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3 years ago

9

You measure a watch's hour and minute hands to be 7.4 mm and 12.1 mm long, respectively. Part A In one day, by how much does the

distance traveled by the tip of the minute hand exceed the distance traveled by the tip of the hour hand?

1 answer:

pychu [463]3 years ago

3 0

Answer:

109.385m

Explanation:

In 1 day, the hour hand travels 2 circles, or 4π rad in angular. The distance it travels is its angle times the radius

7.4 * 4π = 93 mm

In 1 day, the minute hand travels 24*60 = 1440 circles, or 1440 * 2π = 2880π rad in angular. The distance it travels is

12.1 * 2880π = 109478 mm

So the distance traveled by the tip of the minute hand that exceed the distance traveled by the tip of the hour hand is

109478 - 93 = 109385 mm or 109.385 m

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3 years ago

Your 64-cm-diameter car tire is rotating at 3.3 rev/swhen suddenly you press down hard on the accelerator. After traveling 250 m

umka21 [38]

Answer:

0.76 rad/s^2

Explanation:

First, we convert the original and final velocity from rev/s to rad/s:

$v_o = 3.3\frac{rev}{s} * \frac{2\pi rad}{1rev} =20.73 rad/s$

$v_f = 6.4\frac{rev}{s} * \frac{2\pi rad}{1rev}=40.21 rad/s$

Now, we need to find the number of rads that the tire rotates in the 250m path. We use the arc length formula:

$D = x*r \\x = \frac{D}{r} = \frac{250m}{0.64m/2} = 781.25 rads$

Now, we just use the formula:

$w_f^2-w_o^2=2\alpha*x$

$\alpha =\frac{w_f^2-w_o^2}{2x} = \frac{(40.21rad/s)^2-(20.73rad/s)^2}{2*781.25rad} = 0.76 rad/s^2$

6 0

3 years ago

Read 2 more answers

In a game of basketball, a forward makes a bounce pass to the center. The ball is thrown with an initial speed of 4.8 m/s at an

tensa zangetsu [6.8K]

The equation to be used here is the trajectory of a projectile as written below:

y = xtanθ +/- gx²/2v²(cosθ)²
where
y is the vertical distance
x is the horizontal distance
θ is the angle of trajectory or launch angle
g is 9.81 m/s²
v is the initial velcity

Since the angle is below horizontal, let's use the minus equation. Substituting the values:

- 0.8 m = xtan15° - (9.81 m/s²)x²/2(4.8 m/s)²(cos15°)²
Solving for x,
x = 2.549 m

However, we only take half of this distance because it was specified that the distance asked before bouncing. Hence, the horizontal distance is equal to 1.27 m.

5 0

3 years ago

g calculate the effectiveness radiation dosage in sieverts for a 79 kg person who is exposed 6.8x10^9

Elza [17]

Answer:

The answer is " $\bold{dosage = 0.031 rem}$ "

Explanation:

please find the complete question in the attached file.

Given value:

$m = 79\ kg \\\\n = 3.4 \times 10^9 \\\\E = 5.5 \times 10^{-13} \\\\ RBE = 15$

$\to E = n E\\$

$= 3.4 \times 10^9 \times 5.5 \times 10^{-13} \\\\ = 1.87 \times 10^{-3}$

$\to E(absorbed) = 1.87 \times 10^{-3} \times 0.87 = 1.63 \times 10^{-3}$

calculating the radiation absorbed per kg:

$= \frac{1.63 \times 10^{-3}}{79} \\\\ = 2.06 \times 10^{-5} \\\\ = 0.00206 \ rad$

$\to Dosage = 0.00206 \times 15 \\$

$= 0.031 \ \ rem$

4 0

3 years ago

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.200 rev/s . The magnitude

blagie [28]

Answer:

(A). the angular velocity of fan is 0.380 rev/s.

(B). The number of revolution is 0.059 rad.

(C). The tangential speed of the blade is 0.907 m/s.

(D). The tangential acceleration of the blade is 2.10 m/s²

Explanation:

Given that,

Initial angular velocity = 0.200 rev/s

Angular acceleration = 0.883 rev/s²

Diameter = 0.760 m

Time = 0.204 s

(A). We need to calculate the angular velocity of fan

Using equation of angular motion

$\omega_{f}=\omega_{i}+\alpha t$

Put the value into the formula

$\omega_{f}=0.200+0.883\times0.204$

$\omega_{f}=0.380\ rev/s$

(B). We need to calculate the number of revolution

Using formula of angular displacement

$\theta=\omega_{i}t+\dfrac{1}{2}\alpha t^2$

Put the value into the formula

$\theta=0.200\times0.204+\dfrac{1}{2}\times0.883\times(0.204)^2$

$\theta=0.059\ rad$

(C). We need to calculate the tangential speed of the blade

Using formula of speed

$v=\dfrac{d}{2}\omega_{f}$

Put the value into the formula

$v=\dfrac{0.760}{2}\times0.380\times2\pi$

$v=0.907\ m/s$

(D). We need to calculate the tangential acceleration of the blade

Using formula of tangential acceleration

$a_{t}=\alpha r$

Put the value into the formula

$a_{t}=0.883\times0.38\times2\pi$

$a_{t}=2.10\ m/s^2$

Hence, (A). the angular velocity of fan is 0.380 rev/s.

(B). The number of revolution is 0.059 rad.

(C). The tangential speed of the blade is 0.907 m/s.

(D). The tangential acceleration of the blade is 2.10 m/s²

3 0

3 years ago