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djyliett [7]
3 years ago
13

The velocity of a car changes from 20 m/s east to 5 m/s east in 5 seconds. What is the acceleration of the car?

Physics
1 answer:
Klio2033 [76]3 years ago
3 0
acceleration =  \frac{ v_{2}- v_{1}  }{t} = \frac{5-20}{5} =-3m/s^{2}
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4. What is the acceleration of a rock if the net
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Read 2 more answers
A 4.00 m long, massless beam rests horizontally on a support 3.00 m from the left
Rus_ich [418]

If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

  • Length of the massless beam;L = 4.00m
  • Distance of support from the left end; x = 3.00m
  • First mass; m1 = 31.3 kg
  • Distance of beam from  the left end( m₁ is attached to ); x_1 = ?
  • Second mass; m_2 = 61.7 kg
  • Distance of beam from  the right of the support( m₂ is attached to ); x_1 = 0.273m

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, m_1g( x-x_1) = m_2gx_2

we divide both sides by g

m_1( x-x_1) = m_2x_2

Next, we make x_1, the subject of the formula

x_1 = x - [ \frac{m_2x_2}{m_1} ]

We substitute in our given values

x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]

x_1 = 3.00m - 0.538m

x_1 = 2.46m

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

6 0
3 years ago
What is the potential energy for 65kg climber on top of Mount Everest (8,800 m high)
Fed [463]

Answer:

PE= m * g *h

work:

PE= 65kg * 9.8 kg *8,800 m

PE=5605600 m/kg

idk the actual units i forgot

6 0
3 years ago
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