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3 years ago

(a) How many kilograms of water must evaporate from a 60.0-kg woman to lower their body temperature by 0.750ºC?

Physics

1 answer:

Mrrafil [7]3 years ago

5 0

Answer:

69.69 g

Explanation:

Evaporation of water will take out latent heat of vaporization. Let the mass of water be m and latent heat of vaporization of water be 2260000 J per kg

Heat taken up by evaporating water

= 2260000 x m J

Heat lost by body

= mass x specific heat of body x drop in temperature

60 x 3500 x .750 ( specific heat of human body is 3.5 kJ/kg.k)

= 157500 J

Heat loss = heat gain

2260000 m= 157500

m = .06969 kg

= 69.69 g

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You are riding a bicycle. If you apply a forward force of 150 N, and you and the bicycle have a combined mass of 90 kg, what wil

bekas [8.4K]

<h3>Hello There!!</h3>

<h3><u>Given</u>,</h3>

Force(F) = 150N

Mass(m) = 90kg

Acceleration(a) = ?

F= m×a

$150 = 90 \times \text{a} \\ \\ \text{a} = \frac{150}{90} \\ \fbox{cancelling by 3} \\ \\ \text{ a} = \cancel \frac{150}{90} \\ \\ \text{ a} = \frac{5}{3} = 1.67 \text{m/s} {}^{2}$

$\therefore \text{Option A= 1.67 m/s² is the correct answer}$

<h3>Hope this helps</h3>

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2 years ago

A tiger runs at 58 km/h [S]. What is the displacement of the tiger in 38 s?

Artyom0805 [142]

58 K/h = 58000/3600= 16.1 m/s
In 38 s displacement is 38x16.1= 612.2 m

3 0

3 years ago

Help on physics problem (picture above)

natali 33 [55]

The two letters are B and A, in that order.

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2 years ago

Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.

SSSSS [86.1K]

Answer:

a = 9.94 m/s²

Explanation:

given,

density at center= 1.6 x 10⁴ kg/m³

density at the surface = 2100 Kg/m³

volume mass density as function of distance

$\rho(r) = ar^2 - br^3$

r is the radius of the spherical shell

dr is the thickness

volume of shell

$dV = 4 \pi r^2 dr$

mass of shell

$dM = \rho(r)dV$

$\rho = \rho_0 - br$

now,

$dM = (\rho_0 - br)(4 \pi r^2)dr$

integrating both side

$M = \int_0^{R} (\rho_0 - br)(4 \pi r^2)dr$

$M = \dfrac{4\pi}{3}R^3\rho_0 - \pi R^4(\dfrac{\rho_0-\rho}{R})$

$M = \pi R^3(\dfrac{\rho_0}{3}+\rho)$

we know,

$a = \dfrac{GM}{R^2}$

$a = \dfrac{G( \pi R^3(\dfrac{\rho_0}{3}+\rho))}{R^2}$

$a =\pi RG(\dfrac{\rho_0}{3}+\rho)$

$a =\pi (6.674\times 10^{-11}\times 6.38 \times 10^6)(\dfrac{1.60\times 10^4}{3}+2.1\times 10^3)$

a = 9.94 m/s²

7 0

2 years ago

ou are out stargazing with your 13.4-cm telescope. You point your telescope at an interesting formation in the sky, which you th

Alinara [238K]

Answer:

θ = 4.716 10⁻⁶ rad

Explanation:

In order for the releases to be considered separate, they must meet the Rayleigh criterion that establishes that the maximum diffraction of one star must coincide with the first minimum of the diffraction pattern of the second star.

We use the diffraction equation for a slit

a sin θ = m λ

The minimum occurs at m = 1

sin θ = λ / a

Since the angles in these systems are very small, we can approximate the sine to its angle in radians

θ = λ / a

The telescope has a circular aperture whereby polar cords should be used, which introduces a constant number

θ = 1.22 λ / a

Let's calculate

θ = 1.22 518 10⁻⁹ / 13.4 10⁻²

θ = 4.716 10⁻⁶ rad

8 0

2 years ago