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lianna [129]
3 years ago
7

An object of mass 10kg is accelerated upward at 2m/s^2. What force is required

Physics
1 answer:
arsen [322]3 years ago
7 0
Ya the equation F=MA
So you do mass x acceleration
10 x 2 = 20
Force is measured in Newtons which has a unit of N
So the answer is = 20N

Hope that is helpful
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Two identical satellites are in orbit about the earth. One orbit has a radius r and the other 2r. The centripetal force on the s
velikii [3]

Answer:

the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.

Explanation:

Mass of satellite, m

orbit radius of first, r1 = r

orbit radius of second, r2  = 2r

Centripetal force is given by

F= \frac{mv^{2}}{r}

Where v be the orbital velocity, which is given by

v=\sqrt{gr}

So, the centripetal force is given by

F= \frac{mgr}}{r}}=mg

where, g bet the acceleration due to gravity

g=\frac{GM}{r^{2}}

So, the centripetal force

F= \frac{GMm}}{r^{2}}}

Gravitational force on the satellite having larger orbit

F= \frac{GMm}{4r^{2}} .... (1)

Gravitational force on the satellite having smaller orbit

F'= \frac{GMm}{r^{2}} .... (2)

Comparing (1) and (2),

F' = 4 F

So, the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.

8 0
4 years ago
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
Out of these people who r ur favs?
pav-90 [236]

Answer:

Is it a sin to like them all...? haha

4 0
4 years ago
Read 2 more answers
A long, straight wire carrying a current of 2.60 A is placed along the axis of a cylinder of radius 0.500 m and a length of 4.25
shutvik [7]
<h2>Answer:</h2>

0

<h2>Explanation:</h2>

Since the current carrying wire is placed along the axis of the cylinder, according to the right hand rule, the magnetic field will be tangent to the surface of the cylinder. Therefore, there is no magnetic field through the cylinder.

Remember that the magnetic flux through a given area is the total magnetic field passing through that area. Since there is not magnetic field through the cylinder, the total magnetic flux is therefore zero (0).

7 0
3 years ago
Playing near a road construction site, a child falls over a barrier and down onto a dirt slope that is angled downward at 33° to
Debora [2.8K]

Answer:

\mu_{k} \approx 0.719

Explanation:

The equations of equilibrium for the child are: (x' in the direction parallel to slope, y' in the direction perpendicular to slope)

\Sigma F_{x'} = m\cdot g \cdot \sin \theta - \mu_{k}\cdot N = m\cdot a\\\Sigma F_{y'} = N - m\cdot g\cdot \cos \theta = 0

After some algebraic manipulation, an expression for the coefficient of kinetic friction is obtained:

m\cdot g\cdot \sin \theta - \mu_{k}\cdot m \cdot g \cos \theta = m \cdot a

g \cdot (\sin \theta - \mu_{k}\cdot \cos \theta) = a

\mu_{k}\cdot \cos \theta = \sin \theta - \frac{a}{g}

\mu_{k} = \frac{1}{\cos \theta}\cdot (\sin \theta - \frac{a}{g} )

\mu_{k} = \frac{1}{\cos 33^{\textdegree}}\cdot \left(\sin 33^{\textdegree}-\frac{(-0.57\,\frac{m}{s^{2}}) }{9.807\,\frac{m}{s^{2}} } \right)

\mu_{k} \approx 0.719

5 0
4 years ago
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