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S_A_V [24]
3 years ago
11

A 6.9 kg object is suspended by a string from the ceiling of an elevator. The acceleration of gravity is 9.8 m/s 2 . Determine t

he tension in the string if it is accelerating upward at a rate of 2.9 m/s 2 . Answer in units of N.
Physics
1 answer:
lina2011 [118]3 years ago
6 0

Answer:

Tension, T = 87.63  N

Explanation:

Given that,

Mass of the object, m = 6.9 kg

The string is acting in the upward direction, a = 2.9 m/s²

Acceleration due to gravity, g = 9.8 m/s²

As the lift is accelerating upwards, it means the net force acting on it is given by :

T = m(a+g)

= 6.9 (2.9+9.8)

= 6.9(2.7)

= 87.63  N

So, the tension in the string is 87.63  N.

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Explanation:

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v=\sqrt{\frac{MgL}{m} } \\v^{2} =\frac{MgL}{m} \\g=\frac{mv^{2} }{ML}

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3 years ago
At the intersection of Texas Avenue and University Drive,
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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} \ )

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\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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