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Sever21 [200]
3 years ago
10

Two skaters, a man and a woman, are standing on ice. Neglect any friction between the skate blades and the ice. The mass of the

man is 95 kg, and the mass of the woman is 51 kg. The woman pushes on the man with a force of 60 N due east. Determine the acceleration (magnitude and direction) of (a) the man and (b) the woman
Physics
1 answer:
IRISSAK [1]3 years ago
8 0

Answer:

a₁ = 0.63 m/s²  (East)

a₂ = -1.18 m/s²  (West)

Explanation:

m₁ = 95 Kg

m₂ = 51 Kg

F = 60 N

a₁ = ?

a₂ = ?

To get the acceleration (magnitude and direction) of the man we apply

∑Fx = m*a   (⇒)

F = m₁*a₁         ⇒      60 N = 95 Kg*a₁    

⇒  a₁ = (60N / 95Kg) = 0.63 m/s²  (⇒)  East

To get the acceleration (magnitude and direction) of the woman we apply

∑Fx = m*a   (⇒)

F = -m₂*a₂         ⇒      60 N = -51 Kg*a₂    

⇒  a₂ = (60N / 51Kg) = -1.18 m/s²  (West)

For every case we apply Newton’s 3 d Law

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A force F is used to raise a 4-kg mass M from the ground to a height of 5 m. What is the work done by the force F? (Note: sin 60
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Answer:

Answer:196 Joules

Explanation:

Hello

Note:  I think the text in parentheses corresponds to another exercise, or this is incomplete, I will solve it with the first part of the problem

the work  is the product of a force applied to a body and the displacement of the body in the direction of this force

assuming that the force goes in the same direction of the displacement, that is upwards

W=F*D (work, force,displacement)

the force necessary to move the object will be

F=mg(mass *gravity)\\F=4kgm*9.8\frac{m}{s^{2} }\\ F=39.2 Newtons\\replace\\\\W=39.2 N*5m\\W=196\ Joules

Answer:196 Joules

I hope it helps

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A 500 kg motorcycle accelerates at a rate of 2 m/s .how much force was applied to the motorcycle?
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Read 2 more answers
At a certain location, wind is blowing steadily at 9 m/s. Determine the mechanical energy of air per unit mass and the power gen
Misha Larkins [42]

Answer:

  1. The specific mechanical energy of the air in the specific location is 40.5 J/kg.
  2. The power generation potential of the wind turbine at such place is of 2290 kW
  3. The actual electric power generation is 687 kW

Explanation:

  1. The mechanical energy of the air per unit mass is the specific kinetic energy of the air that is calculated using: \frac{1}{2} V^2 where V is the velocity of the air.
  2. The specific kinetic energy would be: \frac{1}{2}(9\frac{m}{s})^2=40.5\frac{m^2}{s^2}=40.5\frac{m^2 }{s^2}\frac{kg}{kg}=40.5\frac{N*m }{kg}=40.5\frac{J}{kg}.
  3. The power generation of the wind turbine would be obtained from the product of the mechanical energy of the air times the mass flow that moves the turbine.
  4. To calculate mass flow it is required first to calculate the volumetric flow. To calculate the volumetric flow the next expression would be: \frac{V\pi D_{blade}^2}{4} =\frac{9\frac{m}{s}\pi(80m)^2}{4} =45238.9\frac{m^3}{s}
  5. Then the mass flow is obtain from the volumetric flow times the density of the air: m_{flow}=1.25\frac{kg}{m^3}45238.9\frac{m^3}{s}=56548.7\frac{kg}{s}
  6. Then, the Power generation potential is: 40.5\frac{J}{kg} 56548.7\frac{kg}{s} =2290221W=2290.2kW
  7. The actual electric power generation is calculated using the definition of efficiency:\eta=\frac{E_P}{E_I}}, where η is the efficiency, E_P is the energy actually produced and, E_I is the energy input. Then solving for the energy produced: E_P=\eta*E_I=0.30*2290kW=687kW
6 0
3 years ago
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