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Tcecarenko [31]
3 years ago
5

A disk, with a radius of 0.25 m, is to be rotated like a merry-go-round through 1000 rad, starting from rest, gaining angular sp

eed at the constant rate through the first 500 rad and then losing angular speed at the constant rate until it is again at rest. The magnitude of the centripetal acceleration of any portion of the disk is not to exceed 400 m/s².
(a) What is the least time required for the rotation?
(b) What is the corresponding value of α₁?

Physics
1 answer:
aniked [119]3 years ago
6 0

Answer:

(a) The least time required for the rotation is 50 seconds

(b) The corresponding value of α₁ is 1.6 rad/s²

Explanation: Please see the attachments below

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Answer:

The potential energy increases and the kinetic energy decreases

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A falling skydiver has a mass of 105 kg. What is the magnitude of the skydiver's acceleration when the upward force of air resis
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F-free = m*g - F_air = m*a

F_air = 1.2 * m

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Consider the three dip1acement vectors A = (3i - 3j) m, B = (i-4j) m, and C = (-2i + 5j) m. Use the component method to determin
tia_tia [17]

Answer with Explanation:

We are given that

A=3i-3j m

B=i-4 j m

C=-2i+5j m

a.D=A+B+C

D=3i-3j+i-4j-2i+5j

D=2i-2j

Compare with the vector r=xi+yj

We get x=2 and y=-2

Magnitude=\mid D\mid=\sqrt{x^2+y^2}=\sqrt{(2)^2+(-2)^2}=2\sqrt 2 units

By using the formula \mid r\mid=\sqrt{x^2+y^2}

Direction:\theta=tan^{-1}\frac{y}{x}

By using the formula

Direction of D:\theta=tan^{-1}(\frac{-2}{2})=tan^{-1}(-1)=tan^{-1}(-tan45^{\circ})=-45^{\circ}

b.E=-A-B+C

E=-3i+3j-i+4j-2i+5j

E=-6i+12j

\mid E\mid=\sqrt{(-6)^2+(12)^2}=13.4units

Direction of E=\theta=tan^{-1}(\frac{12}{-6}=tan^{-1}(-2)=-63.4^{\circ}

4 0
3 years ago
If the temperature of the metal oxide and water solution is increased,this would most likely..
kiruha [24]
Chemical Reaction between metal oxide and water solution
7 0
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How much water will flow in 30 secs through 200 mm of capillary tube of 1.50 mm in diameter, if the pressure difference across t
Paladinen [302]

The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

Qo=1.6 \times 10^{2} \mathrm{~mL}

<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}

Generally, the equation for Rate of flow of Liquid is  mathematically given as

\\$$Q=\frac{\pi r^{4} \times \Delta P}{8 \mu L}

$$

Where dP is pressure difference r is the radius

\mu is the viscosity of water

L is the length of the pipe

Q=\frac{\pi \times\left(\frac{1.5 \times 10^{-3}}{2}\right)^{4} \times 6660}{8 \times 8.01 \times 10^{-4} \times 200 \times 10^{-3}}

Q=5.2 \mathrm{~mL} / \mathrm{s}

In $30s the quantity that flows out of the tube

&Qo=5.2 \times 30 \\&Qo=1.6 \times 10^{2} \mathrm{~mL}

In conclusion, the quantity that flows out of the tube

Qo=1.6 \times 10^{2} \mathrm{~mL}

Read more about the flows rate

brainly.com/question/27880305

#SPJ1

5 0
2 years ago
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