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gavmur [86]
3 years ago
8

if an object is falling on earth at a terminal velocity of 140 km/hr, does that mean that the acceleration is ZERO?

Physics
1 answer:
Simora [160]3 years ago
7 0
If he's falling in a straight line and his speed is not changing, that tells you that his acceleration is zero.

And THAT tells you that the forces on him are balanced, the net force acting on him is zero, and his motion is the same as it would be if there were NO force acting on him.
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The vacuum pressure of a condenser is given to be 80 kpa. if the atmospheric pressure is 98 kpa, what is the gage pressure and a
dimaraw [331]

The absolute pressure is given by the equation,

P_{abs}=P_{atm}-P_{vac}

Here, P_{abs} is absolute pressure,P_{atm} is atmospheric pressure and P_{vac} is vacuum pressure.

Therefore,

P_{abs}=98 kPa-80 kPa=18kPa

The gage pressure is given by the equation,

P_{gage}=P_{abs}-P_{atm}.

Thus,

P_{gage}=18kPa-98 kPa=-80 kPa.

In kn/m^2,

The absolute pressure,

P_{abs}=18kPa(\frac{1kN/m^2}{kPa}) =18\ kN/m^2

The gage pressure,

P_{gage}=-80kPa(\frac{1kN/m^2}{kPa}) =-80\ kN/m^2.

In lbf/in2

The absolute pressure,

P_{abs}=18\ kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=2.6\ lbf/in^2

The gage pressure,

P_{gage}=-80kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=-11.6\ lbf/in^2

In psi,

The absolute pressure,

P_{abs}=18\ kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa})=2.610\ psi.

The gage pressure,

P_{gage}=-80kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa} )=-11.6030\ psi

In  mm Hg

The absolute pressure,

P_{abs}=18kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})= 135\ mm\ of\ Hg

The gage pressure,

P_{gage}=-80kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})=-600\ mm\ of\ Hg



3 0
3 years ago
What is the net charge of a copper atom if it gains 2 electrons?
Alex_Xolod [135]
If an atom gains electrons, it develops a negative charge equal to the number of electrons gained.
So the net charge on the copper atom which gained 2 electrons will be -2.
7 0
3 years ago
What is the force on a person’s hand, which is using a rope to accelerate a 5 kg block upward with an acceleration of 2.2 m/s 2
slavikrds [6]

Answer:

F = 11 N

Explanation:

Given,

Mass of a block, m = 5 kg

Acceleration of the block, a = 2.2 m/s²

We need to find the force on the person's hand. Let it is F. We know that force is given by the product of mass and acceleration as follows :

F = ma

F = 5 kg × 2.2 m/s²

F = 11 N

So, the force on a person's hand is 11 N.

5 0
3 years ago
Object 1 of mass m moves with speed v in the positive direction. Object 2 of mass 3 m moves with speed 4 v in the negative x-dir
Klio2033 [76]

Answer:

This means that the kinetic energy of second object is 48times that of the first object

Explanation:

Kinetic energy is the energy possessed by a body by virtue of its motion e.g motion of an accelerating car. Mathematically,

Kinetic energy = 1/2mv² where;

m is the mass of the object

v is the velocity of the object

If Object 1 of mass m moves with speed v in the positive direction, its kinetic energy will be expressed as;

K1 = 1/2mv²

For Object 2 of mass 3m moving with speed 4v in the negative x-direction, its kinetic energy can be expressed as;

K2 = 1/2(3m)(4v)²

K2 = 1/2(3m)(16v²)

K2 = (3m)(8v²)

K2 = 24mv²

To compare the kinetic energy of both bodies, we will take the ratio of K2:K1 to have;

K2/K1 = 24mv²/(1/2)mv²

K2/K1 = 24/(1/2)

K2/K1 = 48

K2 = 48K1

This means that the kinetic energy of second object is 48times that of the first object and moving in the negative x direction since the body of mass 3m initially moves in the negative x direction.

3 0
3 years ago
A dragster starts from rest and travels 1/4 mi in 6.80 s with constant acceleration. What is its velocity when it crosses the fi
Ahat [919]
<h2>Its velocity when it crosses the finish line is 117.65 m/s</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = ?

        Time, t = 6.8 s    

        Displacement, s = 1/4 mi =    400 meters

     Substituting

                      s = ut + 0.5 at²

                      400 = 0 x 6.8 + 0.5 x a x 6.8²

                      a = 17.30 m/s²

Now we have equation of motion v = u + at

     Initial velocity, u = 0 m/s

     Final velocity, v = ?

     Time, t = 6.8 s

      Acceleration, a = 17.30 m/s²

     Substituting

                      v = u + at  

                      v = 0 + 17.30 x 6.8

                      v = 117.65 m/s

Its velocity when it crosses the finish line is 117.65 m/s

6 0
3 years ago
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