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artcher [175]
3 years ago
15

PLEASE HELP it takes tsooo long! lol ima fail! please help! thanks!

Physics
1 answer:
jenyasd209 [6]3 years ago
8 0

Answer:

Tidal  Turbins

Explanation:

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What if energy, like electricity, could not be converted to other forms like sound, heat, motion, or light?
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Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.0 kg gibbon
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Answer:

241.8 N.

Explanation:

The force on branch provides a reaction to the ape's weight force plus the centripetal force needed to keep the gibbon in a circular motion of radius 0.60 m.

Centripetal force = mv^2/r

F = mg + mv²/r

F = m(g + v²/r)

where,

m = mass

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g = acceleration due to gravity

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v = 3.2 m/s

r = 0.60 m

F = 9 * (9.8 + 3.2²/0.60)

= 241.8 N.

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3 years ago
How to do this question about lens? Please help.
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3 years ago
At the intersection of Texas Avenue and University Drive,
Zielflug [23.3K]

Answer:

  • The initial speed of the truck is 21.93 m/s, and the initial speed of the car is 19.524 m/s  

Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector \hat{i} pointing north and \hat{j} pointing east, the final velocity will be

\vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )

\vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )

The final linear momentum will be:

\vec{P}_f = (m_{car}+ m_{truck}) * V_f

\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) *  ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )

\vec{P}_f = (2.850 \ kg \ ) *  ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )

\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )

As there are not external forces, the total linear momentum must be constant.

So:

\vec{P}_0= \vec{P}_f

As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be

\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} ) 

so:

 \vec{P}_0= \vec{P}_f 

( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )  

so

\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}  } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.

So, for the truck

m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}

1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}

v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}

v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}

v_{truck} = 21.93 \frac{m}{s}

And, for the car

950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}

v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}

v_{car}=19.524 \frac{m}{s}

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3 years ago
What area of science compares and describes quantities and movements of matter and energy?
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