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Alekssandra [29.7K]
3 years ago
8

A solid sphere of mass 8.6 kg, made of metal whose density is 3,400 kg/, hangs by a cord. When the sphere is immersed in a liqui

d of unknown density, the tension in the cord is 38 N. The density of the liquid is closest to which one of the following values?A) 1900 kg/m³B) 1600 kg/m³C) 1700 kg/m³D) 1400 kg/m³E) 1500 kg/m³
Physics
1 answer:
creativ13 [48]3 years ago
4 0

Answer:

A.1900 kg/m^3

Explanation:

We are given that

m=8.6 kg

Density,\rho_s=3400 kg/m^3

Tension,T=38 N

We have to find the density of liquid.

T=mg-\rho_l Vg

g=9.8 m/s^2

Volume,V=\frac{m}{\rho_s}

38=8.6\times 9.8-\rho_l\times \frac{8.6}{3400}\times 9.8

\rho_l\times \frac{8.6}{3400}\times 9.8=8.6\times 9.8-38

\rho_l=\frac{(8.6\times 9.8-38)\times 3400}{8.6\times 9.8}

\rho_l=1867kg/m^3\approx 1900 kg/m^3

Option A is true.

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1 year ago
A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickn
KengaRu [80]

Answer:

88 %

Explanation:

The rate of heat loss by a conducting material of thermal conductivity K, cross-sectional area,A and thickness d with a temperature gradient ΔT is given by

P = KAΔT/d

The total heat lost by the styrofoam wall is P₁ = K₁A₁ΔT₁/d₁ where K₁ =thermal conductivity of styrofoam wall 0.033 W/m-K, A₁ = area of styrofoam wall = 17 m², ΔT₁ = temperature gradient between inside and outside of the wall and d₁ = thickness of styrofoam wall = 0.20 m

The total heat lost by the glass window is P₂ = K₂A₂ΔT₂/d₂ where K₂ =thermal conductivity of glass window pane wall 0.96 W/m-K, A₂ = area of glass window pane = 0.15 m², ΔT₂ = temperature gradient between inside and outside of the window and d₂ = thickness of glass window pane = 7 mm = 0.007 m

The total heat lost is P = P₁ + P₂ = K₁A₁ΔT₁/d₁ + K₂A₂ΔT₂/d₂

Now, since the temperatures of both inside and outside of both window and wall are the same, ΔT₁ = ΔT₂ = ΔT

So, P = K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂

Since P₂ = K₂A₂ΔT₂/d₂ = K₂A₂ΔT/d₂is the heat lost by the window, the fraction of the heat lost by the window from the total heat lost is

P₂/P = K₂A₂ΔT/d₂ ÷ (K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂)

= 1/(K₁A₁ΔT/d₁÷K₂A₂ΔT/d₂ + 1)

= 1/(K₁A₁d₂÷K₂A₂d₁ + 1)

= 1/[(0.033 W/m-K × 17 m² × 0.007 m ÷ 0.96 W/m-K × 0.15 m² × 0.20 m) + 1]

= 1/(0.003927/0.0288 + 1)

= 1/(0.1364 + 1)

= 1/1.1364

= 0.88.

The percentage is thus P₂/P × 100 % = 0.88 × 100 % = 88 %

The percentage of heat lost by window of the total heat is 88 %

6 0
3 years ago
Light striking a metal surface causes electrons to be emitted from the metal via the photoelectric effect.In a particular experi
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The number of electrons emitted from the metal per second increases if the intensity of the incident light is increased.

Answer: Option B

<u>Explanation:</u>

As a result of photoelectric effect, electrons are emitted by the light incident on a metal surface. The emitted electrons count and its kinetic energy can measure as the function of light intensity and frequency. Like physicists, at the 20th century beginning, it should be expected that the light wave's energy (its intensity) will be transformed into the kinetic energy of emitted electrons.

In addition, the electrons count emitting from metal must vary with light wave frequency. This frequency relationship was expected because the electric field oscillates due to the light wave and the metal electrons react to different frequencies. In other words, the number of electrons emitted was expected to be frequency dependent and their kinetic energy should be dependent on the intensity (constant wavelength) of light.

Thus, the maximum in kinetic energy of electrons emitted increases with increase in light's frequency and is experimentally independent of light intensity. So, the number of emitted electrons is proportionate to the intensity of the incident light.

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3 years ago
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Alik [6]

Answer:

Resultant = 3.05N

Explanation:

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