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USPshnik [31]
3 years ago
7

Pls help I will mark brainliest

Physics
2 answers:
Alekssandra [29.7K]3 years ago
6 0
Last one, remember alike repel and opposite attract
DiKsa [7]3 years ago
5 0

Remember that two of the same charge repel each other.

These two balls are repelling each other.  So they must both have the <em>SAME charge</em>. (last choice on the list)

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The charge per unit length on a long, straight filament is -92.0 μC/m. Find the electric field 10.0 cm above the filament.
Pepsi [2]

Answer:

E = 1.655 x 10⁷ N/C towards the filament

Explanation:

Electric field due to a line charge is given by the expression

E = [tex]\frac{\lambda}{2\pi\times\epsilon_0\times r}[/tex]

where λ is linear charge density of line charge , r is distance of given point from line charge and ε₀ is a constant called permittivity and whose value is

8.85 x 10⁻¹².

Putting the given values in the equation given above

E = \frac{92\times10^{-6}}{2\times3.14\times8.85\times10^{-12}\times10^{-1}}

E = 1.655 x 10⁷ N/C

4 0
3 years ago
A 80-kg person sits on a 2.7kg chair. Each leg of the chair makes contact with the floor on a circle that is 1.3 cm in diameter.
Rashid [163]

Answer:

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Explanation:

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6 0
2 years ago
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A 50 kg mass is placed 2 meters from the fulcrum. To balance the lever, a second object is placed 4 meters
julia-pushkina [17]

Answer:

mass =25 kg

using clockwise moment = anticlockwise moment

8 0
3 years ago
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
2 years ago
A 240 kg motorcycle moves with a velocity of 8 m/s. What is its kinetic energy?
QveST [7]
1/2 x 240 x 64 = 120 X 64 = 7680 J
6 0
3 years ago
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