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4 years ago

Can someone please tell me the answer,I can't find the answer XD

Physics

1 answer:

Butoxors [25]4 years ago

7 0

Acid rain is rainfall made sufficiently acidic by atmospheric pollution that it causes environmental harm, typically to forests and lakes. The main cause is the industrial burning of coal and other fossil fuels, the waste gases from which contain sulfur and nitrogen oxides, which combine with atmospheric water to form acids.

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In a classical carnival ride, patrons stand against the wall in a cylindrically shaped room. Once the room gets spinning fast en

g100num [7]

Answer:

- the speed of a person "stuck" to the wall is 14.8 m/s

- the normal force of the wall on a rider of m=54kg is 1851 N

- the minimum coefficient of friction needed between the wall and the person is 0.29

Explanation:

Given information:

the radius of the cylindrical room, R = 6.4 m

the room spin with frequency, ω = 22.1 rev/minutes = 22.1 $\frac{2\pi }{60}$ = 2.31 rad/s

mass of rider, m = 54 kg

the speed of a person "stuck" to the wall

v = ω R

= 2.31 x 6.4

= 14.8 m/s

the normal force of the wall on a rider

F = m a

a = ω^2 R

= $\frac{v^{2} }{R^{2} }$ R

= $\frac{v^{2} }{R}$

F = $\frac{mv^{2} }{R}$

= $\frac{(54)(14.8)^{2} }{6.4}$

= 1851 N

the minimum coefficient of friction needed between the wall and the person

F(friction) = μ N

W = μ N

m g = μ $\frac{mv^{2} }{R}$

g = μ $\frac{v^{2} }{R}$

μ = $\frac{gR}{v^{2} }$

= $\frac{(9.8) (6.4)}{14.8^{2} }$

= 0.29

5 0

3 years ago

A dielectric-filled parallel-plate capacitor has plate area A = 30.0 cm2 , plate separation d = 9.00 mm and dielectric constant

PIT_PIT [208]

Answer:

$9.96\cdot 10^{-10}J$

Explanation:

The capacitance of the parallel-plate capacitor is given by

$C=\epsilon_0 k \frac{A}{d}$

where

ϵ0 = 8.85x10-12 C2/N.m2 is the vacuum permittivity

k = 3.00 is the dielectric constant

$A=30.0 cm^2 = 30.0\cdot 10^{-4}m^2$ is the area of the plates

d = 9.00 mm = 0.009 m is the separation between the plates

Substituting,

$C=(8.85\cdot 10^{-12}F/m)(3.00 ) \frac{30.0\cdot 10^{-4} m^2}{0.009 m}=8.85\cdot 10^{-12} F$

Now we can calculate the energy of the capacitor, given by:

$U=\frac{1}{2}CV^2$

where

C is the capacitance

V = 15.0 V is the potential difference

Substituting,

$U=\frac{1}{2}(8.85\cdot 10^{-12}F)(15.0 V)^2=9.96\cdot 10^{-10}J$

4 0

3 years ago

(a) You short-circuit a 20 volt battery by connecting a short wire from one end of the battery to the other end. If the current

erica [24]

(a) $1.11 \Omega$

When the battery is short-cut, the only resistance in the circuit is the internal resistance of the battery. Therefore, we can apply Ohm's law:

$r=\frac{V}{I}$

where

V = 20 V is the voltage across the internal resistance of the battery

I = 18 A is the current flowing through it

Solving the equation,

$r=\frac{20 V}{18 A}=1.11\Omega$

(b) 360 W

The power generated by the battery is given by the equation

$P=VI$

where

V = 20 V is the voltage of the battery

I = 18 A is the current

Substituting into the formula,

$P=(20 V)(18 A)=360 W$

(c) 360 J

The energy dissipated by the internal resistance is given by

$E=Pt$

where

P = 360 W is the power generated

t = 1 s is the time

Solving the equation, we find

$E=(360 W)(1 s)=360 J$

(d) 1.65 A

The battery is now connected to a $R=11 \Omega$ resistor. This means that the internal resistance of the battery is now connected in series with the other resistor R: so, the total resistance of the circuit is

$R_T = r+R=1.11 \Omega +11 \Omega = 12.11 \Omega$

And so, the current flowing through the circuit is

$I=\frac{V}{R_T}=\frac{20 V}{12.11\Omega}=1.65 A$

(e) 29.9 W

The power dissipated in the external resistor is given by

$P=I^2 R$

where

I = 1.65 A is the current

$R=11 \Omega$ is the resistance

Solving the equation, we find

$P=(1.65 A)^2(11 \Omega)=29.9 W$

(f) 18.17 V

The terminals of the voltmeter are placed at the two end of the battery. The battery provides an emf of 20 V, however due to the internal resistance, some of this voltage is dropped across the internal resistance. Therefore, the actual potential difference that will be read by the voltmeter will be:

$V=\epsilon - Ir =20 V -(1.65 A)(1.11 \Omega)=18.17 V$