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3 years ago

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a. Convert 21.0 cm to inches. Show your dimensional analysis setup. b. Convert 29.7 cm to inches. Show your dimensional analysis

setup. c. State the dimensions, in inches, of the international standard for a letter-sized piece of paper. d. Which piece of paper is longer: a U.S. letter-sized piece of paper, or an international letter-sized piece of paper? e. Suppose the United States adopted the international standard for letter-sized paper. Explain at least two things that might result from this change

1 answer:

never [62]3 years ago

4 0

I don't understand this problem... sorry

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A ball is thrown vertically upward from the top of a building 80 feet tall with an initial velocity of 64 feet per second. The d

-BARSIC- [3]

Answer:

a) $t=6.37s$

b) $t=3.3333s$

Explanation:

The knowable variables are the initial hight and initial velocity

$s_{o}=80ft$

$v_{os}=64ft/s$

The equation that describes the motion of the ball is:

$s=80+64t-16t^{2}$

If we want to know the time that takes the ball to hit the ground, we need to calculate it by doing s=0 that is the final hight.

$0=80+64t-12t^{2}$

a) Solving for t, we are going to have two answers

$t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}$

a=-16

b=64

c=80

$t=-1.045 s$ or $t=6.378s$

<em><u>Since time can not be negative the answer is t=6.378s </u></em>

b) To find the time that takes the ball to pass the top of the building on its way down, we must find how much does it move too

First of all, we need to find the maximum hight and how much time does it take to reach it:

$v_{y}=v_{o}+gt$

at maximum point the velocity is 0

$0=64-32.2t$

Solving for t

$t=1.9875 s$

Now, we must know how much distance does it take to reach maximum point

$s=0+64t-16t^{2} =64(1.9875)-12(1.9875)^{2} =80ft$

So, the ball pass the top of the building on its way down at 160 ft

$160=80+64t-16t^{2}$

Solving for t

$t=2s$ or $t=3.333s$

Since the time that the ball reaches maximum point is almost t=2s that answer can not be possible, so the answer is t=3.333s for the ball to go up and down, passing the top of the building

4 0

3 years ago

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If a load of 5 kg covers a distance of 50m in 2 min, what is the power? (g=10m/s)

eduard

Explanation:

the answer is above with its si unit

3 0

2 years ago

A timline is necessary :

kolezko [41]

Answer:

A. For every goal

Explanation:

A timeline is necessary for every goal set, be it academic, life, high school, personal, and the like. A timeline is an important part of goal setting. It helps determine what should be done by a certain time to ensure that you will reach your goal. Although some goals can take a long time to reach, a timeline will be your constant reminder of what needs to be done.

When setting a timeline, you also need to take in consideration what you want to achieve. You need to set a realistic timeline as many usually underestimate the obstacles they may face in pursuing their goals.

8 0

4 years ago

2. What is the difference between special relativity and general relativity? Briefly describe each theory and cite one piece of

Natali [406]

<span>The Special (as in Limited) Theory of Relativity, is a simplification of the General Theory of Relativity.

Essentially, if you eliminate acceleration, and any significant mass from the General Theory, you get the Special Theory.

Evidence for Special Relativity (solar moons for example), is also evidence for the General Theory. The General Theory is supported by:
- Universal expansion
- the spin down of binary pulsars
- frame dragging
- gravitational lensing
- gravitational time dilation</span><span> </span>

7 0

3 years ago

a disk with a radius of 0.1 m is spinning about its center with a constant angular speed of 10 rad/sec. What are the speed and m

padilas [110]

The speed of the bug at the rim of the circular disk is $\boxed{1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ and the acceleration of the bug is $\boxed{10\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}$ .

Further Explanation:

Given:

The angular velocity of the disk is $10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ .

The radius of the circular disk is $0.1\,{\text{m}}$ .

Concept:

The linear speed of the bug present at the rim of the circular disk is given by.

$v = r \times \omega$

Here, $v$ is the linear speed, $r$ is the radius of the disk and $\omega$ is the angular speed of the disk.

Substitute $0.1\,{\text{m}}$ for $r$ and $10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ for $\omega$ in above expression.

$\begin{aligned}v &= 0.1\,{\text{m}} \times 10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\&= 1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} \\\end{aligned}$

The magnitude of the centripetal acceleration of the bug cling to the rim of the disk is given as.

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

Substitute $1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ for $v$ and $0.1\,{\text{m}}$ for $r$ in above equation.

$\begin{aligned}a_c&=\dfrac{(1\text{m/s})^2}{0.1\text{m}}\\&=\frac{1}{0.1}\text{m/s}^2\\&=10\,\text{m/s}^2\end{aligned}$

Thus, the speed of the bug at the rim of the circular disk is $\boxed{1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ and the acceleration of the bug is $\boxed{10\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}$ .

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Answer Details:

Grade: College

Chapter: Uniform Circular motion

Subject: Physics

Keywords: Circular disk, circular motion, angular speed, linear speed, bug clinging, rim of the disk, acceleration, magnitude, constant, rad/s.

7 0

4 years ago

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