 7.10 \ rad/s^2$

Explanation:

From the question we are told that

The length of the stick is $d = 1.9 \ m$

The mass of the stick is $m = 3.1 \ kg$

The angular displacement is $\theta = 23.4^o$

Generally the torque of this uniform stick after this displacement is mathematically represented as

$\tau = \frac{1}{2} * d * [m*g]* sin (90 - \theta)$

$\tau = \frac{1}{2} * 1.9 * [3.1*9.8]* sin (90 - 23.4)$

$\tau = 26.49 \ kg\cdot m^2 \cdot s^{-2}$

Generally the moment of inertia of the uniform stick is mathematically represented as

$I = \frac{1}{3} * m * d^2$

=> $I = \frac{1}{3} * 3.1 * 1.9 ^2$

=> $I = 3.73 \ kg \cdot m^2$

Generally the angular acceleration is mathematically represented as

$\alpha = \frac{\tau}{I}$

=> $\alpha = \frac{26.49}{3.73}$

=> $\alpha = 7.10 \ rad/s^2$

7 0

3 years ago

On a hot summer day a young girl swings on a rope above the local swimming hole. when she lets go of the rope her initial veloci

weqwewe [10]

Refer to the diagram shown below.
h = height of the girl above water when she lets go of the rope.

The launch velocity is 22.5 m/s at 35° to the horizontal. Therefore the vertical component of the velocity is
v = 22.5 sin(35°) = 12.9055 m/s.

The time of flight is t = 1.10 s before the girl hits the surface of the water at a height of -h.
Therefore
-h = (12.9055 m/s)*(1.10 s) - (1/2)*(9.8 m/s²)*(1.10 s)²
-h = 8.267 m
= 8.3 m (nearest tenth)

Answer:
When the girl let go of the rope, she was about 8.3 m above the surface of the water.

5 0

4 years ago

A moving particle fragments or decays into a particle moving at .53c, mass 135 MeV/c2, and a particle moving at .98c, mass 938 M

Svetllana [295]

Answer:

M = 1073 Mev/c2

u = 0.95 C

Explanation:

given data:

m1 =135 Mev/c2

v1 = 0.53 c

m2 = 938 Mev/c2

v2 = 0.98 c

from conservation of momentum principle we have

$\frac{mu}{\sqrt{1-\frac{u^2}{C^2}}} = \frac{m1v1}{\sqrt{1-\frac{V1^2}{C^2}}} +\frac{m1v1}{\sqrt{1-\frac{V2^2}{C^2}}}$

$\frac{mu}{\sqrt{1-\frac{u^2}{C^2}}} = \frac{135*0.53c}{0.848} +\frac{938*0.98c}{0.2}$

$\frac{mu}{\sqrt{1-\frac{u^2}{C^2}}} = 4680.6 C$ ...............1

Total mass of INITIAL particle M =m1+m2 = 1073 Mev/c2

using equation 1

$\frac{1073 u}{\sqrt{1-\frac{u^2}{C^2}}}= 4680.6C$

solving for u we get

u = 0.95 C

8 0

4 years ago

Which statements describe properties of stars? Check all that apply.

Shtirlitz [24]

One, three and five are correct.

Although if the second statement is saying that stars use gravitational force to support nuclear fusion which in turn produces energy then that would be correct, but I don’t think so :)

7 0

3 years ago

Read 2 more answers

Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

3 0

3 years ago