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Jobisdone
2 months ago
7

A spaceship moves at a speed of 0.95 c away from the Earth. It shoots a star wars torpedo toward the Earthat a speed of 0.90 c r

elative to the ship. What is the velocity of the torpedo relative to the Earth? (The direction in which the spaceship move
Physics
1 answer:
Sav [3.1K]2 months ago
5 0

Answer:

0.345 c

Explanation:

v_{se} = speed of spaceship with respect to earth = 0.95 c

v_{ts} = speed of torpedo with respect to spaceship = - 0.90 c

v_{te} = speed of torpedo with respect to earth

The velocity of the torpedo in relation to earth is calculated as

v_{te}=\frac{v_{se} + v_{ts}}{1 + \frac{v_{se}v_{ts}}{c^{2}}}

v_{te}=\frac{0.95c + (- 0.90 c)}{1 + \frac{(0.95 c)(- 0.90 c))}{c^{2}}}

v_{te} = 0.345 c

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If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2
Maru [3345]

Answer:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

Consequently, the temperature difference across the material will be \Delta T = 375 K

Explanation:

In this case, we apply the Fourier Law of heat conduction expressed by the following equation:

Q = -kA \frac{\Delta T}{\Delta x}   (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 denotes the cross-sectional area

Q= 3KW signifies the heat transfer rate

\Delta T is the temperature difference we need to determine

represents the thickness of the material\Delta x=2.5 cm =0.025 m

To isolate \Delta T from equation (1), we obtain:

\Delta T =\frac{Q \Delta x}{Ak}

Initially, we convert 3KW to W, resulting in:

Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W

With all variables accounted for, we can substitute and calculate:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

Thus, the temperature difference across the material will be \Delta T = 375 K

5 0
2 months ago
A solid conducting sphere carrying charge q has radius a. It is inside a concentric hollow conducting sphere with inner radius b
Softa [3030]

Response:

Clarification:

Refer to the diagram indicating the charges on the specified sphere (see attachment).

The electric field at the stated positions is

E(r) = 0 for r≤a.  Equation 1

E(r) = kq/r² for a<r<b.   Equation 2

E(r) = 0 for b<r<c.      Equation 3

E(r) = kq/r² for r>c.    Equation 4.

We understand that electric potential correlates with the electric field through

V = Ed

A. To compute the potential at the outer surface of the hollow sphere (r=c), we determine that the electric field there is

E = kQ / r²

Then,

V = Ed,

At d = r = c

Thus,

Vc = (kQ / c²) × c

Vc = kQ / c

As a result, the total charge Q consists of +q, -q, and +q

Hence, Q = q - q + q = q

V = kq / c

B. To calculate the potential at the inner surface of the hollow sphere (r=b), we have

V = kQ/r

V = kQ / b,   noting that r = b

So, Q = q

V = kq / b

C. At r = a

Following from equation 1:

E(r) = 0 for r≤a.  Equation 1

The electric field at the surface of the solid sphere is 0, E = 0N/C

Thus,

V = Ed = 0 V

Consequently, the electric potential at the solid sphere's surface is 0.

D. At r = 0

The electric potential can be determined by

V = kq / r

As r approaches 0,

V = kq / 0

V approaches infinity.

8 0
4 months ago
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