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pickupchik [31]
3 years ago
9

An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad / s). if a particular disk is spun at 79

2.7 rad / s while it is being read, and then is allowed to come to rest over 0.234 seconds, what is the magnitude of the average angular acceleration of the disk?
Physics
1 answer:
NeX [460]3 years ago
4 0
(i) |α| = 235.6rad.s / 0.502s = 469 rad/s²
(ii) tang a = α*r = 469rad/s² * 0.12m / 2*11 = 2.56 m/s²
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Determine the values of m and n when the following average distance from the Sun to the Earth is written in scientific notation:
Leviafan [203]

Answer:

150000000000\ m=1.5\times 10^{11}\ m

Explanation:

A number can be written in the form of :

N=m\times 10^n

Where

m is the real number

n is any integer

In this case, the average distance from the Sun to the Earth is given,

d = 150000000000 meters

There are 10 zeroes in this number. We need to write this number in scientific notation. It is given by :

d=1.5\times 10^{11}\ m

So, the average distance from the Sun to the Earth is d=1.5\times 10^{11}\ m. Hence, this is the required solution.

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3 years ago
“Which type of paper, construction paper or notebook paper, will make a paper airplane travel farther?”
DENIUS [597]

Answer:

Notebook paper makes paper airplane fly farther

Explanation:

It's because paper airplane made of notebook is lighter and can fly far.

4 0
1 year ago
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A rock is thrown upward from the level ground in such a way that the maximum height of its flight is equal to its horizontal ran
marshall27 [118]

Answer

a) For the rock

\dfrac{v_t^2sin 2\theta}{g} = \dfrac{v_t^2sin^2\theta}{2g}

2sin\thetacos\theta = \dfrac{sin^2\theta}{2}

2cos\theta = \dfrac{sin\theta}{2}

tan\theta = 4

\theta = tan^{-1} 4

\theta = 76^0

b) \theta = 45^0 for maximum range

\dfrac{d_{max}}{d}=\dfrac{(v_tcos 45^0)(2v_tsin 45^0)g}{(v_tcos 76^0)(2v_tsin 76^0)g}

\dfrac{d_{max}}{d}=\dfrac{0.707\times 0.707)}{0.97\times 0.242}

\dfrac{d_{max}}{d}=2.129

d_{max}=2.129 d

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5 0
3 years ago
3.00 textbook rests on a frictionless, horizontal tabletop surface. A cord attached to the book passes over a pulley whose diame
sammy [17]

Answer:

a1 = 3.56 m/s²

Explanation:

We are given;

Mass of book on horizontal surface; m1 = 3 kg

Mass of hanging book; m2 = 4 kg

Diameter of pulley; D = 0.15 m

Radius of pulley; r = D/2 = 0.15/2 = 0.075 m

Change in displacement; Δx = Δy = 1 m

Time; t = 0.75

I've drawn a free body diagram to depict this question.

Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;

ΣF_x = T1 = m1 × a1

a1 is acceleration and can be calculated from Newton's 2nd equation of motion.

s = ut + ½at²

our s is now Δx and a1 is a.

Thus;

Δx = ut + ½a1(t²)

u is initial velocity and equal to zero because the 3 kg book was at rest initially.

Thus, plugging in the relevant values;

1 = 0 + ½a1(0.75²)

Multiply through by 2;

2 = 0.75²a1

a1 = 2/0.75²

a1 = 3.56 m/s²

6 0
2 years ago
Match each term to its definition.
den301095 [7]
Answer: first one is electrochemical
Second one is combustion
Third one is photosynthesis
Fourth one is respiration
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3 years ago
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