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mariarad [96]
2 years ago
8

El peso normal de un estudiante de secundaria es 725 N y el área de los dos zapatos que usa es de 412 cm2 . La presión medida qu

e sus zapatos ejercen en el suelo es
Physics
1 answer:
kotykmax [81]2 years ago
8 0

Answer:

Presión = 175,97 N/m²

Explanation:

Dados los siguientes datos;

Peso del alumno (fuerza) = 725N

Área de zapatos = 412 cm² a metros cuadrados = 412/100 = 4.12 metros

Para encontrar la presión, usaríamos la siguiente fórmula;

Presión = fuerza / área

Presión = 725 / 4.12

Presión = 175,97 N/m²

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Answer:

1 hour to ride his motorcycle

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2 years ago
An adaptation of an animal in the desert might be to
expeople1 [14]
Keep cool by being active at night, whereas some other desert animals get away from the sun's heat by digging underground burrows.
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A hiker determines the length of a lake by listening for the echo of her shout reflected by a cliff at the far end of the lake.
Nataliya [291]

To solve the exercise it is necessary to take into account the definition of speed as a function of distance and time, and the speed of air in the sound, as well

v=\frac{d}{t}

Where,

V= Velocity

d= distance

t = time

Re-arrange the equation to find the distance we have,

d=vt

Replacing with our values

d= (343)(3.7)

d= 1269.1m

It is understood that the sound comes and goes across the entire lake therefore, the length of the lake is half the distance found, that is

L_{lake} = \frac{d}{2}

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8 0
3 years ago
1.) A stone falls from rest from the top of a cliff.
KengaRu [80]

Answer:

Explanation:

ignore air resistance

Let t be the time of fall for the dropped stone.

½(9.8)t² = 43.12(t - 2.2) + ½(9.8)(t - 2.2)²

4.9t² = 43.12t - 94.864 + 4.9(t² - 4.4t + 4.84)

4.9t² = 43.12t - 94.864 + 4.9t² - 21.56t + 23.716

     0 = 21.56t - 71.148

t = 71.148/21.56 = 3.3 s

h = ½(9.8)3.3² = 53.361 = 53 m

or

h = 43.12(3.3 - 2.2) + ½(9.8)(3.3 - 2.2)² = 53.361 = 53 m

4 0
2 years ago
A 8.0 µF capacitor is initially connected to a 2.0 V battery. Once the capacitor is fully charged the battery is removed and a 6
aalyn [17]

Answer:

I = 0.451 amp

Explanation:

given,

C = 8.0 µF

V = 2 V

resistor connected between two terminal = 6 Ω

current flowing through resistor = 13 µsec

Q = CV

Q = 8 x 2

Q = 16 µC                              

for an RC discharge circuit

V = V_0e^{-\dfrac{t}{RC}}

I = \dfrac{-Q_0}{RC}e^{-\dfrac{t}{RC}}

t =  13 µsec

I = \dfrac{-16}{6\times 2}e^{-\dfrac{13}{6\times 2}}

I = 0.451 amp

neglecting  -ve sign just to show direction.

3 0
3 years ago
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