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schepotkina [342]
4 years ago
6

While calculating electric field should we include the sign of the charge in the formula?

Physics
1 answer:
lidiya [134]4 years ago
3 0
I think that the sign of the charge in the formula would be not needed. T<span>he sign is not important is if you want just the magnitude (the absolute value) of the electric field in a given point in space. Hope this answers the question. Have a nice day.</span>
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A conceptual model is based off of
Klio2033 [76]
<h2>Answer</h2>

Tha conceptual model depends upon the process flow.

<h2>Explanation</h2>

The conceptual model is actually used in the system to stimulate, understand or explore the subject for the better understanding of people. As it consists of different constituents as validation, implementation, accreditation etc. can be affected by the process flow. The process flow has actually consisted of the management, monitoring of objects for specific subject and scales under these monitoring or management are proceedings. As these scales of process flow effects, the objectives of the task impacts.

5 0
3 years ago
If you walk 3 meters north and then 4 meters west, what is the magnitude of your displacement?
ankoles [38]
S^2=3^2+4^2
S^2=9+16
S^2=25
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3 years ago
What are the uses of nuclear fission or fusion and their benefits
tamaranim1 [39]

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6 0
4 years ago
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To develop muscle tone, a woman lifts a 2.4 kg weight held in her hand. She uses her biceps muscle to flex the lower arm through
andrey2020 [161]

Answer:

(a) a=41.1 rad/s^2

(b) W=15.7 J

Explanation:

6 0
4 years ago
Se dispara un proyectil con una velocidad inicial de 50 m/s y un ángulo de 30°, por encima de la horizontal. Calcular: a) Posici
dmitriy555 [2]

Answer:

a) Posición y velocidad después de los 6s

i) Posición = -26.58m

ii) velocidad = -33.86m/s

b) Tiempo para alcanzar la altura máxima

= 2.55s

c) Alcance horizontal

= 220.7m

Explanation:

a) Posición y velocidad después de los 6s

i) Posición

y = (usinθ)t – 1/2 gt²

u = 50m/s

θ = 30°

g = 9.81m/s²

t = 6s

y = (50 × sin 30)6 - 1/2 × 9.81 × 6²

y = 150 - 176.58

y = -26.58m

ii) velocidad

v = u sinθ–gt

u = 50m/s

θ = 30°

g = 9.81m/s²

t = 6s

v = 50 × sin 30 - 9.81 × 6

v = 25 - 58.86

v = -33.86m/s

b) Tiempo para alcanzar la altura máxima

usinθ /g

u = 50m/s

θ = 30°

g = 9.81m/s²

= 50 × sin 30/ 9.81

= 25/9.81

= 2.5484199796s

≈ 2.55s

c) Alcance horizontal

R = u²sin2θ/g

u = 50m/s

θ = 30°

g = 9.81m/s²

R = 50² ×( sin 2 ×30°) /9.81

R = 220.69964419m

≈ 220.7m

7 0
3 years ago
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