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Sunny_sXe [5.5K]
3 years ago
15

A __________ change involves a change in one or more physical properties, but no change in the fundamental components that make

up the substance
Physics
1 answer:
ch4aika [34]3 years ago
3 0
Physical change :) Hope you get it right! :D Good luck! let me know what else I can help you with. Have a great day! :D 
You might be interested in
A proton (mass=1.67x10^-27 kg, charge= 1.60x10^-19 C) moves from point A to point under the influence of an electrostatic force
Tom [10]

Answer:

VB -  VA  =  - 33.4

Explanation:

Generally the workdone in moving the proton is mathematically represented as

     W  =  KE_f  - KE_i

Where KE_i \ and \  KE_f \  are\  the\  initial  \  and  \  final \  kinetic \  energy

So

    KE_i  =  \frac{1}{2} m v_a^2

Here v_a is the velocity at A with value  50 m/s

So

    KE_i  =  \frac{1}{2} (1.67*10^{-27}) * 50^2

    KE_i  = 2.09 *10^{-24} \  J

Also  

     KE_f  =  \frac{1}{2} m v_b^2

Here v_a is the velocity at A with value 80 km/s = 80000 m/s

=>   KE_f  =  \frac{1}{2} (1.67*10^{-27}) * 80000^2

=>   KE_f  = 5.34 *10^{-18} \  J

 So

    W  =   5.34 *10^{-18}  - 2.09 *10^{-24}

     W  =   5.34 *10^{-18}  m/s

Now this workdone is also mathematically represented as

     W =  q *  V

So  

    q *  V =   5.34 *10^{-18}

Here  q =  1.60*10^{-19} C

So

        V =   \frac{5.34 *10^{-18} }{1.60*10^{-19}}

         V =   33.4 \  V

Generally proton movement is in the direction of the electric field it means that  VA>VB

So

    VB -  VA  =  - 33.4

8 0
3 years ago
What is the characteristic that allows an object to have kinetic energy?
Shtirlitz [24]
Kinetic energy is movement, thus movement would be a characteristic that allows an object to have kinetic energy
6 0
3 years ago
A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet
dimaraw [331]

Answer:

v  = 2.8898 \frac{m}{s}

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

E_{ep} = \frac{1}{2} k (\Delta x)^2

where \Delta x is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

\vec{F} = - k \Delta \vec{x}

as we need 9.12 N to compress 4.60 cm, this means:

k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}

k = 198.26 \ \frac{ N}{m}

So, the elastic energy of the compressed spring is:

E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2

E_{ep} = 0.209759 \ Joules

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

\Delta E_{gp} = m g \Delta h

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m

\Delta E_{gp} = 0.00224 \ Joules

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

E_k = \frac{1}{2} m v^2

so, for the pellet will be

E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2

<h3>All together</h3>

By conservation of energy, we know:

E_{ep} = \Delta E_{gp} + E_k

0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2

So

\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2  = 0.209759 \ Joules - 0.00224 \ Joules

\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2  = 0.207519 \ Joules

v  = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }

v  = 2.8898 \frac{m}{s}

7 0
3 years ago
The three components of velocity in a velocity field are given by u = Ax + By + Cz, v = Dx + Ey + Fz, and w = Gx + Hy + Jz. Dete
Alexxandr [17]

Answer:

The relationship is only between the coefficients A, E and J which is:

A + E + J = 0. The remaining coefficients can be anything without any constraints.

Explanation:

Given:

The three components of velocity is a velocity field are given as:

u = Ax + By + Cz\\\\v = Dx + Ey + Fz\\\\w = Gx + Hy + Jz

The fluid is incompressible.

We know that, for an incompressible fluid flow, the sum of the partial derivatives of each component relative to its direction is always 0. Therefore,

\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0

Now, let us find the partial derivative of each component.

\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}(Ax+By+Cz)\\\\\frac{\partial u}{\partial x}=A+0+0=A\\\\\frac{\partial v}{\partial y}=\frac{\partial }{\partial y}(Dx+Ey+Fz)\\\\\frac{\partial v}{\partial y}=0+E+0=E\\\\\frac{\partial w}{\partial z}=\frac{\partial }{\partial z}(Gx+Hy+Jz)\\\\\frac{\partial w}{\partial z}=0+0+J=J

Hence, the relationship between the coefficients is:

A+E+J=0

There is no such constraints on other coefficients. So, we can choose any value for the remaining coefficients B, C, D, F, G and H.

6 0
3 years ago
A recent home energy bill indicates that a household used 475475 kWh (kilowatt‑hour) of electrical energy and 135135 therms for
faltersainse [42]

Answer:

Explanation:

Given that,.

A house hold power consumption is

475 KWh

Gas used is

135 thermal gas for month

Given that, 1 thermal = 29.3 KWh

Then,

135 thermal = 135 × 29.3 = 3955.5 KWh

So, total power used is

P = 475 + 3955.5

P =4430.5 KWh

Since 1 hr = 3600 seconds

So, the energy consumed for 1hr is

1KW = 1000W

P = energy / time

Energy = Power × time

E = 4430.5 KWhr × 1000W / KW × 3600s / hr

E = 1.595 × 10^10 J

So, using Albert Einstein relativity equation

E = mc²

m = E / c²

c is speed of light = 3 × 10^8 m/s

m = 1.595 × 10^10 / (3 × 10^8)²

m = 1.77 × 10^-7 kg

Then,

1 kg = 10^6 mg

m = 1.77 × 10^-7 kg × 10^6 mg / kg

m = 0.177mg

m ≈ 0.18 mg

5 0
3 years ago
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