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Alinara [238K]
2 years ago
14

I'm feeling nice so.... F.r.e.e POINTS (pls follow me) ​

Physics
2 answers:
NISA [10]2 years ago
6 0

Answer:

ty :,)

Explanation:

ziro4ka [17]2 years ago
3 0

Answer: ty

Explanation:

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Initially, a particle is moving at 5.25 m/s at an angle of 35.5° above the horizontal. Three seconds later, its velocity is 6.0
ivolga24 [154]

Answer:

 a =( -0.32 i ^ - 2,697 j ^)  m/s²

Explanation:

This problem is an exercise of movement in two dimensions, the best way to solve it is to decompose the terms and work each axis independently.

Break down the speeds in two moments

initial

  v₀ₓ = v₀ cos θ

  v₀ₓ = 5.25 cos 35.5

v₀ₓ = 4.27 m / s

   v_{oy} = v₀ sin θ

 v_{oy}= 5.25 sin35.5

v_{oy} = 3.05 m / s

Final

vₓ = 6.03 cos (-56.7)

vₓ = 3.31 m / s

v_{y} = v₀ sin θ

v_{y} = 6.03 sin (-56.7)

v_{y} = -5.04 m / s

Having the speeds and the time, we can use the definition of average acceleration that is the change of speed in the time order

    a = (v_{f} - v₀) /t

    aₓ = (3.31 -4.27)/3

    aₓ = -0.32 m/s²

    a_{y} = (-5.04-3.05)/3

   a_{y} =  -2.697 m/s²

6 0
3 years ago
Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m
aev [14]

Answer:

Explanation:

Given

charge of first body q_1=2.5\ mu C

charge of second body q_2=-3.5\ mu C

Particle 1 is at origin and particle 2 is at x=0.6\ m

third Particle which charge +q must be placed left of 2.5\mu C because it will repel the q charge while -3.5\mu C will attract it

suppose it is placed at a distance of x m

F_{1q}=\frac{kq(2.5)}{x^2}

F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}

F_{1q}+F_{2q}=0

\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0

\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}

\frac{0.6+x}{x}=(\frac{3.5}{2.5})^{0.5}

0.6+x=1.1832x

x=3.27\ m

5 0
3 years ago
Learning Goal:
enot [183]

Answer:

A. U_0 = \dfrac{\epsilon_0 A V^2}{2d}

B. U_1 = \dfrac{\epsilon_0 A V^2}{6d}

C. U_2 = \dfrac{K\epsilon_0 A V^2}{2d}

Explanation:

The capacitance of a capacitor is its ability to store charges. For parallel-plate capacitors, this ability depends the material between the plates, the common plate area and the plate separation. The relationship is

C=\dfrac{\epsilon A}{d}

C is the capacitance, A is the common plate area, d is the plate separation and \epsilon is the permittivity of the material between the plates.

For air or free space, \epsilon is \epsilon_0 called the permittivity of free space. In general, \epsilon=\epsilon_r \epsilon_0 where \epsilon_r is the relative permittivity or dielectric constant of the material between the plates. It is a factor that determines the strength of the material compared to air. In fact, for air or vacuum, \epsilon_r=1.

The energy stored in a capacitor is the average of the product of its charge and voltage.

U = \dfrac{QV}{2}

Its charge, Q, is related to its capacitance by Q=CV (this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for U,

U = \dfrac{CV^2}{2}

A. Substituting for C in U,

U_0 = \dfrac{\epsilon_0 A V^2}{2d}

B. When the distance is 3d,

U_1 = \dfrac{\epsilon_0 A V^2}{2\times3d}

U_1 = \dfrac{\epsilon_0 A V^2}{6d}

C. When the distance is restored but with a dielectric material of dielectric constant, K, inserted, we have

U_2 = \dfrac{K\epsilon_0 A V^2}{2d}

6 0
3 years ago
1. A woman driving in a car at 21 m/s pushes down on the gas pedal. 6 seconds later, she is moving at 36 m/s. What was her avera
Furkat [3]

a=Vf-Vi/t=36-21/6=15/6=2.5m/s^2

5 0
3 years ago
Review:
Tresset [83]

Answer:

1. 1. A quantity is completely described by magnitude alone. A quantity Is completely described by a magnitude with a direction.

[a]. scalar, vector

b. vector, scalar

2.2. Speed is a velocity is a quantity and quantity.

a. scalar, vector

[b]. vector, scalar

3 0
2 years ago
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