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Alinara [238K]
2 years ago
12

Describing a Physical Change

Physics
1 answer:
babymother [125]2 years ago
6 0

Answer:

physical change is the temporary change or riversible change here the physical properties r only changed

for example when water is cooled its get freezed Nd becomes ice similarly wen ice is heated again then it becomes water so here it's not changed permanently

I hope my ans is comprehensive

plz add me in brainliest Nd plz plz plz follow me I request u

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The pressure and temperature at the beginning of compression of an air-standard Diesel cycle are 95 kPa and 300 K, respectively.
Liula [17]

Answer:

A.33.01

B.2.081

C.66%

Explanation:

See attached file pls

4 0
3 years ago
Two small conducting point charges, separated by 0.4 m, carry a total charge of 200 C. They repel one another with a force of 12
Lunna [17]

Answer:

200 C

Explanation:

Let C1 and C2 be their charges. According to Coulomb's law

F_C = k\frac{C_1C_2}{R^2}

where k = 8.99\times10^9 nm^2/C^2 is the constant, R = 0.4m is the distance between them, F = 120 N is their resulting charge force

120 = 8.99\times10^9\frac{C_1C_2}{0.4^2}

C_1C_2 = \frac{120*0.4^2}{8.99\times10^9} = 2.13\times10^{-9}

Since their total charge is 200C:

C_1 + C_2 = 200 or C_1 = 200 - C_2

We can substitute the above equation

C_1C_2 = (200 - C_2)C_2 = 2.13\times10^{-9}

-C_2^2 +200C_2 - 2.13\times10^{-9} = 0

C= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

C= \frac{-200\pm \sqrt{(200)^2 - 4*(-1)*(-0.00000000213)}}{2*(-1)}

C= \frac{-200\pm200}{-2}

C = 1.06 \times 10^{-11} or C \approx 200

So the larger charge is C = 200 C

8 0
3 years ago
Three forces act on a moving object. One force has a magnitude of 83.7 N and is directed due north. Another has a magnitude of 5
LekaFEV [45]

Answer:

  • |\vec{F}_3| = 102.92 \ N
  • \theta = 57 \° 24 ' 48''

Explanation:

For an object to move with constant velocity, the acceleration of the object must be zero:

\vec{a} = \vec{0}.

As the net force equals acceleration multiplied by mass , this must mean:

\vec{F}_{net} = m \vec{a} = m * \vec{0} = \vec{0}.

So, the sum of the three forces must be zero:

\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{0},

this implies:

\vec{F}_3  = - \vec{F}_1 - \vec{F}_2.

To obtain this sum, its easier to work in Cartesian representation.

First we need to define an Frame of reference. Lets put the x axis unit vector \hat{i} pointing east,  with the y axis unit vector \hat{j} pointing south, so the positive angle is south of east. For this, we got for the first force:

\vec{F}_1 = 83.7 \ N \ (-\hat{j}),

as is pointing north, and for the second force:

\vec{F}_2 = 59.9 \ N \ (-\hat{i}),

as is pointing west.

Now, our third force will be:

\vec{F}_3  = - 83.7 \ N \ (-\hat{j}) - 59.9 \ N \ (-\hat{i})

\vec{F}_3  =  83.7 \ N \ \hat{j}  + 59.9 \ N \ \hat{i}

\vec{F}_3  =  (59.9 \ N , 83.7 \ N )

But, we need the magnitude and the direction.

To find the magnitude, we can use the Pythagorean theorem.

|\vec{R}| = \sqrt{R_x^2 + R_y^2}

|\vec{F}_3| = \sqrt{(59.9 \ N)^2 + (83.7 \ N)^2}

|\vec{F}_3| = 102.92 \ N

this is the magnitude.

To find the direction, we can use:

\theta = arctan(\frac{F_{3_y}}{F_{3_x}})

\theta = arctan(\frac{83.7 \ N }{ 59.9 \ N })

\theta = 57 \° 24 ' 48''

and this is the angle south of east.

7 0
3 years ago
I need help answering this question in the photo (middle)
GalinKa [24]
I believe it is because of weight if Timmy is larger and bigger than Maria that would mean he would stop slower just because of his bodyweight pushing on the back of the skateboard while Maria is all those skinny and she doesn’t have as much weight as she can go farther
3 0
3 years ago
A 25-kg child sits at the top of a 4-meter slide. After sliding down, the child is traveling at 5 m/s. How much PE does he start
Semmy [17]

Daniddmelo says it right there, don't know why he got reported.

The potential energy (PE) is mass x height x gravity. So it would be 25 kg x 4  m x 9.8 = 980 joules. The child starts out with 980 joules of potential energy. The kinetic energy (KE) is (1/2) x mass x velocity squared. KE = (1/2) x 25 kg x 5 m/s2 = 312.5 joules. So he ends with 312.5 joules of kinetic energy. The Energy lost to friction =  PE - KE. 980- 312.5 = 667.5 joules of energy lost to friction.

Please don't just copy and paste, and thank you Dan cause you practically did it I just... elaborated more? I dunno. 

4 0
3 years ago
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