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3 years ago

You are taking a turn at 30.0 m/s on a ramp of radius 25.0 m. What is your acceleration?

Physics

1 answer:

Nat2105 [25]3 years ago

6 0

Acceleration = v^2 / r

Acceleration = 30^2/25

Acceleration = 900/25

Acceleration = 36 m/s^2

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f the magnitude of the acceleration of a propeller blade's tip exceeds a certain value amaxamax, the blade tip will fracture. If

Luden [163]

Answer:

The angular velocity is $w= \sqrt[4]{\frac{a_{max}^2}{r^2} - \alpha ^2}$

Explanation:

Generally the acceleration experienced by the propeller blade's is broken down into

The Radial acceleration which is mathematically represented as

$a_r = \frac{v^2}{r} = w^2r$

And the Tangential acceleration which is mathematically represented as

$a_r = \alpha r$

The net acceleration is evaluated as

$a = \sqrt{a_r^2 + a_t^2}$

Now since angular speed varies directly with angular acceleration so when acceleration is maximum the angular velocity is maximum also and this point if the propeller blade's tip exceeds it the blade would fracture

So at maximum angular acceleration we a have

$a_{max} = \sqrt{a_r^2 + a_t^2}$

$a_{max}^2 = a_r^2 + a_t^2$

$a_{max}^2 = (w^2r)^2 + (\alpha r)^2$

$a_{max}^2 = r^2 w^4 + r^2 \alpha ^2$

$a_{max}^2 = r^2 (w^4 + \alpha^2 )$

$w^4 +\alpha ^2 = \frac{a_{max}^2}{r^2}$

$w^4 = \frac{a_{max}^2}{r^2} - \alpha ^2$

$w= \sqrt[4]{\frac{a_{max}^2}{r^2} - \alpha ^2}$

3 0

3 years ago

Prove that..<br>please help<br>

GaryK [48]

$\large{ \boxed{ \bf{ \color{red}{Universal \: law \: of \: gravitation}}}}$

Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The forces along the line joining the centre of the two objects.

❍ Let us consider two masses m1 and m2 line at a separation distance d. Let the force of attraction between the two objects be F.

According to universal law of gravitation,

$\large{ \longrightarrow{ \rm{F \propto m_1 m_2}}}$

Also,

$\large{ \longrightarrow{ \rm{ F \propto \dfrac{1}{ {d}^{2} } }}}$

Combining both, We will get

$\large{ \longrightarrow{ \rm{F \propto \dfrac{ m_1 m_2}{ {d}^{2}}}}}$

Or, We can write it as,

$\large{ \longrightarrow{ \rm{F \propto \: G \dfrac{ m_1 m_2}{ {d}^{2} }}}}$

Where, G is the constant of proportionality and it is called 'Universal Gravitational constant'.

☯️ Hence, derived !!

<u>━━━━━━━━━━━━━━━━━━━━</u>

8 0

3 years ago

If a rigid body rotates about a fixed axis passing through its center of mass, the body's linear momentum is

dexar [7]

Answer:

The linear momentum is zero

Explanation:

Because

When a rigid body is rotating about a fixed axis passing through point O, the body’s linear momentum given as L = mvG

But VG= 0 so

Linear momentum is zero

6 0

3 years ago

What happens if you drop a chicken wing from the top of the Eiffel Tower

horsena [70]

Answer:

It either goes WEEEEEEE. Or it just breaks apart.

Explanation:

4 0

2 years ago

Read 2 more answers

- Explain why an aircraft will tend to lose height

olganol [36]

Answer:

the planes wings are lifting at an angle to gravity so the plane isn't lifting as much against gravity when it banks. some of the wing lift is going into turning the plane. :) so it needs more lift to bank and stay up

Explanation:

8 0

2 years ago