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Nezavi [6.7K]
3 years ago
9

What is the benefit of developing the atomic mass unit as a standard unit of mass?

Physics
1 answer:
Alexeev081 [22]3 years ago
3 0

Answer:

The atomic mass unit is 1/12 of an atom of carbon 12, and is a very small amount to represent in kilograms:

1m_{u}=1.66054x10^{-27}kg

m_{u} is atomic mass unit.

This is why the benefits of the atomic mass unit is that it makes the representation of atomic masses easier in terms of the simplicity of the numbers that are used to represent the masses. Also using the atomic mass unit it is easier to compare the masses of different atoms, These numbers would be very small and would require negative powers of 10 to represent them, so it is more convenient to use the atomic mass unit.

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Inside a freely falling runaway elevator, your
nasty-shy [4]

The answer is apparent weight is zero.

You are still accelerating downwards at 9.8m/s^2 (if you are on Earth).

You still are being affected by the Earth's gravity.

Not all because of the previous two statements.

Not none because apparent weight is zero as you are falling.

7 0
3 years ago
Read 2 more answers
An object of mass 6 kg. is resting on a horizontal surface. A horizontal force
son4ous [18]

Answer:

a) The work done by the applied force is 1500 joules.

b) The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) 300 joules of energy are lost during motion.

Explanation:

a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force (W), measured in joules, is defined by the following expression:

W = F\cdot \Delta x (1)

Where:

F - Force, measured in newtons.

\Delta x - Distance, measured in meters.

If we know that F = 15\,N and \Delta x = 100\,m, then the work done by the force exerted on the object is:

W = (15\,N)\cdot (100\,m)

W = 1500\,J

The work done by the applied force is 1500 joules.

b) At first we need to calculate the net acceleration of the object (a), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:

\Delta x = v_{o}\cdot t +\frac{1}{2}\cdot a\cdot t^{2} (2)

Where v_{o} is the initial velocity of the object, measured in meters per second.

We clear the acceleration within the equation above:

\frac{1}{2}\cdot a \cdot t^{2} =  \Delta x-v_{o}\cdot t

a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}

If we know that \Delta x = 100\,m, v_{o} = 0\,\frac{m}{s} and t = 10\,s, then the net acceleration experimented by the object is:

a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}

a = 2\,\frac{m}{s^{2}}

By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:

\Sigma F = F - f = m\cdot a (3)

Where:

F - External force exerted on the object, measured in newtons.

f - Kinetic friction force, measured in newtons.

If we know that F = 15\,N, m = 6\,kg and a = 2\,\frac{m}{s^{2}}, the kinetic friction force is:

f = F-m\cdot a

f = 15\,N-(6\,kg)\cdot \left(2\,\frac{m}{s^{2}} \right)

f = 3\,N

The work done by friction (W'), measured in joules, is:

W' = f\cdot \Delta x (4)

W' = (3\,N) \cdot (100\,m)

W' = 300\,J

And the net work experimented by the object is:

\Delta W = 1500\,J - 300\,J

\Delta W = 1200\,J

By the Work-Energy Theorem we understand that change in translational kinetic energy (\Delta K), measured in joules, is equal to the change in net work. That is:

\Delta K = \Delta W (5)

If we know that \Delta W = 1200\,J, then the change in translational kinetic energy is:

\Delta K = 1200\,J

The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.

7 0
3 years ago
The great limestones caverns were formed by dripping water. If water droplets of 10 ml fall from a height of 5 m at a rate of 10
loris [4]

The average force of the water droplets is the force given by the impact

per second of the droplets on the limestone floor.

  • The average force exerted on the limestone floor is approximately <u>1.6013 × 10⁻² N</u>

Reasons:

The given parameters are;

Volume of a droplet = 10 ml = 1 × 10⁻⁵ m³

Height from which the water falls, <em>h </em>= 5 meters

Rate at which the water falls = 10 per minute

Required:

The average force exerted on the floor by the water droplets.

Solution:

According to Newton's Second Law of motion, we have;

Force = Rate of change of momentum

Momentum = Mass × Velocity

Mass of a droplet of water = Volume × Density

Density of water = 997 kg/m³

Mass of a droplet = 1 × 10⁻⁵ m³ × 997 kg/m³ = 0.00997 kg

The velocity just before the droplet reaches the ground, v = √(2·g·h)

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Which gives;

v = √(2 × 9.81 m/s² × 5 m) ≈ 9.905 m/s

The rate of change in momentum per minute = 1

Therefore;

\displaystyle The \ rate \ of \ change \ in \ momentum = Average \ force = \mathbf{\frac{\Delta Momentum }{\Delta Time}}

ΔMomentum = Mass × ΔVelocity

Considering the 10 drops per minute, we have;

ΔMomentum = 10 × 0.0097 kg × 9.905 m/s = 0.960785 kg·m/s

ΔTime = 1 minute = 60 seconds

Therefore;

\displaystyle Average \ force, \, F_{ave}  \frac{0.960785 \, kg\cdot m/s }{60 \, s} \approx =\mathbf{1.6013 \times 10^{-2} \, N}

  • The average force exerted on the limestone floor by the droplets of water is F_{ave} ≈ <u>1.6013 × 10⁻² N</u>

Learn more about Newton's Second Law of motion and force exerted water here:

brainly.com/question/3999427

brainly.com/question/4197598

3 0
2 years ago
A woman falls to the ground while wearing a parachute. The air resistance on the parachute of the parachute is 500N. If the woma
scoundrel [369]

The gravitational force on the woman is A) 500 N

Explanation:

There are two forces acting on the woman during her fall:

  • The force of gravity, F_G, acting downward
  • The air resistance, F_D, acting upward

According to Newton's second law, the net force acting on the woman is equal to the product between the woman's mass and her acceleration:

\sum F=ma

where m is the mass of the woman and a her acceleration.

The net force can be written as

\sum F = F_G - F_D

Also, we know that the woman falls at a constant velocity (5 m/s), this means that her acceleration is zero:

a=0

Combining the equations together, we get:

F_G-F_D = 0

which means that the magnitude of the gravitational force is equal to the magnitude of the air resistance:

F_G=F_D=500 N

Learn more about forces and Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

5 0
3 years ago
Read 2 more answers
35 POINTS! Say If You Drop A Ball from 100 Centimeters. When The ball Bounces, The Ball Does Not Bounce Back Up To 100 Centimete
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some ball when you bounce it it comes back up but according to gravity the energy goes away

8 0
3 years ago
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