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11 days ago

0.A 20-g bullet moving at 1 000 m/s is fired through a one-kg block of wood emerging at a speed of 100 m/s. What is the kinetic

energy of the block that results from the collision if the block had not been moving prior to the collision and was free to move?

Physics

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As a rough approximation, the human body may be considered to be a cylinder of length L=2.0m and circumference C=0.8m. (To simpl

inna [3103]

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.

4 0

2 months ago

Wire A has the same length and twice the radius of wire B. Both wires are made of the same material and carry the same current.

serg [3582]

Answer:

$V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}$

$V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}$

Consequently, we find that:

$V_A = \frac{1}{4} V_B$

Thus, the most suitable answer would be:

a. vA = vB/4

Explanation:

In this situation, we can establish the following conditions:

$L_A = L_B =L$ both wires share the same length

both wires carry an identical current $I_A = I_B =I$

Both wires are constructed of the same material, indicating that the electron density (n) remains constant across both wires

$n_A = n_B =n$

We also know that $r_A = 2 r_B$ where r signifies the radius.

Given that wires are cylindrical in shape, we can determine the area for each case:

$A_A= \pi r^2_A = \pi (2r_B)^2 = 4 \pi r^2_B= 4 A_B$

$A_B = \pi r^2_B$

Thus, we conclude that

$A_A = 4 A_B$

Now we are aware that the drift velocity of an electron in a wire can be described by:

$v_d = \frac{I}{neA}$

Where I denotes the current, n is the electron density, e represents the electron charge, and A signifies the area.

By substituting, we arrive at:

$V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}$

$V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}$

So we observe that:

$V_A = \frac{1}{4} V_B$

Thus, the most fitting answer is:

a. vA = vB/4

6 0

1 month ago

A standard baseball has a circumference of approximately 23 cm . if a baseball had the same mass per unit volume as a neutron or

Softa [3030]

Answer:

Baseball mass: $m_b=3.992*10^{14}kg$

Explanation:

Circumference of a baseball is calculated using 2πr = 23 cm

Thus, the radius comes out to be 3.66 cm, which equals $3.66*10^{-2}$ m

The mass density of the baseball matches that of a neutron or proton.

Proton mass = $10^{-27}$ kg

Proton diameter = $10^{-15}$ m

Proton radius = $5*10^{-16}$ m

Volume of the baseball is $\frac{4}{3} \pi r^3$

Now by substituting all values into the mass per unit volume equation for the baseball, we get:

$\frac{m_b}{\frac{4}{3}\pi *(3.66*10^{-2})^3} =\frac{10^{-27}}{\frac{4}{3}\pi *(5*10^{-16})^3}$

$\frac{m_b}{(3.66*10^{-2})^3} =\frac{10^{-27}}{(5*10^{-16})^3}$

$m_b=3.992*10^{14}kg$

Therefore, the baseball mass amounts to $m_b=3.992*10^{14}kg$

5 0

2 months ago

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A constant torque of 260 Nm acts on a flywheel. If the flywheel makes 25 complete revolutions in 2 minutes, what is the power ex

ValentinkaMS [3465]

I don't know that; sorry, I should just be removed from here.

7 0

2 months ago

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