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HACTEHA [7]
3 years ago
12

(7%) Problem 5: A thermos contains m1 = 0.73 kg of tea at T1 = 31° C. Ice (m2 = 0.095 kg, T2 = 0° C) is added to it. The heat ca

pacity of both water and tea is c = 4186 J/(kg⋅K), and the latent heat of fusion for water is Lf = 33.5 × 104 J/kg. [email protected]
Physics
1 answer:
klio [65]3 years ago
8 0
Could you explain a little more?
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A square plate of copper with 55.0 cm sides has no net charge and is placed in a region of uniform electric field of 82.0 kN/C d
Alex73 [517]

Answer

given,

Side of copper plate, L = 55 cm

Electric field, E = 82 kN/C

a) Charge density,σ = ?

  using expression of charge density

 σ = E x ε₀

ε₀ is Permittivity of free space = 8.85 x 10⁻¹² C²/Nm²

now,

 σ = 82 x 10³ x 8.85 x 10⁻¹²

 σ = 725.7 x 10⁻⁹ C/m²

 σ = 725.7 nC/m²

change density on the plates are 725.7 nC/m² and -725.7 nC/m²

b) Total change on each faces

   Q = σ  A

   Q = 725.7 x 10⁻⁹ x 0.55²

   Q = 219.52 nC

Hence, charges on the faces of the plate are 219.52 nC and -219.52 nC

7 0
3 years ago
A wind turbine works by slowing the air that passes its blades and converting much of the extracted kinetic energy to electric e
ddd [48]

Answer:

2649600 Joules

Explanation:

Efficiency = 40%

m = Mass of air = 92000 kg

v = Velocity of wind = 12 m/s

Kinetic energy is given by

K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 92000\times 12^2\\\Rightarrow K=6624000\ J

The kinetic energy of the wind is 6624000 Joules

The wind turbine extracts 40% of the kinetic energy of the wind

E=0.4\times K\\\Rightarrow E=0.4\times 6624000\\\Rightarrow E=2649600\ J

The energy extracted by the turbine every second is 2649600 Joules

8 0
3 years ago
Assuming that the limits of the visible spectrum are approximately 380 and 700 nm, find the angular range of the first-order vis
ch4aika [34]

Answer:

angular range is ( 0.681 rad , 0.35 rad )

Explanation:

given data

wavelength λ = 380 nm = 380 × 10^{-9} m

wavelength λ  = 700 nm =  700 × 10^{-9} m

to find out

angular range of the first-order

solution

we will apply here slit experiment equation that is

d sinθ = m λ    ...........1

here m is 1 for single slit and d is = \frac{1}{900*10^3 m}

so put here value in equation 1 for 380 nm

we get

d sinθ = m λ

\frac{1}{900*10^3} sinθ = 1 × 380 × 10^{-9}

θ = 0.35 rad

and for 700 nm

we get

d sinθ = m λ

\frac{1}{900*10^3} sinθ = 1 × 700 × 10^{-9}

θ = 0.681 rad

so angular range is ( 0.681 rad , 0.35 rad )

3 0
3 years ago
Some homes that use baseboard heating use copper tubing. hot water runs through and heats the copper tubing, which in turn heats
AlekseyPX
When you heat a certain substance with a difference of temperature \Delta T the heat (energy) you must give to it is
E(=Q) =mc\Delta T
where c is the specific heat of that substance (given in J/(g*Celsius))
In this case
E=645*0.3850*(28.22-13.20) =3729.8 (Joule)

Observation: the specific heat of a substance is given in J/(g*Celsius) or J/(g*Kelvin)  because on the temperature scale a difference of 1 degree Celsius = 1 degree Kelvin
7 0
3 years ago
Ian walks 2 km to his best friends house, then walks 0.5 km to the library. He then makes 2.5 km walk home. The entire walk took
allochka39001 [22]
1. 2+0.5+2.5= 3. 2km/hr average




2. 14-6=4seconds. 8m/s in 4s = 2m/s acceleration


3. 15m/s divided by 2.5 = 6m/s acceleration
5 0
3 years ago
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