You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves along a frictionless track. however, af

According to Newton's first law, an object remains at rest until an external force acts upon it, or an object in motion continues to move at a constant speed without accelerating.

Thus, x can solely represent a body coming to a halt. Accordingly, the last option is the most correct.

Part a) The connection between the electric field and the magnetic field in an electromagnetic wave is

where
E signifies the strength of the electric field
B indicates the strength of the magnetic field
c represents the speed of light
Using the equation, we determine:


Part b) The text does not clarify the orientation of the magnetic field on the y-axis: I speculate it points in the y+ direction.
The direction of the electric field can be established using the right-hand rule, which states:
- the index finger shows the direction of E
- the middle finger indicates the orientation of B
- the thumb reveals the propagation direction of the wave
Because the wave propagates in the x+ direction, and the magnetic field in the y+ direction, we conclude that the electric field direction (index finger) must be z-.

Response:

83%

Clarification:

At the surface, the weight can be expressed as:

W = GMm / R²

where G denotes the gravitational constant, M represents the Earth's mass, m signifies the shuttle's mass, and R is the Earth's radius.

When in orbit, the weight is given by:

w = GMm / (R+h)²

where h indicates the shuttle's altitude above Earth's surface.

The weight ratio is as follows:

w/W = R² / (R+h)²

w/W = (R / (R+h))²

For R = 6.4×10⁶ m and h = 6.3×10⁵ m:

w/W = (6.4×10⁶ / 7.03×10⁶)²

w/W = 0.83

Thus, the shuttle maintains 83% of its weight as it orbits.

Here is an image displaying the correct answers.

Answer:

a, 71.8° C, 51° C

b, 191.8° C

Explanation:

Given the data:

D(i) = 200 mm

D(o) = 400 mm

q' = 24000 W/m³

k(r) = 0.5 W/m.K

k(s) = 4 W/m.K

k(h) = 25 W/m².K

The heat generation formula can be articulated as follows:

q = πr²Lq'

q = π. 0.1². L. 24000

q = 754L W/m

Thermal conduction resistance, R(cond) = 0.0276/L

Thermal conduction resistance, R(conv) = 0.0318/L

Applying the energy balance equation,

Energy In = Energy Out

This equates to q, which is 754L

From the initial analysis, the temperature at the interface between the rod and sleeve is found to be 71.8° C

Additionally, the outer surface temperature records as 51° C

Furthermore, based on the second analysis, the calculated temperature at the center of the rod is determined to be 191.8° C