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Anna11 [10]
2 years ago
6

What are the like terms in the expression? 4m-9+3n+2

Physics
2 answers:
mash [69]2 years ago
8 0
The like terms in this expression are, -9 and 2, but that is it
wel2 years ago
5 0
-9 and +2
Hope I helped :)

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Which sequence shows electromagnetic waves arranged in a decreasing order of their wavelengths?
Hunter-Best [27]

Answer: Gamma rays, x-rays, ultraviolet rays, visible light, and infrared rays.

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The acceleration of an object depends on the ____________ exerted on it and it's_________ this is to do with newton's seconed la
MArishka [77]
Force is the first and mass is the second
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3 years ago
A closed container initially holds 50 monatomic Aparticles that have a combined energy of 480 units. After 100 monatomic B parti
Molodets [167]

Answer:

"8 units" is the appropriate answer.

Explanation:

According to the question,

Throughout equilibrium all particles are of equivalent intensity, and as such the integrated platform's total energy has been uniformly divided across all individuals.

Now,

The total energy will be:

= 480+720

= 1200 \ units

The total number of particles will be:

= 50+100

= 150

hence,

Energy of each A particle or each B particle will be:

= \frac{1200}{150}

= 8 \ units

5 0
2 years ago
If the resultant force acting on a 2.0 kg object is equal to (3.0î + 4.0ĵ) N, what is the change in kinetic energy as the object
12345 [234]

Answer:

ΔK = 24 joules.

Explanation:

ΔK = Work done on the object

Work is equal to the dot product of force supplied and the displacement of the object.

W = F * Δs

Δs can be found by subtracting the vectors (7.0, -8.0) and (11.0, -5.0), which is written as Δs = (11.0 - 7.0, -5.0 - -8.0) which equals (4.0, 3.0).

This gives us

W = < 3, 4 > * < 4, 3 > = (3*4)+(4*3) = 24 J

3 0
1 year ago
A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.
AveGali [126]

Answer:

The magnitude of electric force is  7.2\times10^{-3} N

Explanation:

Coulomb's Law:

The force of attraction or repletion is

  • directly proportional to the products of charges i.e F\propto q_1q_2
  • inversely proportional to the square of distance i.e F\propto \frac{1}{r^2}

\therefore F\propto \frac{q_1q_2}{r^2}

\Rightarrow F=k \frac{q_1q_2}{r^2}    [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force  be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5

we know,

cos \theta = \frac{base }{hypotenuse}

\Rightarrow cos \theta = \frac{4 }{r}

Total force F_Q = 2.F_1 cos\theta \hat{i}

                         =2k\frac{Qq_1}{r^2} cos\theta \hat i

                         =2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i

                         =8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i     [ r=5]

                         =7.2\times10^{-3}\hat i   N

The magnitude of electric force is  7.2\times10^{-3} N

                         

3 0
3 years ago
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