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3 years ago

5

You lower a 2.50 kg textbook (remember when textbooks used to be made out of paper instead of being digital?) from a height of 1

.85 m to 1.50 m. What is its change in potential energy?

1 answer:

aev [14]3 years ago

6 0

Answer:

Change in potential energy = 8.58 J

Explanation:

Potential energy = Mass x Acceleration due to gravity x Height

Potential energy = mgh

m = 2.50 kg

g = 9.81 m/s²

Initial height, h₁ = 1.85 m

Final height, h₂ = 1.50 m

Change in potential energy = mgh₁ - mgh₂

Change in potential energy = 2.5 x 9.81 x 1.85 - 2.5 x 9.81 x 1.50

Change in potential energy = 8.58 J

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El tubo de entrada que suministra presión de aire para operar un gato hidráulico tiene 2 cm de diámetro. El pistón de salida es

Bess [88]

Answer:

La presión neumática para levantar un automóvil de 17,640 newtons es 220,500 pascales.

Explanation:

Asumiendo que la presión ( $P$ ), medida en pascales, tiene una distribución uniforme sobre la superficie del pistón, se calcula a partir de la siguiente expresion:

$P = \frac{F}{A}$

Donde:

$F$ - Fuerza motriz, medida en newtons.

$A$ - Área del pistón, medida en metros cuadrados.

La fuerza motriz es equivalente al peso del automóvil. El área del pistón ( $A$ ), medido en metros cuadrados, es determinado por:

$A=\frac{\pi}{4}\cdot D^{2}$

Donde $D$ es el diámetro del pistón, medido en metros.

Si $D = 0.32\,m$ y $F =17,640\,N$ , entonces la presión neumática es:

$A = \frac{\pi}{4}\cdot (0.32\,m)^{2}$

$A \approx 0.080\,m^{2}$

$P = \frac{17,640\,N}{0.080\,m^{2}}$

$P = 220,500\,Pa$

La presión neumática para levantar un automóvil de 17,640 newtons es 220,500 pascales.

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3 years ago

A 4.00-kg particle moves along the x axis. Its position varies with time according to x 5 t 1 2.0t 3, where x is in meters and t

White raven [17]

Answer:

Explanation:

Given equation is ,

x =t + 2 t³ ,

dx/dt = velocity ( v ) = 1 + 6 t²

a) kinetic energy = 1/2 m v² = .5 x 4 ( 1 + 6 t² )² = 2 ( 1 + 6 t²)²

b ) Acceleration = dv /dt = 12 t .

force( F ) = mass x acceleration = 4 x 12 t = 48 t

Power = force x velocity = 48 t x ( 1 + 6 t²). = 48 t + 288 t³ )

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3 years ago

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Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces

Mandarinka [93]

Answer: $0.81\times 10^{16}$ beta particles

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass = 14.0 g

Molar mass = 137 g/mol

$\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles$

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 mole of cesium contains atoms = $6.023\times 10^{23}$

0.102 moles of cesium contains atoms = $\frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}$

The relation of atoms with time for radioactivbe decay is:

$N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}$

Where $N_t$ =atoms left undecayed

$N_0$ = initial atoms

t = time taken for decay = 3 minutes

${t_{\frac{1}{2}}}$ = half life = 30.0 years = $1.577\times 10^7$ minutes

The fraction that decays : $1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}$

Amount of particles that decay is = $0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}$

Thus $0.81\times 10^{16}$ beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

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3 years ago

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rodikova [14]

Answer:

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Explanation:

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3 years ago

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Fiesta28 [93]

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Explanation:

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