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Citrus2011 [14]
3 years ago
7

What is your mass in kilograms if you weigh 120 pounds

Physics
1 answer:
schepotkina [342]3 years ago
6 0

It depends where you are.

-- If you weigh 120 pounds on the Moon,
then your mass is  329.1 kilograms.

-- If you weigh 120 pounds on Mars,
 then your mass is  143.8 kilograms.

-- If you weigh 120  pounds on the Earth,
then your mass is  54.4 kilograms.

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A 1250 kg car starts are rest then speeds up to 15 m/s. What is the impulse of the car?
Masteriza [31]

Answer:

B)18750

Explanation:

it because

1250 \times 15 = 18750

7 0
3 years ago
Read 2 more answers
A high-speed K0 meson is traveling at β = 0.90 when it decays into a π + and a π − meson. What are the greatest and least speeds
san4es73 [151]

Answer:

greatest speed=0.99c

least speed=0.283c

Explanation:

To solve this problem, we have to go to frame of center of mass.

Total available energy fo π + and π - mesons will be difference in their rest energy:

E_{0,K_{0} }-2E_{0,\pi }  =497Mev-2*139.5Mev\\

                       =218 Mev

now we have to assume that both meson have same kinetic energy so each will have K=109 Mev from following equation for kinetic energy we have,

K=(γ-1)E_{0,\pi }

K=E_{0,\pi}(\frac{1}{\sqrt{1-\beta ^{2} } } -1)\\\frac{1}\sqrt{1-\beta ^{2} }}=\frac{K}{E_{0,\pi}}+1\\   {1-\beta ^{2}=\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\beta ^{2}=1-\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\beta = +-\sqrt{\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\\\\beta =+-\sqrt{1-\frac{1}{(\frac{109Mev}{139.5Mev+1)^2}}

u'=+-0.283c

note +-=±

To find speed least and greatest speed of meson we would use relativistic velocity addition equations:

u=\frac{u'+v}{1+\frac{v}{c^{2} } } u'\\u_{max} =\frac{u'_{+} +v_{} }{1+\frac{v}{c^{2} } } u'_{+} \\u_{max} =\frac{0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } 0.828\\ u_{max} =0.99c\\u_{min} =\frac{u'_{-} +v_{} }{1+\frac{v}{c^{2} } } u'_{-}\\u_{min} =\frac{-0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } -0.828c\\u_{min} =0.283c

6 0
3 years ago
Shinra wants to find and save his younger brother Sho. In order to do this
Debora [2.8K]

Answer:

My answer is 250m/sec

Explanation:

the formula is clearly

Speed = <u>Distance</u>

Time

Speed = <u>5</u><u>0</u><u>0</u><u>m</u><u>e</u><u>t</u><u>e</u><u>r</u><u>s</u><u> </u> (Divide the two)

2seconds

Speed = 250m/sec

4 0
3 years ago
Read 2 more answers
An oscillator consists of a block of mass 0.373 kg connected to a spring. When set into oscillation with amplitude 33 cm, the os
aniked [119]

Answer:

(a)  T = 0.412s

(b)  f = 2.42Hz

(c)  w = 15.25 rad/s

(d)  k = 86.75N/m

(e)  vmax = 5.03 m/s

Explanation:

Given information:

m: mass of the block = 0.373kg

A: amplitude of oscillation = 22cm = 0.22m

T: period of oscillation = 0.412s

(a) The period is the time of one complete oscillation = 0.412s

The period is 0.412s

(b) The frequency is calculated by using the following formula:

f=\frac{1}{T}=\frac{1}{0.412s}=2.42Hz

The frequency is 2.42 Hz

(c) The angular frequency is:

\omega=2\pi f=2\pi (2.42Hz)=15.25\frac{rad}{s}

The angular frequency is 15.25 rad/s

(d) The spring constant is calculated by solving the following equation for k:

\omega=\sqrt{\frac{k}{m}}\\\\k=m\omega^2=(0.373kg)(15.25rad/s)^2=86.75\frac{N}{m}

The spring constant is 86.75N/m

(e) The maximum speed is:

v_{max}=\omega A=(15.25rad/s)(0.33m)=5.03\frac{m}{s}

(f) The maximum force applied by the spring if for the maximum elongation, that is, the amplitude:

F=kA=(86.75N/m)(0.2m)=17.35N

The maximum force that the spring exerts on the block is 17.35N

3 0
4 years ago
What’s the answer to the summary ?
Talja [164]

Because of the different speeds..

7 0
3 years ago
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