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3 years ago

Eric threw a baseball 20 meters in 0.5 seconds. What was the average speed of the baseball to the nearest hundredths of a m/sec.

Physics

2 answers:

anzhelika [568]3 years ago

4 0

Answer:

40m/sec

Explanation:

speed = distance ÷ time

speed = 20 ÷ 0.5

speed = 40 m/sec

UkoKoshka [18]3 years ago

3 0

Answer:

40 m/s

Explanation:

Speed or velocity is distance over time. Therefore,

20 meters/0.5 seconds

= 40 m/s

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What constant acceleration (in ft/s2) is required to increase the speed of a car from 27 mi/h to 50 mi/h in 5 seconds

SIZIF [17.4K]

Answer:

6.746 ft/s^2

Explanation:

v(t)=50

v(0)=27

t=5/3600 = 1/720 hours

v(t)-v(0)= a(t-0)

50-27= a(1/720)

a= 23*720= 16560 mi/h^2

16560mi/h^2 * 5280/3600^2 (ft/s^2) =6.746 ft/s^2

4 0

3 years ago

A flexible shaft consists of a 3 mm diameter steel wire in a flexible hollow tube which imposes a frictional torque of 0.04 N m

Rom4ik [11]

Answer:

2.5m

Explanation:

Torque is defined as the rotational effect of a force on a body.

The torque T for the maximum shear stress is given as 0.1 Nm

Frictional torque is the torque caused by a frictional force

The frictional torque F is given as 0.04 Nm/m

The maximum length of the shaft is thus given as

L = T / F

= 0.1/0.04

L= 2.5 m

6 0

4 years ago

to 10 Hz. Superimposed on this signal is 60-Hz noise with an amplitude of 0.1 V. It is desired to attenuate the 60-Hz signal to

givi [52]

Answer:

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

Explanation:

For this case we can use the formula for the Butterworth filter gain given by:

[tec] G = \frac{1}{\sqrt{1 +(\frac{f}{f_c})^{2n}}}[/tex]

Where:

G represent the transfer function and we want that G =0.1 since the desired signal is less than 10% of it's value

$f_c = 10 Hz$ represent the corner frequency

$f= 60 Hz$ represent the original frequency

n represent the filter order and that's the variable that we need to find

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

7 0

3 years ago

A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 13° above the horizontal. (a) if

Blababa [14]

<span>Answer: Therefore, x component: Tcos(24Â°) - f = 0 y component: N + Tsin(24Â°) - mg = 0 The two equations I get from this are: f = Tcos(24Â°) N = mg - Tsin(24Â°) In order for the crate to move, the friction force has to be greater than the normal force multiplied by the static coefficient, so... Tcos(24Â°) = 0.47 * (mg - Tsin(24Â°)) From all that I can get the equation I need for the tension, which, after some algebraic manipulation, yields: T = (mg * static coefficient) / (cos(24Â°) + sin(24Â°) * static coefficient) Then plugging in the values... T = 283.52. Reference https://www.physicsforums.com/threads/difficulty-with-force-problems-involving-friction.111768/</span>

7 0

3 years ago

A simple experiment to measure the speed of sound doesn't involve a stopwatch. You can fill up along tube with water and put a t

Serhud [2]

Answer:

Explanation:

In order to answer this problem you have to know the depth of the column, we say R, this information is important because allows you to compute some harmonic of the tube. With this information you can compute the depth of the colum of air, by taking tino account that the new depth is R-L.

To find the fundamental mode you use:

$f_n=\frac{nv_s}{4L}$