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dlinn [17]
3 years ago
12

An object of mass m is sliding with speed vi at some instant across a level tabletop, with which its coefficient of kinetic fric

tion is µ. It then moves through a distance d and comes to rest. Which of the following equations for the speed vi is reasonable?
A) vi = v2µd
B) vi = v2µmgd
C) vi = v-2µmgd
D) vi = v-2µgd
E) vi = v2µgd
Physics
1 answer:
Mnenie [13.5K]3 years ago
5 0

Answer:

Explanation:

Given

initial speed =v_i

coefficient of kinetic friction=\mu

It moves a distance d after that it comes to rest

acceleration offered by table top=\mu g

v^2-u^2=2as

0-(v_i)^2=2(-\mu g)d

v_i=\sqrt{2(\mu g)d}

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A que profundidad esta nadando una persona dentro de una alberca si la presión absoluta sobre ésta es de 156kPa?
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Answer:

La persona está nadando en la alberca a una profundida de 5.575 metros.

Explanation:

La presión absoluta (P_{tot}) experimentada por la persona es la suma de la presión atmosférica (P_{atm}) y la presión hidrostática de la columna de agua de la alberca (P_{h}), medidas en kilopascales. Es decir,

P_{tot} = P_{atm}+P_{h} (1)

P_{tot} = P_{atm} + \frac{\rho\cdot g \cdot z}{1000} (2)

Donde:

\rho - Densidad del fluido de la alberca, medida en kilogramos por metro cúbico.

g - Aceleración gravitacional, medida en metros por segundo al cuadrado.

z - Profundidad de la persona en la alberca, medida en metros.

Si sabemos que P_{atm} = 101.325\,kPa, \rho = 1000\,\frac{kg}{m^{3}}, g = 9.807\,\frac{m}{s^{2}} y P_{tot} = 156\,kPa, entonces la profundidad de la persona en la alberca es:

156 = 101.325 +\frac{(1000)\cdot (9.807)\cdot z}{1000}

54.675 = 9.807\cdot z

z = 5.575\,m

La persona está nadando en la alberca a una profundida de 5.575 metros.

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A gas cylinder contains argon atoms (m=40.0 u). The temperature is increased from 286 K (13°C) to 362 K (89°C) (a) What is the c
Rama09 [41]

Answer:

(a). The change in the average kinetic energy per atom is 1.57\times10^{-21}\ eV.

(b). The change in vertical position is 2413 m.

Explanation:

Given that,

Mass = 40.0 u

The increased temperature from 286 K to 362 K.

(a). We need to calculate the change in the average kinetic energy per atom

Using formula of kinetic energy

\Delta K.E=\dfrac{3}{2}k\Delta t

Put the value into the formula

\Delta K.E=\dfrac{3}{2}\times1.38\times10^{-23}\times(362-286)

\Delta K.E=1.57\times10^{-21}\ eV

(b). The change in potential energy of the container due to change in the vertical position

We need to calculate the change in vertical position

Using formula of potential energy

\Delta U=mg\Delta h

\Delta h =\dfrac{\Delta U}{mg}

\Delta h=\dfrac{1.57\times10^{-21}}{40.0\times1.66\times10^{-27}\times9.8}

\Delta h=2412.7=2413\ m

Hence, (a). The change in the average kinetic energy per atom is 1.57\times10^{-21}\ eV.

(b). The change in vertical position is 2413 m.

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