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13 days ago

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3. A 3.0 × 10−3 kg copper penny drops a distance of 50.0 m to the ground. If 65 percent of the initial potential energy goes int

o increasing the internal energy of the penny, determine the magnitude of that increase.

1 answer:

Keith_Richards [1K]13 days ago

5 0

Response:

The increase in energy amounts to 0.9555J

Details:

Potential Energy (PE) can be calculated as mgh

where mass (m) is 3×10^-3kg, gravitational acceleration (g) is 9.8m/s^2, and height (h) is 50m

This results in PE = 3×10^-3 × 9.8 × 50 = 1.47J

The rise in internal energy is calculated as 65% of 1.47J, which gives 0.65 × 1.47J = 0.9555J

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Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller

Softa [913]

Answer:

σ₁ = $3.167 * 10^{-6}$ C/m²

σ₂ = $7.6 * 10 ^{-6}$ C/m²

Explanation:

Provided Information:

i) Smaller sphere's radius ( r ) = 5 cm.

ii) Larger sphere's radius ( R ) = 12 cm.

iii) Electric field at the larger sphere's surface ( E₁ ) = 358 kV/m, which is equivalent to 358 * 1000 v/m

$E_{1} = (\frac{1}{4\pi\epsilon }) (\frac{Q_{1} }{R^{2} } )$

$358000 = 9 * 10^{9 } *\frac{Q_{1} }{0.12^{2} }$

Charge (Q₁) = 572.8 $* 10^{-9}$ C

Since the electric field inside a conductor is zero, the electric potential ( V ) remains constant.

V = constant

∴ $\frac{Q_{1} }{R} = \frac{Q_{2} }{r}$

$Q_{2} = \frac{r}{R} *Q_{1}$

$Q_{2} = \frac{5}{12} *572.8*10^{-9}$ = $238.666 *10^{-9}$ C

Surface charge density ( σ₁ ) for the larger sphere.

Calculated Area ( A₁ ) = 4 * π * R² = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁ = $\frac{Q_{1} }{A_{1} }$ = $\frac{572.8 *10^{-9} }{0.180864}$ = $3.167 * 10^{-6}$ C/m².

Surface charge density ( σ₂ ) for the smaller sphere.

Calculated Area ( A₂ ) = 4 * π * r² = 4 * 3.14 * 0.05² = 0.0314 m².

σ₂ = $\frac{Q_{2} }{A_{2} }$ = $\frac{238.66 *10^{-9} }{0.0314}$ = $7.6 * 10 ^{-6}$ C/m²

8 0

4 days ago

the flow energy of 124 L/min of a fluid passing a boundary to a system is 108.5 kJ/min. Determine the pressure at this point

serg [1198]

Answer:

The pressure measured at this moment is 0.875 mPa

Explanation:

Given that,

Flow energy = 124 L/min

Boundary to system P= 108.5 kJ/min

$P=1.81\ kW$

We are tasked with finding the pressure here

Applying the pressure formula

$P=F\times v$

$P=A_{1}P_{1}\times v_{1}$

Here, $A_{1}v_{1}=Q_{1}$

Where, v refers to velocity

Insert the values into the equation

$1.81 =P_{1}\times0.124\times\dfrac{1}{60}$

$P_{1}=\dfrac{1.81\times60}{0.124}$

$P_{1}=875.80\ kPa$

$P_{1}=0.875\ mPa$

Therefore, the pressure at this moment is 0.875 mPa

5 0

11 days ago

A very small round ball is located near a large solid sphere of uniform density. The force that the large sphere exerts on the b

Softa [913]

Respuesta:

Opción e

Explicación:

La Ley de Gravitación Universal indica que toda masa puntual atrae a otra masa puntual en el universo con una fuerza que se dirige en línea recta entre los centros de masa de ambos, siendo esta fuerza proporcional a las masas de los objetos y inversamente proporcional a su separación. Esta fuerza atractiva siempre es dirigida del uno hacia el otro. La ley es aplicable a objetos de cualquier masa, sin importar su tamaño. Dos objetos grandes pueden ser considerados masas puntuales si la distancia entre ellos es considerablemente mayor que sus dimensiones o si presentan simetría esférica. En tales casos, la masa de cada objeto puede ser modelada como una masa puntual en su centro de masa.

La misma fuerza actúa sobre ambas bolas.

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10 days ago

A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vect

Softa [913]

Answer:

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

Explanation:

Consider the following:

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field measured at point P

$E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta$

$sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}$

$r^2=d^2+x^2$

Thus,

$E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}$

Now, by applying integration to the equation above

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

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5 days ago

The maximum mass the can be hung vertically from a string without breaking the string is 10kg. A length of this string that is 2

serg [1198]

Respuesta:

Explicación:

Al analizar esta pregunta, considera el movimiento circular. Primero, determina la máxima fuerza que puede aplicarse al hilo. F = mg, entonces F = (10)(10) = 100 N. Luego, calcula la aceleración centrípeta de la masa de 0.5 kg, a = F/m, así que a = 100/.5 = 200 m/s². En la hoja de ecuaciones, usa la fórmula a (aceleración centrípeta) = v²/r, por lo que 200 = v²/2; por consiguiente, v = 20 m/s. ¡Espero que esto sea útil!

0 0

13 days ago