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bulgar [2K]
3 years ago
11

A big kid pushes a little kid and the little kid falls backward. The big kid remains relatively stationary. What best explains t

his
Physics
2 answers:
postnew [5]3 years ago
7 0
The big kid is the strongest ?
charle [14.2K]3 years ago
7 0

Answer:

Becase he pushed the little kid not himself

Explanation:

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Another circuit question. So far I've incorrectly done this one 3 out of 5 times lol
marta [7]
Hope this helps you.

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4 years ago
Why does a vibrating simple Pendulum not produce any sound????
Yuri [45]
Human ear can hear only the sounds having frequency from 20 Hz-20,000Hz but the simple pendulum produces a sound having frequency much less than 20Hz so we cannot hear it.
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4 years ago
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A brick is dropped from rest from a height of 4.9 m. How long does it take for the brick to
noname [10]

Answer:

t=1s

Explanation:

Using the formula

s=ut+1/2at^2

4.9 = (0)(t) + 1/2(9.8)(t^2)

4.9 = 4.9(t^2)

t^2 = 1

t = 1

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3 years ago
Why do most carnivores live in forest? given reason ​
kotegsom [21]

\huge\purple{Hi!}

most carnivore live in forest because

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  3. Carnivores help in maintaining the herbivore population at a healthy level, by hunting.
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3 0
2 years ago
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In a rectangular coordinate system, a positive point charge q = 6.50 nC is placed at the point x=0.190 m ,y=0, and an identical
son4ous [18]

Answer:

A) x and y components of the electric field  (Ep) at the origin.

Epx = -1620.5 N/C

Epy = -1620.5 N/C

Ep = (1620.5(-i)+1620.5(-j)) N/C

B) Magnitude of the electric field  (Ep) at the origin.

   Ep= 2291.7 N/C

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Equivalence  

1nC= 10⁻⁹C

Data

K= 9x10⁹N*m²/C²

q₁ = q₂= +6.5nC=+6.5 *10⁻⁹C

d₁=d₂=0.190m

Graphic attached

The attached graph shows the field due to the charges:

Ep₁ : Electric Field at point P  (x=0, y=0) due to charge q₁. As the charge q₁ is positive (q₁+) ,the field leaves the charge.

Ep₂: Electric Field at point  P (x=0, y=0) due to charge q₂. As the charge q₂  is positive (q₂+) ,the field leaves the charge

Ep: Total field at point P due to charges q₁ and q₂.

Because q₁ = q₂ and d₁ = d₂, then, the magnitude of Ep₁ is equal to the magnitude of Ep₂

Ep₁ = Ep₂ = k*q/d² = 9*10⁹*6.5*10⁻⁹/0.190m² = 1620.5 N/C

Look at the attached graphic :

Epx = Ep₁= -1620.5 N/C

Epy = Ep₂= -1620.5 N/C

A) x and y components of the electric field  (Ep) at the origin.

Ep = (1620.5(-i)+1620.5(-j)) N/C

B) Magnitude of the electric field  (Ep) at the origin.

E_{p} =\sqrt{1620.5^{2}+1620.5^{2} } = 2291.7 \frac{N}{C}

8 0
3 years ago
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