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GenaCL600 [577]

3 years ago

7

A pitcher throws a baseball that reaches the catcher in 0.75 s. The ball curves because it is spinning at an average angular vel

ocity of 230 rev/min (assumed constant) on its way to the catcher. a) Determine the angular velocity in rad/sec. b) Determine the angular displacement spend by the ball from the pitcher to the catcher.

1 answer:

7nadin3 [17]3 years ago

4 0

The change in angular displacement as a function of time is the definition given for angular velocity, this is mathematically described as

$\omega = \frac{\theta}{t}$

Here,

$\theta$ = Angular displacement

t = time

The angular velocity is given as

$\omega = 230rev/min$

PART A) The angular velocity in SI Units will be,

$\omega = 230rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega = \frac{23}{3}\pi rad/s \approx 24.08rad/s$

PART B) From our first equation we can rearrange to find the angular displacement then

$\theta = \omega t$

Replacing,

$\theta = (24.08)(0.75)$

$\theta = 18.06 rad$

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A helicopter is ascending vertically with a speed of 5.10m/s. At a height of 105m above the Earth, a package is dropped from a w

Luden [163]

Here it is given that initial speed of the package will be same as speed of the helicopter

$v_i = 5.10 m/s$

displacement of the package as it is dropped on ground

$d = -105 m$

acceleration is due to gravity

$a = -9.8 m/s^2$

now by kinematics

$y = v* t + \frac{1}{2}at^2$

$-105 = 5.1 * t - \frac{1}{2}*9.8*t^2$

$4.9 t^2 - 5.1 t - 105 = 0$

by solving above equation we have

$t = 5.2 s$

so it will take 5.2 s to reach the ground

7 0

4 years ago

A car, initially traveling 28.0ft/s, steadily speeds up to 50.0ft/s in 7.40s. Determine all unknowns and answer the following qu

nataly862011 [7]

Explanation:

Given:

$v_0 = 28.0\:\text{ft/s}$

$v = 50.0\:\text{ft/s}$

$t = 7.40\:\text{s}$

First, we calculate the acceleration of the car during this time:

$v = v_0 + at \Rightarrow a = \dfrac{v - v_0}{t}$

Plugging in the given values, we get

$a = \dfrac{50.0\:\text{ft/s} - 28.0\:\text{ft/s}}{7.40\:\text{s}} = 2.97\:\text{ft/s}^2$

Now that we have the value for the acceleration, we can solve for the distance traveled during the time t:

$x = v_0t + \frac{1}{2}at^2$

$\:\:\:\:=(28.0\:\text{ft/s})(7.40\:\text{s})$

$\:\:\:\:\:\:\:\:\:\:\:\:+ \frac{1}{2}(2.97\:\text{ft/s}^2)(7.40\:\text{s})^2$

$\:\:\:\:= 289\:\text{ft}$

7 0

3 years ago

Answer these questions about a light year.?

aliina [53]

These are five questions and five answers:

a. State the speed in m/sec using scientific notation

i) Find the place value of the digit with the highest place value (the digit most to the left) in the number 300,000,000: hundred millions.

ii) Determine the number of zeros corresponding to that place value: 9

ii) Write the digit with the highest place value, mulitplied by the power of ten with exponent equal to the number of zeros corresponding to the place of such digit less 1 (9 - 1 = 8): 3 × 10⁸

iii) Complete with the units: 3 × 10⁸ m/s ← answer

b. State the speed in km/sec using scientific notation

i) State the conversion factor: 1 km = 1000 m = 10³ m

⇒ 1 = 1 km / 10³ m

ii) Multiply by the conversion factor:

3 × 10⁸ m/s × 1 km / (10³ m)

iii) Simplify (cancel the units that are common in the numerator and denominator)

3 × (10⁸ / 10³) km/s = 3 × 10⁵ km/s ← answer

c. State the number of seconds in one year using scientific notation

i) Find the place value of the digit with the highest place value (the digit most to the left) in the number 32,000,000: ten millions.

ii) Determine the number of zeros corresponding to that place value: 8

ii) Write the digit with the highest place value, as integer, add the other significant figures as decimals, mulitplied by the power of ten with exponent equal to the number of zeros corresponding to the place of the digit with highest place value less 1 (8 - 1 = 7): 3.2 × 10⁷

iii) Complete with the units: 3.2 × 10⁷ s ← answer

d. Calculate the distance in meters of one light year

i) distance formula: distance = speed × time

ii) Replace with the numbers in scientific notation:

distance = 3 × 10⁸ m/s × 3.2 × 10⁷ s

iii) Simplify (due the operations and cancel the units that are common in the numerator and denominator)

3 × 10⁸ m/s × 3.2 × 10⁷ s = 9.6 × 10 ¹⁵ m ← answer

e. State this distance in centimeters

i) State the conversion factor: 1m = 100 cm = 10² cm

⇒ 1 = 10² cm / 1 m

ii) Multiply by the conversion factor: 9.6 × 10 ¹⁵ m × 10² cm / 1 m

iii) Simplily: 9.6 × 10⁷ cm ← answer

4 0

3 years ago

Read 2 more answers

A flag is waved 3.2 m above the surface of a flat pool of water. When viewed from under the water, what is the magnification of

jek_recluse [69]

Answer:

1.33

Explanation:

For an optical instrument, the magnification ratio of the apparent diameter of the image to that of the object.

Mathematically, from the given information;

Magnification $= \dfrac{n_{water}}{n_{air}}$

where;

$n_{water} =1.33\\ \\ n_{air} = 1.00$

$= \dfrac{1.33}{1.00} \\ \\= \mathbf{1.33}$

5 0

3 years ago

A toy car moves around a circular track at constant speed. It suddenly doubles its speed — a change of a factor of 2. As a resul

Zigmanuir [339]

Answer:

option B

Explanation:

given,

toy car is moving in circular track

speed is doubled— a change of a factor of 2

to find change in factor of acceleration

radius doesn't change

centripetal acceleration formula

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

velocity is change in factor of 2

so acceleration will be change at the factor of

= $\dfrac{(2v)^2}{r}$

= 2² = 4

so the correct answer is option B

4 0

4 years ago