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Vera_Pavlovna [14]

3 years ago

5

What will be the pressure exerted by the object if 5,000 N of force area of 200 cm²?

1 answer:

devlian [24]3 years ago

3 0

Answer:

$250000 \frac{n}{m {}^{2} }$

Explanation:

pressure is the force applied perpendicular to the surface area of an object per unit area.

for the question,

Force (F) = 5000N

$Area = 200cm {}^{2} = 0.02m {}^{2} \\ Pressure = \frac{force}{area} \\ Pressure = \frac{5000}{0.02} \\ Pressure = 250000 \frac{n}{m {}^{2} }$

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What can be the maximum value of the original kinetic energy of disk AA so as not to exceed the maximum allowed value of the the

timurjin [86]

The question is incomplete. The complete question is :

In your job as a mechanical engineer you are designing a flywheel and clutch-plate system. Disk A is made of a lighter material than disk B, and the moment of inertia of disk A about the shaft is one-third that of disk B. The moment of inertia of the shaft is negligible. With the clutch disconnected, A is brought up to an angular speed ?0; B is initially at rest. The accelerating torque is then removed from A, and A is coupled to B. (Ignore bearing friction.) The design specifications allow for a maximum of 2300 J of thermal energy to be developed when the connection is made. What can be the maximum value of the original kinetic energy of disk A so as not to exceed the maximum allowed value of the thermal energy?

Solution :

Let M.I. of disk A = $I_0$

So, M.I. of disk B = $3I_0$

Angular velocity of A = $\omega_0$

So the kinetic energy of the disk A = $\frac{1}{2}I_0\omega^2$

After coupling, the angular velocity of both the disks will be equal to ω.

Angular momentum will be conserved.

So,

$I_0\omega_0 = I_0 \omega + 3I_0 \omega$

$I_0\omega_0 = 4I_0 \omega$

$\omega = \frac{\omega_0}{4}$

Now,

$K.E. = \frac{1}{2}I_0\omega^2+ \frac{1}{2}3I_0\omega^2$

$K.E. = \frac{1}{2}I_0\frac{\omega_0^2}{16}+ \frac{1}{2}3I_0\frac{\omega_0^2}{16}$

$K.E. = \frac{1}{2}I_0\omega_0^2 \left(\frac{1}{16}+\frac{3}{16}\right)$

$K.E. = \frac{1}{2}I_0\omega_0^2\times \frac{1}{4}$

$\Delta K = \frac{1}{2}I_0\omega_0^2 - \frac{1}{2}I_0\omega_0^2 \times \frac{1}{4}$

$2300=\frac{3}{4}\left(\frac{1}{2}I_0\omega_0^2\right)$

$\frac{1}{2}I_0\omega_0^2=2300 \times \frac{4}{3 } \ J$

Therefore, the maximum initial K.E. = 3066.67 J

3 0

3 years ago

Which of the following are the two factors used to calculate average

kow [346]

Answer:

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Explanation:

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A 26.2-kg dog is running northward at 3.21 m/s, while a 5.30-kg cat is running eastward at 2.64 m/s. Their 67.2-kg owner has the

Elza [17]

Answer:

v=1.26 m/s

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Explanation:

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$p=mv$

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v = 0.20m/s i + 1.25 m/s j

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3 years ago

Science assessment help pls

DENIUS [597]

Answer:

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AysviL [449]

Mass is converted into energy<span>, in the form of light, heat, kinetic energy. </span>

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4 years ago

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