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Serga [27]
3 years ago
5

If the ankylosaurs starts running and accelerates at 1.3m/s^2 for 3 seconds how far does he make it before the velociraptor catc

hes him and bad things happened?
Physics
1 answer:
sveta [45]3 years ago
7 0

Acceleration is the change of velocity, and velocity is the change of distance. The opposite of finding change, or differentiation, is integration.

Acceleration = 1.3 m/s²

Velocity: ∫ 1.3 dx = 1.3x + c m/s

Distance: ∫ 1.3x dx = 1.3x²/2 + c m

Distance run: 1.3*3²/2 = 5.85 m

<em>What</em><em> </em><em>bad</em><em> </em><em>thing</em><em> </em><em>happened</em><em>?</em>

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2 years ago
Explain the relationship between mass and gravity
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Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr
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Given:\\T=29.46y\approx 9.29\cdot 10^8s\\M_S\approx2.0\cdot 10^{30}kg\\G=6.67\cdot 10^{-11} \frac{m^3}{kg\cdot s^2} \\\\Find:\\R=?\\\\Solution:\\\\F_g= G\frac{mM_s}{R^2} \\\\F_c= \frac{mv^2}{R} \\\\F_g=F_c\\\\G\frac{mM_s}{R^2}=\frac{mv^2}{R} \Rightarrow G\frac{M_s}{R^2}=\frac{v^2}{R}\\\\v=\omega r\\\\G\frac{M_s}{R^2}= \frac{\omega^2R^2}{R}\Rightarrow G\frac{M_s}{R^2}=\omega^2R \\\\\omega= \frac{2 \pi }{T} \\\\G\frac{M_s}{R^3}= \frac{4 \pi ^2}{T^2}

GM_ST^2=4 \pi ^2R^3\Rightarrow R= \sqrt[3]{ \frac{GM_ST^2}{4 \pi ^2} }\\\\\\R= \sqrt[3]{ \frac{6.67\cdot 10^{-11} \frac{m^3}{kg\cdot s^2}\cdot2.0\cdot 10^{30}kg( 9.29\cdot 10^8s)^2}{4\cdot 3.14^2} }  \approx 1.42\cdot 10^{12}m

3 0
3 years ago
You set a tuning fork into vibration at a frequency of 723 Hz and then drop it off the roof of the Physics building where the ac
zaharov [31]

Answer:

Explanation:

Given

Original Frequency f=723\ Hz

apparent Frequency f'=697\ Hz

There is change in frequency whenever source move relative to the observer.

From Doppler effect we can write as

f'=f\cdot \frac{v-v_o}{v+v_s}

where  

f'=apparent frequency  

v=velocity of sound in the given media

v_s=velocity of source

v_0=velocity of observer  

here v_0=0

697=723\cdot (\frac{343-0}{343+v_s})

v_s=(\frac{f}{f'}-1)v

v_s=(\frac{723}{697}-1)\cdot 343

v_s=12.79\approx 12.8\ m/s

i.e.fork acquired a velocity of 12.8 m/s

distance traveled by fork is given by

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

v_s^2-0=2\times 9.8\times s

s=\frac{12.8^2}{2\times 9.8}

s=8.35\ m

                                       

5 0
3 years ago
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