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2 years ago

3

A soap bubble of radius R is situated in a uniform electric field of magnitude E. The electric flux through the surface of the s

oap bubble is

1 answer:

uranmaximum [27]2 years ago

8 0

Answer:

<em>The electric flux around the soap bubble will be </em>

<em>Ф = q/ε</em>

<em></em>

Explanation:

The radius of the soap bubble = R

The electric field around the soap bubble = E

The electric flux = ?

The soap can be approximated to be a sphere, so we find the surface area of the sphere

For the soap bubble, the surface area will be

A = $4\pi R^{2}$

Recall that electric flux is given as

Ф = EA

substituting value of A from above, we'll have

Ф = E $4\pi R^{2}$ ..... equ 1

Also recall that the electric field E is given as

E = q/(4πε $R^{2}$ )

substitute the value of E into equ 1, to get

Ф = q/(4πε $R^{2}$ ) * $4\pi R^{2}$

The electric flux around the soap bubble reduces to

<em>Ф = q/ε</em>

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If you find an igneous rock which has 450 radioactive isotopes and 3,150 stable daughter isotopes, how many half-lifes of this i

slavikrds [6]

Answer:

$3t_{1/2}$

Explanation:

To find the half-lifes of the isotope we need to use the following equation:

$N_{t} = N_{0}2^{-\frac{t}{t_{1/2}}}$ (1)

<em>where Nt: is the amount of the isotope that has not yet decayed after a time t, N₀: is the initial amount of the isotope, t: is the time and </em> $t_{1/2}$ <em>: is the half-lifes.</em>

By solving equation (1) for t we have:

$\frac{t}{t_{1/2}} = - \frac{Ln(Nt/N_{0})}{Ln(2)}$

<u>Having that:</u>

Nt = 450

N₀ = 3150 + 450 = 3600,

The half-lifes of the isotope is:

$t = - \frac{Ln(450/3600)}{Ln(2)} \cdot t_{1/2} = 3t_{1/2}$

Therefore, 3 half-lives of the isotope passed since the rock was formed.

I hope it helps you!

3 0

2 years ago

A 2200 kg truck has put its front bumper against the rear bumper of a 2400 kg SUV to give it a push. With the engine at full pow

leonid [27]

Answer:

a) The maximum possible acceleration the truck can give the SUV is 7.5 meters per second squared

b) The force of the SUV's bumper on the truck's bumper is 18000 newtons

Explanation:

a) By Newton's second law we can find the relation between force and acceleration of the SUV:

$F=ma$

With F the maximum force the truck applies to the SUV, m the mass of the SUV and a the acceleration of the SUV; solving for a:

$a=\frac{F}{m}=\frac{18000}{2400}\approx7.5\,\frac{m}{s^{2}}$

b) Because at this acceleration the truck's bumper makes a force of 18000 N on the SUV’s bumper by Third Newton’s law the force of the SUV’s bumper on the truck’s bumper is 18000 N too because they are action-reaction force pairs.

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2 years ago

2.27 A gas is compressed from V1= 0.3 m3, p1=1 bar to V2= 0.1 m3, p2 =3 bar. The pressure and

Georgia [21]

Answer:

-40 kJ

80 kJ

Explanation:

Work is equal to the area under the pressure vs volume graph.

W = ∫ᵥ₁ᵛ² P dV

2.27) Pressure and volume are linearly related. When we graph P vs V, the area under the line is a trapezoid. So the work is:

W = ½ (P₁ + P₂) (V₂ − V₁)

W = ½ (100 kPa + 300 kPa) (0.1 m³ − 0.3 m³)

W = -40 kJ

2.29) Pressure and volume are inversely proportional:

pV = k

The initial pressure and volume are 500 kPa and 0.1 m³. So the constant is:

(500) (0.1) = k

k = 50

The final pressure is 100 kPa. So the final volume is:

(100) V = 50

V = 0.5

The work is therefore:

W = ∫ᵥ₁ᵛ² P dV

W = ∫₀₁⁰⁵ (50/V) dV

W = 50 ln(V) |₀₁⁰⁵

W = 50 (ln 0.5 − ln 0.1)

W ≈ 80 kJ

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2 years ago

4. Dr. Copus is in charge of the cognition department at the University of Wisconsin-Madison. A new drug named Mem-Reen has beco

MrMuchimi

Answer:

See the answer below

Explanation:

<u>Independent variable</u>: Type of drug (Mem-Reen or placebo)

<u>Dependent variable</u>: memories

<u>Experimental group</u>: The group that was given Mem-Reen

<u>Control group</u>: The group that was given placebo

<u>Constants</u>: Food, hours of sleep, memory test procedures.

The independent variable is an input variable that produces effects on the dependent variable. As the variable is changed, it produces different effects on the dependent variable.

The dependent variable is the actual variable that is measured during an experiment. It is the main purpose of setting-up of an experiment.

The experimental group is also referred to as the treatment group while the control group is the group that does not receive treatment at all or they receive fake treatment/placebo.

Constants are unchanging variables included in experiments. They remain unchanged both in the treatment and the control group, otherwise, the outcome of the experiment will be unreliable.

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2 years ago

A stationary particle of charge q = 2.1 × 10-8 c is placed in a laser beam (an electromagnetic wave) whose intensity is 2.9 × 10

alisha [4.7K]

(a) The intensity of the electromagnetic wave is related to the amplitude of the electric field by
$I= \frac{1}{2} c \epsilon_0 E^2$
where
I is the intensity
c is the speed of light
$\epsilon_0$ is the electric permittivity
E is the amplitude of the electric field

By substituting the numbers of the problem and re-arranging the equation, we can find E:
$E= \frac{2 I}{c \epsilon_0} = \frac{2 ( 2.9 \cdot 10^3 Wm^{-2})}{(3 \cdot 10^8 m/s)(8.85 \cdot 10^{-12} Fm^{-1})} =2.2 \cdot 10^6 N/C$

Now that we have the intensity of the electric field, we can calculate the electric force on the charge:
$F=qE=(2.1 \cdot 10^{-8} C)(2.2 \cdot 10^6 N/C)=0.046 N$

(b) We can calculate the amplitude of the magnetic field starting from the amplitude of the electric field:
$B= \frac{E}{c}= \frac{2.2 \cdot 10^6 N/C}{3 \cdot 10^8 m/s}=7.3 \cdot 10^{-3} T$

The magnetic force is given by
$F=qvB \sin \theta$
where v is the particle's speed, B the magnetic field intensity and $\theta$ the angle between B and v.
In this case the charge is stationary, so v=0, and so the magnetic force is zero: F=0.

(c) The electric force has not changed compared to point (a), because it does not depend on the speed of the particle, so we have again F=0.046 N.

(d) This time, the particle is moving with speed $v=3.7 \cdot 10^4 m/s$ , in a direction perpendicular to the magnetic field (so, the angle $\theta$ is $90^{\circ}$ ), and so by using the intensity of the magnetic field we found in point (b), we can calculate the magnetic force on the particle:
$F=qvB \sin \theta = (2.1 \cdot 10^{-8}C)(3.7 \cdot 10^4 m/s)(7.3 \cdot 10^{-3} T)(\sin 90^{\circ} )=$
$=5.7 \cdot 10^{-6} N$

5 0

2 years ago