A 12 kg box sliding on a horizontal floor has an initial speed of 4.0 m/s. The coefficient of friction bctwecn thc box and the f

loor is 0.20. The box moves n distance of 4.0 m in 2.0 s. The magnitude of the change in momentum of the box during this time is most nearly (A) 12 kg-m/s (B) 48 kg-m/s (C 60 kg-m/s (D) 96 kg-m/s

Physics

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A fan that is switched on for 1 minute uses 500 W usefully but also wastes 300 W through the emission of sound and heat. What's

Ostrovityanka [3204]

Explanation:

Efficiency can be expressed as the ratio of useful output to the total power consumed. $E = \frac{useful}{total}$

The fan delivers 500W as useful output while wasting 300W. Thus, the overall power consumed equals 800W (500 + 300).

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The energy difference between the 5d and the 6s sublevels in gold accounts for its color. If this energy difference is about 2.7

Yuliya22 [3333]

The light's wavelength absorbed during the transition is 459 nm. Energy difference between the 5-d and the 6-s sub-levels in gold is expressed as ΔE. Let the wavelength associated with the electron's transition from the 5-d to the 6-s state be λ. The relationship that describes the connection between energy and wavelength is defined as: E = hc/λ, where E stands for photon energy, h represents Planck's constant, c is the speed of light, and λ denotes the wavelength of the photon. Therefore, the absorption wavelength in this transition stands at 459 nm.

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A Federation starship (8.5 ✕ 106 kg) uses its tractor beam to pull a shuttlecraft (1.0 ✕ 104 kg) aboard from a distance of 14 km

Keith_Richards [3271]

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)