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4 years ago

Electromagnets in radios can be used to A) play a CD. B) create sound. C) change the station. D) make a remote control.

Physics

2 answers:

SSSSS [86.1K]4 years ago

5 0

Change the station....

Vesna [10]4 years ago

5 0

The answer is change the station

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A 7.8 µF capacitor is charged by a 9.00 V battery through a resistance R. The capacitor reaches a potential difference of 4.20 V

Vaselesa [24]

Answer:

655128 ohm

Explanation:

C = Capacitance of the capacitor = 7.8 x 10⁻⁶ F

V₀ = Voltage of the battery = 9 Volts

V = Potential difference across the battery after time "t" = 4.20 Volts

t = time interval = 3.21 sec

T = Time constant

R = resistance

Potential difference across the battery after time "t" is given as

$V = V_{o} (1-e^{\frac{-t}{T}})$

$4.20 = 9 (1-e^{\frac{-3.21}{T}})$

T = 5.11 sec

Time constant is given as

T = RC

5.11 = (7.8 x 10⁻⁶) R

R = 655128 ohm

3 0

3 years ago

A laser beam of unknown wavelength passes through adiffraction grating having 5510 lines/cm after striking itperpendicularly. Ta

Ivahew [28]

Answer:

Explanation:

distance between two slit d = $\frac{1 \times 10^{-2}}{5510}$

d = 18.15 x 10⁻⁷ m

Let wave length of light λ

formula for position of first pair of bright spot

Tanθ = λ / d , λ is wave length of light and d is distance between two slit .

Tan 15.4 = $\frac{\lambda}{18.15\times10^{-7}}$

λ = Tan 15.4 x 18.15 x 10⁻⁷

=5 x 10⁻⁷ m

If θ be the position of next bright spot

Tanθ = 2 λ / d

= $\frac{2 \times\ 5\times\ 10^ {-7}}{18.15\times10^{-7}}$

= $\frac{2\times5}{18.5}$

θ = 28.4 degree .

7 0

4 years ago

Newton's law of universal gravitation states that

ladessa [460]

Newtons law of the inversion gravitation state that every object in the universe attracts ever other object.

3 0

3 years ago

Read 2 more answers

How much heat must be removed from 456 g of water at 25.0°C to change it into ice at - 10.0°C ? The specific heat of ice is 2090

cupoosta [38]

Answer:209.98 kJ

Explanation:

mass of water $m=456 gm$

Initial Temperature of Water $T_i=25^{\circ}C$

Final Temperature of water $T_f=-10^{\circ}C$

specific heat of ice $c=2090 J/kg-K$

Latent heat $L=33.5\times 10^4 J/kg$

specific heat of water $c_{water}=4.184 KJ/kg-K$

Heat require to convert water at $T=25^{\circ}C$ to $T=0^{\circ}C$

$Q_1=0.456\times 4.184\times (25-0)=47.69 kJ$

Heat require to convert water at $T=0^{\circ}$ to ice at $T=0^{\circ}$

$Q_2=m\times L=0.456\times 33.5\times 10^4=152.76 kJ$

heat require to convert ice at $T=0^{\circ} C\ to\ T=-10^{\circ} C$

$Q_3=0.456\times 2090\times (0-(-10))=9.53 kJ$

Total heat $Q=Q_1+Q_2+Q_3$

$Q=47.69+152.76+9.53=209.98 kJ$

7 0

3 years ago

A tank having a volume of 0.85 m 3 initially contains water as a two-phase liquid-vapor mixture at 260 o C and a quality of 0.7.

Keith_Richards [23]

Answer:=14,160 kJ

Explanation: Let m1 and m2 be the initial and final amounts of mass within the tank, respectively. The steam properties are listed in the table below

Specific Internal SpecificTemp Pressure Volume Energy Enthalpy Quality Phase

C MPa m^3/kg kJ/kg kJ/kg

1 260 4.689 0.02993 2158 2298 0.7 Liquid Vapor Mixture

2 260 4.689 0.0422 2599 2797 1 Saturated Vapor

The mass initially contained in the tank is m1 = V/v1

m1 =0.85 m^3 /0.02993 m^3 /kg

= 28.4 kg

The mass finally contained in the tank is

m2 =V2/v

= 0.85 m^3 /0.0422 m^3 /kg

= 20.14 kg

The heat transfer is then

Qcv = m2u2 − m1u1 − he(m2 − m1)

Qcv = (20.14)(2599) − (28.4)(2158) − (2797)(20.14 − 28.4) = 14,160 kJ

4 0

4 years ago