Answer:
The resistance of the skin is 98 kΩ
Explanation:
Given:
Resistivity
Ωm
Thickness
m
Resistivity of the skin:

With assumed radius for the worker's palm,
m
Area of the worker's palm,



Thus the resistance of palm is,

Ω
Consequently, the resistance of the skin is 98 kΩ
Answer:
1. The sun is moved away from the Earth by an equal force exerted by the Earth.
Answer:
the temperature on the left side is 1.48 times greater than that on the right
Explanation:
GIVEN DATA:

T1 = 525 K
T2 = 275 K
It is known that


n and v are constant on both sides. Therefore we have

..............1
let the final pressure be P and the temperature 

..................2
similarly
.............3
divide equation (2) by equation (3)
![\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}](https://tex.z-dn.net/?f=%5Cfrac%7B21%7D%7B11%7D%5E%7B-2%2F3%7D%20%5Cfrac%7B21%7D%7B11%7D%5E%7B5%2F3%7D%20%3D%20%5B%5Cfrac%7BT_1%20%7Bf%7D%7D%7BT_2%20%7Bf%7D%7D%5D%5E%7B5%2F3%7D)

thus, the left side temperature equals 1.48 times the right side temperature
Answer:
20 cm
Explanation:
The electric potential energy U is calculated with the formula U = kq₁q₂/r, where q₁ = 5 nC (5 × 10⁻⁹ C) and q₂ = -2 nC (-2 × 10⁻⁹ C) and r is determined as √(x - 2)² + (0 - 0)² + (0 - 0)² = x - 2. This leads to U = -0.5 µJ (-0.5 × 10⁻⁶ J), where k = 9 × 10⁹ Nm²/C².
Thus, solving for r gives us r = kq₁q₂/U
which leads to x - 2 = kq₁q₂/U
Then, rearranging gives x = 0.02 + kq₁q₂/U m
So, x = 0.02 + 9 × 10⁹ Nm²/C² × 5 × 10⁻⁹ C × -2 × 10⁻⁹ C/-0.5 × 10⁻⁶ J
Resulting in x = 0.02 - 90 × 10⁻⁹ Nm²/-0.5 × 10⁻⁶ J
This simplifies to x = 0.02 + 0.18 = 0.2 m, or 20 cm
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.