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3 years ago

The total electric flux from a cubical box 26.0 cm on a side is 1840 N m2/C. What charge is enclosed by the box?

Physics

1 answer:

vitfil [10]3 years ago

7 0

The expression of the electric flux is

$\Phi = \frac{Q}{\epsilon_0}$

Here,

Q = Total charge enclosed in the closed surface

$\epsilon_0$ = Permittivity due to free space

Rearranging to find the charge,

$Q = \epsilon_0 \Phi$

Replacing with our values we have finally

$Q = (8.85*10^{-12}F\cdot m^{-1})(1.84*10^3 N\cdot m^2/C)$

$Q = 1.6284*10^{-8} C$ $(\frac{10^9nC}{1C})$

$Q = 0.1684nC$

The charge enclosed by the box is 0.1684nC

The sign of the charge can be decided by using the direction of the flux. The charge enclosed by the cube can be calculated by using the electric flux and the permitivity of free space.

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A car's bumper is designed to withstand a 5.04 km/h (1.4-m/s) collision with an immovable object without damage to the body of t

emmasim [6.3K]

Answer:

the magnitude of the average force on the bumper is 3189.8 N

Explanation:

Given the data in the question;

In terms of force and displacement, work done is;

W = $F^>$ × $x^>$

W = $Fxcos\theta$ ------- let this be equation 1

where F is force applied, x is displacement and θ is angle between force and displacement.

Now, since the displacement of the bumper and force acting on it is in the same direction,

hence, θ = 0°

we substitute into equation 1

W = $Fxcos($ 0° $)$

W = $Fx$ ------- let this be equation 2

Now, using work energy theorem,

total work done on the system is equal to the change in kinetic energy of the system.

$W_{net$ = ΔKE

= $\frac{1}{2}$ mv² - $\frac{1}{2}$ mu² --------- let this be equation 3

where m is mass of object, v is final velocity, u is initial velocity.

from equation 2 and 3

$Fx$ = $\frac{1}{2}$ mv² - $\frac{1}{2}$ mu²

we make F, the subject of formula

F = $\frac{m}{2x}$ ( v² - u² )

given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s

so we substitute

F = $\frac{830}{(2)(0.255)}$ ( (0)² - (1.4)² )

F = 1627.45098 ( 0 - 1.96 )

F = 1627.45098 ( - 1.96 )

F = -3189.8 N

The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.

Therefore, the magnitude of the average force on the bumper is 3189.8 N

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2 years ago

You want to estimate the height of the Empire State Building. You start at its base and walk 15 m away. Then you approximate the

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Answer:

429 m

Explanation:

In this exercise you are suppose to simulate a right triangle, the catheti are the distance you walked and the height of the empire state building.

tan(88) = (height) /(distance you walked)

28.64 = height / 15

height = 429 m

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3 years ago

A)what are oxidation numbers? b) what useful functions do oxidation numbers serve

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<span>A) indicate the general distribution of electrons among bonded atoms in molecular compounds
B)</span>useful in naming compounds, writing formulas, and balancing chemical equations
<span>may be used to determine the simplest chemical formula </span>

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3 years ago

A 5.30kg block hangs from a spring with a spring constant 1700 N/m. The block is pulled down 4.50cm from the equilibrium positio

Andru [333]

To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)

PART A) By definition the frequency in a spring is given by the equation

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

Where,

m = mass

k = spring constant

Our values are,

k=1700N/m

m=5.3 kg

Replacing,

$f = \frac{1}{2\pi} \sqrt{\frac{1700}{5.3}}$

$f=2.85 Hz$

PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.

$\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2} kY^2$