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cluponka [151]
3 years ago
6

Which will have a larger momentum when moving at the same speed: a 2,000-kg truck or a 1,000-kg sedan

Physics
1 answer:
Alina [70]3 years ago
5 0

Answer:

2000 kg

Explanation:

Given that Which will have a larger momentum when moving at the same speed: a 2,000-kg truck or a 1,000-kg sedan

According to the definition of momentum, momentum is the product of mass and velocity.

That is,

Momentum = mass × velocity

Since velocity or speed is the same, then, the one of higher mass will have a greater momentum.

Therefore, the 2000 kg truck will have the greater momentum.

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For 50 coil, with increase of 1.5 V battery voltage, the electromagnet picks about 15 more clips. So, for a 7.5 V battery, it would pick about 30 paper clips.

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Which one of the following is a decomposition reaction?
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What is a cost of urbanization
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4 0
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
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hram777 [196]
OK what is the hole answer i can help you
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