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3 years ago

A point at the end of a propeller has a centripetal acceleration of 532.6 m/s^2.

Physics

2 answers:

irinina [24]3 years ago

4 0

Answer:

C. 2.3 m

Explanation:

Centripetal acceleration is:

a = v² / r

Given a = 532.6 m/s² and v = 35 m/s:

532.6 m/s² = (35 m/s)² / r

r = 2.3 m

Vesnalui [34]3 years ago

3 0

Answer:

r = 2.3 meters

Explanation:

Solution,

Given that,

Centripetal acceleration of the propeller, $a=532.6\ m/s^2$

Velocity of the propeller, v = 35 m/s

To find,

Length of the propeller blade.

Solution,

The centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

r is the radius of propeller

$r=\dfrac{v^2}{a}$

$r=\dfrac{35^2}{532.6}$

r = 2.3 meters

So, the length of the propeller is 2.3 merets.

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3 years ago

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ICE Princess25 [194]

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3 years ago

The specific heat of substance A is greater than that of substance B. Both A and B are at the same initial temperature when equa

Sonja [21]

Answer:

$m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)$

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

Explanation:

For this case we know the following info:

$c_{pA} > c_{pB}$

Where c means specific heat for the substance A and B.

We also know that the initial temperatures for both sustances are equal:

$T_{iA}= T_{iB}$

We assume that we don't have melting or vaporization in the 2 substances. So we just have presence of sensible heat given by this formula:

$Q = m c_p \Delta T$

And for this case we know that Both A and B are at the same initial temperature when equal amounts of energy are added to them, so then we have this:

$Q_A = Q_B$

And if we replace the formula for sensible heat we got:

$m_A c_{pA} \Delta T_A = m_B c_{pB} \Delta T_B$

And if we replace for the change of the temperature we got:

$m_A c_{pA} (T_{fA} -T_{iA}) = m_B c_{pB} (T_{fB}- T_{iB})$

And since $T_{iA}= T_{iB}= T$ we have this:

$m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)$

d) More information is needed

(The relation between the masses is not given)

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4 years ago

give an example of motion in which displacement of the particle is zero but acceleration is not zero in journey?

BartSMP [9]

Motion of a ball thrown by a person upwards and caught after some time is an example of motion in which displacement of the particle is zero but acceleration is not zero in journey.

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3 years ago

Projectile Motion—A tennis ball is thrown out a window 28 m above the ground at an initial velocity of 15.0 m/s and 20.0° below

NNADVOKAT [17]

Answer:

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