A 45.0-kg sample of ice is at 0.00°C. How much heat is needed to melt it? For water, Lf=334 kJ/kg and Lv=2257 kJ/kg 

Answer:

She will be located in a loud area.

Explanation:

To evaluate whether Susan is situated in a quiet or loud area, we must first ascertain the difference in the sound waves reaching her from the two speakers.

If we refer to d₁ as the distance to speaker A, which measures 19.5 m, then d₂ to the other speaker is calculated as follows:

d₂ = 42.0 m - 19.5 m = 22.5 m

This reveals that the path difference between both speakers is:

d = d₂-d₁ = 22.5 m -19.5 m = 3.0 m

Next, we relate this distance to the sound wave's wavelength; for constructive interference, the path difference must equal an even multiple of half the wavelength:

d = (2n)*(λ/2)

To determine λ, we know that for any wave, the relationship between speed, frequency, and wavelength holds:

v = λ*f

<pwhere v="345" m="" and="" f="115" we="" can="" calculate="">

λ = v/f = 345 m/s / 115 (1/s) = 3.0 m.

Since the difference between the distances to Susan from both speakers corresponds exactly to one wavelength, this indicates that both waves arrive at the same phase, resulting in constructive interference, thus labeling it as a loud area.

</pwhere>

solution:

the spring force applied by a spring with spring constant k can be expressed as

where k acts as the spring constant

and x indicates the spring's deformation

to determine the work completed by the spring

the amount of work done by the spring when moving from x=0 to x=L

substituting the limits x=0 and x=L

we derive the work done in terms of k and L

Answer:

A. a collection of mathematical topics that are pertinent to basic physics.

Explanation:

The physics primer is not the same as the comprehensive online mathematics textbooks. Instead, it comprises topics in mathematics that challenge students and are noteworthy.

Therefore, it can be understood as the framework for resolving physics-related problems. Thus, mathematical skills are integral within physics courses, serving as a preparatory tool for success.

In summary, it represents a compilation of mathematical subjects that are relevant to foundational physics.

Thermal Power is approximately 460W. According to the Stephan-Boltzmann Law Formula: P = єσT⁴A, where: P = radiation energy, σ = Stefan-Boltzmann Constant, T = absolute temperature in Kelvin, є = emissivity of the material, and A = the surface area. Given that σ = 5.67 x 10^(-8), ε = 0.6, and T = 30°C which converts to Kelvin as 303K, with the human body dimensions of 2m length and 0.8m circumference leading to an area of 1.6m², so thermal power equals 0.6 x 5.67 x 10^(-8) x 303⁴ x 1.6 = 458.8W. Rounding gives about 460W.

Answer:

3.4 x 10⁴ m/s

Explanation:

Analyze the circular path of the electron

B = magnetic field = 80 x 10⁻⁶ T

m = mass of an electron = 9.1 x 10⁻³¹ kg

v  = speed in the radial direction

r = radius of the circular trajectory = 2 mm = 0.002 m

q = charge of an electron = 1.6 x 10⁻¹⁹ C

For the electron’s circular movement

qBr = mv

(1.6 x 10⁻¹⁹) (80 x 10⁻⁶) (0.002) = (9.1 x 10⁻³¹) v

v = 2.8 x 10⁴ m/s

Now, consider the electron's movement in a straight line:

v' = speed in linear motion

x = distance traveled horizontally = 9 mm = 0.009 m

t = duration = = = 4.5 x 10⁻⁷ sec

Using the formula

x = v' t

0.009 = v' (4.5 x 10⁻⁷)

v' = 20000 m/s

v' = 2 x 10⁴ m/s

The resultant speed is given by

V = sqrt(v² + v'²)

V = sqrt((2.8 x 10⁴)² + (2 x 10⁴)²)

v = 3.4 x 10⁴ m/s