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olga2289
1 month ago
12

A factory robot drops a 10 kg computer onto a conveyor belt running at 3.1 m/s. The materials are such that μs = 0.50 and μk = 0

.30 between the belt and the computer. How far is the computer dragged before it is riding smoothly on the belt?
Physics
1 answer:
inna [3.1K]1 month ago
8 0

Response:

x = 1.63 m

Details:

mass (m) = 10 kg

μk = 0.3

velocity (v) = 3.1 m/s

Assuming that the weight of the computer is largely applied to the belt instantaneously, we can implement the constant acceleration equation below

x = v^{2}/2a

where a = μk.g, thus

x = v^{2}/2μk.g

x = (3.1 x 3.1)/(2 x 0.3 x 9.8)

x = 1.63 m

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A puck rests on a horizontal frictionless plane. A string is wound around the puck and pulled on with constant force. What fract
Ostrovityanka [3204]

Answer:

Explicación:

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La energía cinética (lineal) = 1/2 mv²

La energía cinética rotacional = 1/2 I ω²

I = 1/2 m r² (donde m y r son la masa y el radio del disco)

La energía cinética rotacional = 1/2 x1/2 m r² ω²

= 1/4 m v² (v = r ω)

Energía total

= Energía cinética (lineal) + Energía cinética rotacional

= 1/2 mv² + 1/4 mv²

= 3/4 mv²

La relación de K E rotacional / K E total = 1/4 m v² / 3/4 mv²

= 1 /3

Por lo tanto, 1 /3 de la energía total se debe a la energía cinética rotacional.

3 0
2 months ago
Raphael refers to a wave by noting its wavelength. lucinda refers to a wave by noting its frequency. which student is correct an
Keith_Richards [3271]
Both students are correct because they highlight different aspects of waves. Raphael describes the wave in terms of its wavelength, while Lucinda focuses on its frequency, showing that both measurements are valid.
7 0
2 months ago
A fire engine is moving south at 35 m/s while blowing its siren at a frequency of 400 Hz.
ValentinkaMS [3465]

To tackle this issue, we will apply the principles linked to the Doppler effect. This effect refers to the alteration in the perceived frequency of any wave when there is relative motion between the source of the waves and the observer. It can be mathematically explained as

f_d= f_s \frac{(v+v_d)}{(v-v_s)}

In this case,

f_d=frequency detected by the receiver

f_s=frequency emitted by the source

v_d=speed of the detector

v_s=speed of the source

v=speed of sound waves

Substituting values yields,

f_d = 400(\frac{(343+18)}{(343-35)})

f_d=422 Hz

Thus, the frequency that passengers would hear is 422Hz

8 0
1 month ago
When a mass m1is placed on top of a spring and allowed to come to rest, the spring iscompressed by an amount x1. How much would
Keith_Richards [3271]
x₂ = 16 g m₂ / k. The spring's behavior adheres to Hooke's law, expressed as F = k x. For equilibrium illustrated in Newton's diagram, F - W = 0 and k x₁ = m₁ g, leading to k = g m₁ / x₁. In an elevator moving upward with acceleration, the relation adjusts to F - W = m a, which gives F = m (g + a). Compression becomes K x₂ = 4 m (g + 3g), simplifying to x₂ = 4m / k (4g) and ultimately x₂ = 16m2 / k g.
4 0
1 month ago
A machine has an efficiency of 20 percent. Find the input work if the output work is 140 Joules
Maru [3345]

Answer:

X= 700 Joules

Explanation:

The inquiry pertained to the calculation of work efficiency.

The efficiency formula is stated as: Efficiency = (Useful output / input work) * 100%

The useful output specified in the problem is 140J, while input work is denoted as X. The efficiency rate provided is 20%.

Applying the efficiency formula yields

20 = (140/X) * 100

Thus, we can proceed to solve the above equation.

X= 140*100/20

X= 700 Joules

6 0
1 month ago
Read 2 more answers
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