Which of these substances will form a basic solution in water? A. Ca(OH)2 B. H3PO4 C. H2SO4 D. HCl

Predict using Boyle's law, what will happen to a balloon that an ocean diver takes to a pressure of 202 kPa.

kobusy [5.1K]

The volume of the balloon will halve

Explanation:

Boyle's law states that for an ideal gas kept at constant temperature, the pressure of the gas is proportional to its volume. Mathematically,

$pV=const.$

where

p is the gas pressure

V is the volume

The equation can also be rewritten as

$p_1 V_1 = p_2 V_2$

And if we apply it to the gas inside the balloon in this problem (assuming its temperature is constant), we have:

$p_1 = 101 kPa$ is the initial pressure at sea level (the atmospheric pressure)

$V_1$ is the initial volume

$p_2 = 202 kPa$ is the final pressure

$V_2$ is the final volume

Substituting into the equation, we find:

$V_2 = \frac{p_1 V_1}{p_2}=\frac{(101)V_1}{202}=\frac{V_1}{2}$

Which means that the volume of the balloon will halve.

Learn more about ideal gases:

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4 years ago

A 0.2-kg steel ball is dropped straight down onto a hard, horizontal floor and bounces straight up. The ball's speed just before

nadya68 [22]

Answer:

Explanation:

Given

mass of steel ball $m=0.2\ kg$

initial speed of ball $u=10\ m/s$

Final speed of ball $v=-10\ m/s$ (in upward direction)

Impulse imparted is given by change in the momentum of object

therefore impulse J is given by

$J=\Delta P$

$\Delta P=m(v-u)$

$\Delta P=0.2(-10-10)$

$\Delta =-4\ N-s$

so magnitude of Impulse =4 N-s

7 0

3 years ago

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3. The expression 0.62 x10^3 is equivalent to...

Korolek [52]

$\\ \sf\longmapsto 0.62\times 10^3$

$\\ \sf\longmapsto 62\times 10^{-2}\times 10^3$

$\\ \sf\longmapsto 62\times 10^{-2+3}$

$\\ \sf\longmapsto 62\times 10^1$

$\\ \sf\longmapsto 62\times 10$

$\\ \sf\longmapsto 620$

5 0

3 years ago

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A small loop of area A = 5.5 mm2 is inside a long solenoid that has n = 919 turns/cm and carries a sinusoidally varying current

Tema [17]

Answer: $410.90\times 10^{-6} V$

Explanation:

Given

$A=5.5 mm^2$

$n=919 turns/cm\approx 85400 turns/m$

$\omega =226 rad/s$

According to the Faraday law of induction, induced emf is given by

$E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}$

Magnetic field $\phi _B=B\cdot A$

$B=\mu _0ni=\mu _0ni_osin\left ( \omega t\right )$

$B=4\pi \times 10^{-7}\times 85400\times 3.08\times \sin (226t)$

$B=0.3305\sin (226t)$

$\phi _{B}=0.3305\times \sin (226t)\times 5.5\times 10^{-6}$

$\phi _{B}=1.818\times 10^{-6}\sin (226t)$

$E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}$

$E=-1.818\times 10^{-6}\times 226\cos (226t)$

$E=-410.90\times 10^{-6}\cos (226t)$

Amplitude of EMF induced $=410.90\times 10^{-6} V$

8 0

4 years ago

A bimetallic strip (brass/steel), which is straight at room temperature, will be immersed in boiling water and allowed to equili

lakkis [162]

The thermal expansion of the materials allows to find the deflection of the bimetallist strip is Δy = 3.48 cm

given paramers

* Bimetallic brass / steel tape

* Initial temperature, room temperature T = 20ºC

* Final temperature, boiling water = 100ºC

* initial length L₀ = 222mm (1cm / 10mm) = 22.2cm

* thickness of bimetallic tape e = 0.036 inch (2.54 cm/1 inch) = 0.0914 cm

to find

* perpendicular deviation or deflection (Δy)

Thermal expansion is the phenomenon of change in the length of a body due to the change in temperature, due to the increase in the length of the atomic and molecular bonds, macroscopically it is described by

ΔL = α L₀ ΔT

ΔL and ΔT are the variation of the length and temperature respectively, L₀ is the initial length and α the coefficient of expansion ends.

In this case we have a strip formed by two materials with different coefficient of thermal expansion,

Brass α_{brass} = 19 10⁻⁶ ºC⁻¹

Steel α_{steel} = 11 10⁻⁶ ºC⁻¹

In the attached we can see a diagram of the process, as the temperature increases, the material with greater thermal expansion lengthens more, so the system must curve towards the center of the material with less

thermal expansion. Let's find the length of the strip for each material

brass L_{f brass} - L₀ = α_{brass} L₀ ΔT

Steel L_{f steel} - L₀ = \alpha_{steel} L₀ ΔT

Note that the initial length is the same for the two materials and that the strip is in thermal equilibrium at room temperature.

If we assume that we have an arc of circumference, we can write the length of the arc

θ = L / r

where θ is the angle in radines, L the length of the arc and r the radius of curvature, let's write this equation for each material

brass L_{f \ brass} =θ r₁

steel L_{f \ steel} = θ r₂

we substitute in our equations

θ r₁ - L₀ = α_{brass} L₀ ΔT

θ r₂ - L₀ = α_{steel} L₀ ΔT

let's subtract the two equations

θ (r₁- r₂) = L₀ ΔT (α_{brass} - α_{steel})

the thickness of the strip is

e = r₁ -r₂

θ = $Lo \ \Delta T \ \frac{\alpha_{brass} - \alpha_{steel}}{e}$

we calculate

θ = $22.2 \ (100-20) \ \frac{(19-11) \ 10^{-6}}{0.0914}$

θ = 0.155 rad

Let's use trigonometry to find the perpendicular deflection

tan θ = Δy / L₀

Δy = L₀ tan θ

Δy = 22.2 tan 0.155

Δy = 3.48 cm

Using the thematic expansion of the two materials we find the deflection of the bimetallist strip is 3.38 cm

Learn more about thermal expansion here: brainly.com/question/18717902

7 0

3 years ago