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soldier1979 [14.2K]
2 years ago
14

Sí un auto viaja a 8m/s determine Tiempo en llegar a 200km de distancia Distancia que recorre en 40 minutos

Physics
1 answer:
GrogVix [38]2 years ago
7 0

Answer:

a. Tiempo = 25000 segundos

b. Distancia = 19200 metros

Explanation:

Dados los siguientes datos;

Velocidad = 8 m/s

Distancia = 200 km a metros = 200 * 1000 = 200,000

Para encontrar el tiempo para cubrir la distancia anterior;

Tiempo = distancia/velocidad

Tiempo = 200000/8

Tiempo = 25000 segundos

b. Para encontrar la distancia recorrida en 40 minutos;

Tiempo = 40 minutos a segundos = 40 * 60 = 2400 segundos

Distancia = velocidad * tiempo

Distancia = 8 * 2400

Distancia = 19200 metros

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Question: A loader sack of total mass

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Answer:

19.6 Joules

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Applying

W = mgh........................ Equation 1

Where W = Workdone by gravity on the load, m = mass of the loader sack, h = height, g = acceleration due to gravity

From the question,

Given: m = 1000 grams = (1000/1000) kilogram = 1 kg, h = 200 cm = 2 m

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3 years ago
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Simple Circuit and Ohm's Law Check-for-Understanding
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3 years ago
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 How does ​energy change (transforms) as the mass is dropping?
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Answer:

Mass has total mechanical energy, which is the sum of kinetic and potential energy. as the mass is dropping, potential energy is converted into kinetic energy so mechanical energy is preserved If there is no friction. If there is friction, some of the mechanical energy is lost as heat energy so it changes.

Explanation:

4 0
3 years ago
A 5.00-A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5 * 1028 free ele
evablogger [386]

Answer:

a)n= 3.125 x 10^{19 electrons.

b)J= 1.515 x 10^{6 A/m²

c)V_{d =1.114 x 10^{4m/s

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Explanation:

Current 'I' = 5A =>5C/s

diameter 'd'= 2.05 x 10^{-3 m

radius 'r' = d/2 => 1.025 x 10^{-3 m

no. of electrons 'n'= 8.5 x 10^{28}

a) the amount of electrons pass through the light bulb each second can be determined by:

I= Q/t

Q= I x t => 5 x 1

Q= 5C

As we know that: Q= ne

where e is the charge of electron i.e 1.6 x 10^{-19C

n= Q/e => 5/ 1.6 x 10^{-19

n= 3.125 x 10^{19 electrons.

b)  the current density 'J' in the wire is given by

J= I/A => I/πr²

J= 5 / (3.14 x (1.025x 10^{-3)²)

J= 1.515 x 10^{6 A/m²

c) The typical speed'V_{d' of an electron is given by:

V_{d = \frac{J}{n|q|}

    =1.515 x 10^{6 / 8.5 x 10^{28} x |-1.6 x 10^{-19|

V_{d =1.114 x 10^{4m/s

d) According to these equations,

J= I/A

V_{d = \frac{J}{n|q|} =\frac{I}{nA|q|}

If you were to use wire of twice the diameter, the current density and drift speed will change

Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.

Also drift velocity will decrease as it is inversely proportional to the area

 

5 0
2 years ago
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