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3 years ago

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You are a pirate working for dread pirate roberts. you are in charge of a cannon that exerts a force 10000 n on a cannon ball wh

ile the ball is in the barrel of the cannon. the length of the cannon barrel is 2.44 m and the cannon is aimed at a 45◦ angle from the

1 answer:

KATRIN_1 [288]3 years ago

7 0

<span>Answer: Let m = mass of cannon Then 10000 = ma a = 10000/m v^2 = u^2 + 2as v^2 = 0 + 2as 84^2 = 2(2.21)(10000/m) 84^2 m = 4.42(10000) m = 6.264172336 = 6.26 kg Part 2 Range = u^2sin(2x38)/g = 84^2sin(76)/9.8 = 698.6129229 = 698.6 m</span>

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An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e

White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

$x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)$

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

$\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}$

with this value you can compute the frequency:

a)

$f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz$

b)

the mass of the block is given by the formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg$

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

$\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s$

Finally, the amplitude is:

$x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m$

5 0

2 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

VMariaS [17]

At point C because it is at the lowest position.

7 0

3 years ago

Calculate the wavelength of each frequency of electromagnetic radiation: a. 100.2 MHz (typical frequency for FM radio broadcasti

Natalka [10]

Answer:

a). 100.2 MHz (typical frequency for FM radio broadcasting)

The wavelength of a frequency of 100.2 Mhz is 2.99m.

b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)

The wavelength of a frequency of 1070 khz is 280.3 m.

c. 835.6 MHz (common frequency used for cell phone communication)

The wavelength of a frequency of 835.6 Mhz is 0.35m.

Explanation:

The wavelength can be determined by the following equation:

$c = \lambda \cdot \nu$ (1)

Where c is the speed of light, $\lambda$ is the wavelength and $\nu$ is the frequency.

Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave.

<em>a). 100.2 MHz (typical frequency for FM radio broadcasting)</em>

Then, $\lambda$ can be isolated from equation 1:

$\lambda = \frac{c}{\nu}$ (2)

since the value of c is $3x10^{8}m/s$ . It is necessary to express the frequency in units of hertz.

$\nu = 100.2 MHz . \frac{1x10^{6}Hz}{1MHz}$ ⇒ $100200000Hz$

But $1Hz = s^{-1}$

$\nu = 100200000s^{-1}$

Finally, equation 2 can be used:

$\lambda = \frac{3x10^{8}m/s}{100200000s^{-1}}$

$\lambda = 2.99 m$

Hence, the wavelength of a frequency of 100.2 Mhz is 2.99m.

<em>b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)</em>

<em> </em>

$\nu = 1070kHz . \frac{1000Hz}{1kHz}$ ⇒ $1070000Hz$

But $1Hz = s^{-1}$

$\nu = 1070000s^{-1}$

Finally, equation 2 can be used:

$\lambda = \frac{3x10^{8}m/s}{1070000s^{-1}}$

$\lambda = 280.3 m$

Hence, the wavelength of a frequency of 1070 khz is 280.3 m.

<em>c. 835.6 MHz (common frequency used for cell phone communication) </em>

$\nu = 835.6MHz . \frac{1x10^{6}Hz}{1MHz}$ ⇒ $835600000Hz$

But $1Hz = s^{-1}$

$\nu = 835600000s^{-1}$

Finally, equation 2 can be used:

$\lambda = \frac{3x10^{8}m/s}{835600000s^{-1}}$

$\lambda = 0.35 m$

Hence, the wavelength of a frequency of 835.6 Mhz is 0.35m.

6 0

3 years ago

44. A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down

Vladimir [108]

Answer:

18.4 m

Explanation:

(a)

The known variables in this problem are:

u = 1.40 m/s is the initial vertical velocity (we take downward direction as positive direction)

t = 1.8 s is the duration of the fall

a = g = 9.8 m/s^2 is the acceleration due to gravity

(b)

The vertical distance covered by the life preserver is given by

$d=ut + \frac{1}{2}at^2$

If we substitute all the values listed in part (a), we find

$d=(1.40 m/s)(1.8 s)+\frac{1}{2}(9.8 m/s^2)(1.8 s)^2=18.4m$

8 0

3 years ago

A certain shade of blue has a frequency of 7.15 × 1014 hz. what is the energy of exactly one photon of this light?

Ket [755]

The energy carried by a single photon of frequency f is given by:
$E=hf$
where $h=6.6 \cdot 10^{-34} m^2 kg s^{-1}$ is the Planck constant. In our problem, the frequency of the photon is $f=7.15 \cdot 10^{14}Hz$ , and by using these numbers we can find the energy of the photon:
$E=(6.6\cdot 10^{-34}m^2 kg s^{-1})(7.15 \cdot 10^{14}Hz)=4.7 \cdot 10^{-19}J$

4 0

3 years ago