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marusya05 [52]
3 years ago
12

HELLO .................. OVER HERE ................ HELP NEEDED ............... The mass of 7.5 kg has weight of 30 N on a certa

in planet. Calculate the acceleration due to gravity on this planet.
Physics
1 answer:
sveticcg [70]3 years ago
3 0

Answer:

4 kg

Explanation:

Equation for kg to N

7.5 * x = 30

Solve for x

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Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resi
nirvana33 [79]

Answer:

a) F = 882.63\,N, b) \dot W= 4413.15\,W, c) \eta = 15.216\,\%.

Explanation:

a) Let assume that car travel on a horizontal surface. The equations of equilibrium of the car are:

\Sigma F_{x} = F - \mu_{r}\cdot N = 0

\Sigma F_{y} = N - m\cdot g = 0

After some algebraic handling, the following expression for the propulsion force is constructed:

F = \mu_{r}\cdot m \cdot g

F = (0.06)\cdot (1500\,kg)\cdot (9.807\,\frac{m}{s^{2}} )

F = 882.63\,N

b) The power require to move the car at a speed of 5 meters per second is:

\dot W = F\cdot v

\dot W = (882.63\,N)\cdot (5\,\frac{m}{s} )

\dot W= 4413.15\,W

c) The efficiency of the car is:

\eta = \frac{(882.63\,N)\cdot (15\,mi)\cdot (\frac{1609\,m}{1\,mi} )}{(1.4\times 10^{8}\,J)} \times 100\,\%

\eta = 15.216\,\%

3 0
3 years ago
If the mass of a material is 99 grams and the volume of the material is 22 cm3, what would the density of the material be?
Kazeer [188]
You are asked to give the answer in <span>g/cm3. So without knowing any single formulae you can just divide grams by cm3. 

</span>\frac{99 grams}{22 cm^{3} }

 = 4.5 g/cm3
7 0
3 years ago
What three things can happen to water when it falls on earth's surface
777dan777 [17]
It can be stored on the land surface as ice and snow...it can seep into the earth and be stored as surface water...it can flow in the surface of lands.
3 0
3 years ago
A brick of mass 5 kg is released from rest at a height of 3 m. How fast is it going when it hits the ground? Acceleration due to
sineoko [7]

Taking into account the definition of kinetic, potencial and mechanical energy, when the brick hits the ground, it has a speed of 7,668 m/s.

<h3>Kinetic energy</h3>

Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.

Kinetic energy is defined as the amount of work necessary to accelerate a body of a given mass and at rest, until it reaches a given speed. Once this point is reached, the amount of accumulated kinetic energy will remain the same unless there is a change in speed or the body returns to its state of rest by applying a force.

The kinetic energy is represented by the following expression:

Ec= ½ mv²

Where:

  • Ec is the kinetic energy, which is measured in Joules (J).
  • m is the mass measured in kilograms (kg).
  • v is the speed measured in meters over seconds (m/s).

<h3>Potential energy</h3>

On the other hand, potential energy is the energy that measures the ability of a system to perform work based on its position. In other words, this is the energy that a body has at a certain height above the ground.

Gravitational potential energy is the energy associated with the gravitational force. This will depend on the relative height of an object to some reference point, the mass, and the force of gravity.

So for an object with mass m, at height h, the expression applied to the gravitational energy of the object is:

Ep= m×g×h

Where:

  • Ep is the potential energy in joules (J).
  • m is the mass in kilograms (kg).
  • h is the height in meters (m).
  • g is the acceleration of fall in m/s².
<h3>Mechanical energy</h3>

Finally, mechanical energy is that which a body or a system obtains as a result of the speed of its movement or its specific position, and which is capable of producing mechanical work. Then:

Potential energy + kinetic energy = total mechanical energy

<h3>Principle of conservation of mechanical energy </h3>

The principle of conservation of mechanical energy indicates that the mechanical energy of a body remains constant when all the forces acting on it are conservative (a force is conservative when the work it does on a body depends only on the initial and final points and not the path taken to get from one to the other.)

Therefore, if the potential energy decreases, the kinetic energy will increase. In the same way, if the kinetics decreases, the potential energy will increase.

<h3>This case</h3>

A brick of mass 5 kg is released from rest at a height of 3 m. Then, at this height, the brick of mass has no speed, so the kinetic energy has a value of zero because it depends on the speed or moving bodies. But the potential energy is calculated as:

Ep= 5 kg× 9.8 \frac{m}{s^{2} }× 3 m

Solving:

<u><em>Ep= 147 J</em></u>

So, the mechanical energy is calculated as:

Potential energy + kinetic energy = total mechanical energy

147 J +  0 J= total mechanical energy

147 J= total mechanical energy

The principle of conservation of mechanical energy  can be applied in this case. Then, when the brick hits the ground, the mechanical energy is 147 J. In this case, considering that the height is 0 m, the potential energy is zero because this energy depends on the relative height of the object. But the object has speed, so it will have kinetic energy. Then:

Potential energy + kinetic energy = total mechanical energy

0 J +  kinetic energy= 147 J

kinetic energy= 147 J

Considering the definition of kinetic energy:

½  5 kg×v²= 147 J

v=\sqrt{\frac{2x147 J}{5 kg} }

v=7.668 m/s

Finally, when the brick hits the ground, it has a speed of 7,668 m/s.

Learn more about mechanical energy:

brainly.com/question/17809741

brainly.com/question/14567080

brainly.com/question/12784057

brainly.com/question/10188030

brainly.com/question/11962904

#SPJ1

6 0
1 year ago
A uniform magnetic field passes through a horizontal circular wire loop at an angle 19.5 ∘ from the vertical. The magnitude of t
nlexa [21]

To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,

\Phi = BA Cos \theta

Here,

\theta = Angle between areal vector and magnetic field direction.

According to Faraday's law, induced emf in the loop is,

\epsilon= -N \frac{d\Phi }{dt}

\epsilon = -N \frac{(BAcos\theta)}{dt}

\epsilon = -NAcos\theta \frac{dB}{dt}

\epsilon = -N\pi r^2 cos\theta \frac{d}{dt} ( ( 3.75 T ) + ( 3.05T/s ) t + ( -6.95 T/s^2 ) t^2)

\epsilon = -N\pi r^2 cos\theta( (3.05T/s)-(13.9T/s)t )

At time t = 5.71s,  Induced emf is,

\epsilon = -(1) \pi (0.220m)^2 cos(19.5\°)(  (3.05T/s)-(13.9T/s)(5.71s))

\epsilon = 10.9V

Therefore the magnitude of the induced emf is 10.9V

4 0
3 years ago
Read 2 more answers
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