The position of an object at any time t is given by s(t)=3t4−40t3+126t2−9. i. Determine the velocity of the object at any time t

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)

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.