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Anna11 [10]
3 years ago
15

A student is holding a stone at a certain height. The stone has 50 joules of potential energy and 0 joules of kinetic energy. Th

e stone is released from that height. How much energy does the stone have
right before it hits the ground?
Select one:
a) The stone will have no energy
b) The stone will have about 25 joules each of kinetic and potential energy
c) The stone will have about 50 joules of potential energy and ovogres van kinetic
energy
d) The stone will have about 50 joules of kinetic energy and ovoules of potencia
energy​
Physics
1 answer:
Hitman42 [59]3 years ago
6 0

Answer:

d) The stone will have about 50 joules of kinetic energy and 0 joules of potential energy​ .

Explanation:

Given :

Initial Potential energy , P_i=50\ J .

Initial Kinetic energy , K_i=0\ J . ( because ball is in rest )

Now , we know , kinetic energy is maximum when an object reaches ground .

Also , potential energy is zero when an object is in ground .

We know , by conservation of energy :

Initial total energy = Final total energy

P_i+K_i=P_f+K_f\\\\50+0=0+K_f\\\\K_f=50 \ J

Therefore , option d) is correct .

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Un neumático sin cámara, soporta una presión de 1.5 atm cuando la temperatura ambiente es de 300°K. ¿Qué presión llegará a sopor
arlik [135]

Answer:

El neumático soportará una presión de 1.7 atm.

Explanation:

Podemos encontrar la presión final del neumático usando la ecuación del gas ideal:

PV = nRT

En donde:

P: es la presión

V: es el volumen

n: es el número de moles del gas

R: es la constante de gases ideales

T: es la temperatura

Cuando el neumático soporta la presión inicial tenemos:

P₁ = 1.5 atm

T₁ = 300 K

V_{1} = \frac{nRT_{1}}{P_{1}}  (1)  

La presión cuando T = 67 °C es:

P_{2} = \frac{nRT_{2}}{V_{2}}   (2)

Dado que V₁ = V₂  (el volumen del neumático no cambia), al introducir la ecuación (1) en la ecuación (2) podemos encontrar la presión final:

P_{2} = \frac{nRT_{2}}{V_{2}} = \frac{nRT_{2}}{\frac{nRT_{1}}{P_{1}}} = \frac{P_{1}T_{2}}{T_{1}} = \frac{1.5 atm*(67 + 273)K}{300 K} = 1.7 atm  

Por lo tanto, si en el transcurso de un viaje las ruedas alcanzan una temperatura de 67 ºC, el neumático soportará una presión de 1.7 atm.

Espero que te sea de utilidad!

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3 years ago
En el mar, la luz visible alcanza una profundidad de aproximadamente 200 metros (zona fotica). ¿De dónde obtienen energías las p
Marizza181 [45]

Answer:

Explanation:

En la zona apótica (profundidad inferior a 200 m); todo lo que queda de la luz solar es una luz tenue, opaca, azul-verde, demasiado impotente para siquiera considerar permitir que ocurra la fotosíntesis. Sin embargo, hay comida para tener; basura, trozos de plantas podridas y derroche de criaturas caen desde arriba para cuidar a los seres vivos en la zona apótica.

Las formas de vida a una profundidad inferior a 200 m dependen de los productos químicos que salen de los respiraderos; el procedimiento que utilizan para hacer los alimentos se llama quimiosíntesis en lugar de fotosíntesis.

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Does the ke of a car change more when it accelerates from 11 km/h to 21 km/h or when it accelerates from 21 km/h to 31 km/h?
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3 years ago
By what factors does the speed of the elctron exceed that of the proton?<br>​
____ [38]

Answer:

» An electron is lighter than a proton.

<u>explanation</u><u>:</u>

{  =  \: \sf{an \: electron \: has \: formula \:  \: }}{ \bf{ {}^{0}_{  - 1}e }}

hence it's mass number is zero

{  =  \: \sf{an \: electron \: is \: helium \: particle \:  \: }}{ \bf{ {}^{4} _{2}He  }}

hence it's mass number is 4

<u>Therefore</u><u>,</u><u> </u><u>proton</u><u> </u><u>is</u><u> </u><u>heavier</u><u> </u><u>than</u><u> </u><u>electron</u>

» An electron has a small charge magnitude than a proton.

<u>Explanation</u><u>:</u>

An electron has charge of -1 while proton has charge of +2, therefore electron is less deflected by any energetic fields than a proton

8 0
3 years ago
A spring stretches by 0.0208 m when a 3.39-kg object is suspended from its end. How much mass should be attached to this spring
sashaice [31]

Answer:

COMPLETE QUESTION

A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?

Explanation:

Given that,

Extension of spring

x = 0.0208m

Mass attached m = 3.39kg

Additional mass to have a frequency f

Let the additional mass be m

Using Hooke's law

F= kx

Where F = W = mg = 3.39 ×9.81

F = 33.26N

Then,

F = kx

k = F/x

k = 33.26/0.0208

k = 1598.84 N/m

The frequency is given as

f = ½π√k/m

Make m subject of formula

f² = ¼π² •(k/m

4π²f² = k/m

Then, m4π²f² = k

So, m = k/(4π²f²)

So, this is the general formula,

Then let use the frequency above

f = 3Hz

m = 1598.84/(4×π²×3²)

m = 4.5 kg

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3 years ago
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