A positive point charge q is placed at the center of an uncharged metal sphere insulated from the ground. The outside of the sph

Consider the diagram below.

m₁ = 8.5 x 10⁶ kg, the starship's mass
m₂ = 10⁴ kg, the shuttlecraft's mass
a₁ =  acceleration of the starship
a₂ = the acceleration of the shuttle
F = 4 x 10⁴ N, the force exerted

Let y represent the distance covered by the starship
Let x denote the distance covered by the shuttlecraft
If t indicates the travel time, then
y = 0.5a₁t²                  (1)
x = 0.5a₂t²                  (2)

F = m₁a₁ = m₂a₂         (3)
Additionally,
x + y = 14000 m          (4)

From (2), we derive
a₁ = (4 x 10⁴ N)/(8.5 x 10⁶ kg) = 4.706 x 10⁻³ m/s²
a₂ = (4 x 10⁴ N)/(10⁴ kg) = 4 m/s²

From (1), (2) and (4), we find
0.5*(t s)²*(4 + 4.706 x 10⁻³ m/s²) = 14000 m
2.002353t² = 14000
t² = 6991.774 s²
t = 83.617 s

Thus
x = 0.5*4*6991.774 = 13984 m = 13.984 km
y = 0.5*4.706 x 10⁻³*6991.774 = 16.452 m

The starship moves roughly 16.5 meters while towing the shuttlecraft by 13.98 kilometers.

Result: The starship shifts by 16.5 m (to the nearest tenth)

Answer:

Power output, P = 924.15 watts

Explanation:

We have the following parameters:

Length of the ramp, l = 12 m

Weight of the individual, m = 55.8 kg

Incline angle with respect to the horizontal,

Elapsed time, t = 3 s

Let h represent the vertical height of the hill:

h = 5.07 m

Power P required for a person to ascend the hill can be expressed as:

P = 924.15 watts

This indicates that a minimum average power output of 924.15 watts is essential for an individual to ascend this elevation. Thus, this is the answer sought.

J(r) = Br. We know that the area of a small segment, dA, is represented as 2 π dr. Thus, I = J A and dI = J dA. Plugging in the values gives us dI = B r. 2 π dr which simplifies to dI= 2π Br² dr. Now, integrating the above equation: Given that B= 2.35 x 10⁵ A/m³, with r₁ = 2 mm and r₂ equal to 2 + 0.0115 mm, or 2.0115 mm.

Utiliza Scoratic, funciona con cualquier tema.

1) The buoyant force acting on an object submerged in a fluid can be described as:

where indicates the fluid's density, represents the volume of the fluid displaced, and signifies the gravitational acceleration.

2) To determine the volume of the displaced fluid, we note that the titanium object is entirely submerged in the fluid (air), thus this volume matches the volume of 1 Kg of titanium, which has a density of . Using the correlation between density, volume, and mass, we derive


3) We can now revisit the equation in step 1) to compute the buoyant force. Given that the air density is , this provides us with


4) The weight of 1 Kg of titanium is:

Therefore, the buoyant force is negligible when compared to the weight.