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2 years ago

What is the mass of a dog that weighs 382 N?(unit=kg)

Physics

1 answer:

Luba_88 [7]2 years ago

6 0

Answer:

The answer to your question is: mass = 38.93 kg

Explanation:

Data

mass = ?

Weight = 382 N

gravity = 9.81 m/s2

Formula

Weight = mass x gravity

mass = weight / gravity

mass = 382 / 9.81 substitution

mass = 38.93 kg result

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The human nervous system can propagate nerve impulses at about 102 m>s. Estimate the time it takes for a nerve impulse to tra

Olin [163]

Answer:

t = 0.196 s

Explanation:

The speed of a pulse is determined by the characteristics of the medium, its density and its resistance to stress, as long as these remain the speed will be constant for which we can use the kinetic expressions of the uniform movement

v = x / t

t = x / v

calculate

t = 2/102

t = 0.196 s

4 0

2 years ago

A Christmas light is made to flash via the discharge of a capacitor. The effective duration of the flash is 0.25 s (which you ca

Sonbull [250]

Answer:

The correct solution is:

(a) $1.375\times 10^{-2} \ J$

(b) $4.43\times 10^{-3} \ C$

(c) $1.42\times 10^{-3} \ F$

(d) $178.57 \ \Omega$

Explanation:

The given values are:

Effective duration of the flash,

ζ = 0.25 s

Average power,

$P_{avg}=55 \ mW$

$=55\times 10^{-3} \ W$

Average voltage,

$V_{avg}=3.1 \ V$

Now,

(a)

⇒ $E=P_{avg}\times \zeta$

On substituting the values, we get

⇒ $=55\times 10^{-3}\times 0.25$

⇒ $=1.375\times 10^{-2} \ J$

(b)

⇒ $E=Q\times V_{avg}$

then,

⇒ $Q=\frac{E}{V_{avg}}$

On substituting the values, we get

⇒ $=\frac{1.375\times 10^{-2}}{3.1}$

⇒ $=4.43\times 10^{-3} \ C$

(c)

⇒ $C=\frac{Q}{V}$

⇒ $=\frac{4.43\times 10^{-3}}{3.1}$

⇒ $=1.42\times 10^{-3} \ F$

(d)

As we know,

⇒ $R=\frac{1}{4C}$

⇒ $=\frac{1}{4\times 1.42\times 10^{-3}}$

⇒ $=\frac{1000}{5.6}$

⇒ $=178.57 \ \Omega$

5 0

2 years ago

Star temperature is indicated by?

astraxan [27]

A star's temperature is most likely indicated by the color of it. The hotter the star, the bluer it is. The colder the star, the redder it is.

3 0

3 years ago

Which pulls harder the earth or the student

Nana76 [90]

Answer:

wha

Explanation:

5 0

3 years ago

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!

4 0

3 years ago