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3 years ago

Graph are pictorical representations of

2 answers:

8 0

They are pictorial representation of a chart or a list made into a graph to make things easier for the person answering an equation

inn [45]3 years ago

6 0

Answer:

Graphs are pictorial representations of relationships.

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The electric flux through a square-shaped area of side 5 cm near a very large, thin, uniformly-charged sheet is found to be 3\ti

deff fn [24]

Answer:

Explanation:

Given

side of square shape $a=5\ cm$

Electric flux $\phi =3\times 10^{-5}\ N.m^2/C$

Permittivity of free space $\epsilon_0=8.85\times 10^{-12} \frac{C^2}{N.m^2}$

Flux is given by

$\phi =EA\cos \theta$

where E=electric field strength

A=area

$\theta$ =Angle between Electric field and area vector

$E=\frac{\phi }{A\cos (0)}$

$E=\frac{3\times 10^{-5}}{25\times 10^{-4}\times \cos(0)}$

$E=0.012\ N/C$

and Electric field by a uniformly charged sheet is given by

$E=\frac{\sigma }{2\epsilon_0}$

where $\sigma$ =charge density

$=\frac{\sigma }{\epsilon_0}$

$\sigma =0.012\times 8.85\times 10^{-12}$

$\sigma =2.12\times 10^{-13}\ C/m^2$

5 0

3 years ago

The Pacific Giant Kelp grows at a rate of 18 in/day. How many centimeters per hour is this?

Airida [17]

For this case we must do a conversion. By definition we have to:

1 inch equals 2.54 centimeters

1 day equals 24 hours.

So, we have: $18 \frac {in} {day} * \frac {1} {24} \frac {day} {h} * \frac {2.54} {1} \frac {cm} {in} =$

Canceling units we have:

$\frac {18 * 1 * 2.54} {24 * 1} \frac {cm} {h} =\\\frac {45.72} {24}\frac {cm} {h} =$

$1,905 \frac {cm} {h}$

Answer:

$1,905 \frac {cm} {h}$

5 0

3 years ago

Neon signs require about 12,000 V for their operation. Consider a neon-sign transformer that operates off 120- V lines. How many

Ugo [173]

I can't give you the actual number of turns, because it's the RATIO
that counts.

However many turns the primary has, the secondary should have
about TEN TIMES that number. Then the transformer will multiply
the primary voltage by 10 ... 120 volts of AC at the primary will
become 1,200 volts of AC at the secondary.

Note that it HAS TO be AC. If the transformer is supplied with DC,
then 120 volts at the primary becomes zero volts at the secondary
and a big cloud of stinky smoke in the room.

8 0

3 years ago

A small sphere of

mixas84 [53]

Answer:

θ = 39.7º

Explanation:

In this exercise we must use Newton's second law for the sphere, at the equilibrium point we write the equations in each exercise; we will assume that plate 1 is on the left

Y Axis

$T_{y}$ -W = 0

$T_{y}$ = W

X axis

- $F_{e1}$ <u> - F_{e2} + Tₓ = 0 </u>

let's use trigonometry to find the components of the tension, we measure the angle with respect to the vertical

sin θ = Tₓ / T

cos θ = T_{y} / t

Tₓ = T sin θ

T_{y} = T cos θ

let's use gauss's law to find the electric field of each leaf; We define a Gaussian surface formed by a cylinder, so the component of the field perpendicular to the base of the cylinder is the one with electric flow.

F = ∫ E. dA = $q_{int}$ / ε₀

in this case the scalar product is reduced to the algebraic product, the flow is towards both sides of the plate

F = 2E A = q_{int} / ε₀

let's use the concept of surface charge density

σ = q_{int} / A

we substitute

2E A = σ A /ε₀

E = σ / 2ε₀

this is the field created by each plate. The electric force is

$F_{e}$ = q E

for plate 1 with σ₁ = -30 10⁻⁶ C / m²

F_{e1} = q σ₁ /2ε₀

for plate 2 with s2 = ab 10⁻⁶ C / m², for the calculations a value of this charge density is needed, suppose s2 = 10 10⁻⁶ C / m²

F_{e2} = q σ₂ /2ε₀

we substitute and write the system of equations

T cos θ = mg

- q σ₁ / 2ε₀ - q σ₂ /2ε₀ + T sinθ = 0

we introduce t in the second equations

- q /2 ε₀ (σ₁ + σ₂) + (mg / cos θ) sin θ = 0

mg tan θ = q /2ε₀ (σ₁ + σ₂)

θ = tan -1 (q / 2ε₀ mg (σ₁ + σ₂)

data indicates the mass of 0.25 g = 0.25 10⁻³ kg

give the charge density on plate 2, suppose ab = 10 10⁻⁶ C / m²

let's calculate

θ = tan⁻¹ (9.0 10⁻¹⁰ (30 + 10) 10⁻⁶ / (2 8.85 10⁻¹² 0.25 10⁻³ 9.8))

θ = tan⁻¹ 8.3 10⁻¹)

θ = 39.7º

5 0

3 years ago

John and Tau take their toys down to a nearby pond to play. John has a toy boat made from wood. He places the wooden boat on the

Charra [1.4K]

Answer:

The smallest amount of lead that needs to be attached to John's boat in order to sink it has to have a mass slightly greater than the mass of the boat.

Explanation:

A body floats in a fluid when its density is less than the density of the fluid.

The body sinks when its density is more than the density of the fluid.

Density of the pond = density of pure water = 1 g/cm³

Specific gravity of an object = (density of object)/(density of water)

0.5 = (density of John's boat)/1

Density of the John's boat = 0.5 g/cm³

And density is given as

Density = (mass/volume)

0.5 = (mass of John's boat)/(volume of John's boat)

Let the volume of John's boat be v and the mass of John's boat be m

0.5 = (m/v)

v = (m/0.5) = 2m

To sink the boat, we need the total mass on the boat to increase the density to a value greater than 1.

Let the minimum mass of lead required for this to be M

The volume of the boat remains the same, but the total mass on the boat is now (m+M)

1 < (m + M)/v

M + m > v

Recall, v = 2m

M + m > 2m

M > 2m - m

M > m

Hence, the smallest amount of lead that needs to be attached to John's boat in order to sink it has to have a mass slightly greater than the mass of the boat.

Hope this Helps!!!!

4 0

3 years ago