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solmaris [256]
2 years ago
8

A 6.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.4 cm if the marble is to just reac

h a target 22 m above the marble's position on the compressed spring. (a) What is the change ΔUg in the gravitational potential energy of the marble-Earth system during the 22 m ascent? (b) What is the change ΔUs in the elastic potential energy of the spring during its launch of the marble? (c) What is the spring constant of the spring?
Physics
1 answer:
RoseWind [281]2 years ago
4 0

Answer:

a) \Delta U_{g} = 12.945\,J, b) \Delta U_{k} = 12.945\,J, c) k = 2930.059\,\frac{N}{m}

Explanation:

a) The change in the gravitational potential energy of the marble-Earth system is:

\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)

\Delta U_{g} = 12.945\,J

b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

\Delta U_{k} = 12.945\,J

c) The spring constant of the gun is:

\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}

k = \frac{2\cdot \Delta U_{k}}{x^{2}}

k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}

k = 2930.059\,\frac{N}{m}

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Answer:c

Explanation:

The buoyancy force on toy depends upon the volume of toy under pool water.

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Buoyancy force is given by

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6 0
3 years ago
Calculate the height of a column of carbon tetrachloride, CCl4(l), with a density of 1.59 g/mL that exerts the same pressure as
julia-pushkina [17]

Answer:

1.30 m

Explanation:

The pressure (P) exerted by a column of a liquid can be calculated with the following expression:

P=\rho \times g \times h

where,

ρ is the density of the liquid

g is the gravity

h is the height of the column

If both liquids exert the same pressure:

PHg = PCCl_{4}\\\rho Hg \times g \times hHg = \rho CCl_{4} \times g \times hCCl_{4}\\\rho Hg \times hHg = \rho CCl_{4} \times hCCl_{4}\\hCCl_{4 = \frac{\rho Hg \times hHg}{\rho CCl_{4}} =\frac{13.6g/mL \times 15.2cm}{1.59g/mL} = 130cm=1.30m

6 0
3 years ago
3. A Gas has a volume of 35 Liters at 10 atm pressure. To expand the volume to 80 Liters, what will the pressure be?
satela [25.4K]

Answer:

3) using Robert Boyles's law the pressure will be 4.375 atm

4) using Charle's law the volume will be 340 Liters

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8 0
2 years ago
Estimate the peak wavelength for radiation from ice at 273 k.
Andrews [41]
<h2>Answer: 10615 nm</h2>

Explanation:

This problem can be solved by the Wien's displacement law, which relates the wavelength  \lambda_{p} where the intensity of the radiation is maximum (also called peak wavelength) with the temperature T of the black body.

In other words:

<em>There is an inverse relationship between the wavelength at which the emission peak of a blackbody occurs and its temperature.</em>

Being this expresed as:

\lambda_{p}.T=C    (1)

Where:

T is in Kelvin (K)

\lambda_{p} is the <u>wavelength of the emission peak</u> in meters (m).

C is the <u>Wien constant</u>, whose value is 2.898(10)^{-3}m.K

From this we can deduce that the higher the black body temperature, the shorter the maximum wavelength of emission will be.

Now, let's apply equation (1), finding \lambda_{p}:

\lambda_{p}=\frac{C}{T}   (2)

\lambda_{p}=\frac{2.898(10)^{-3}m.K}{273K}  

Finally:

\lambda_{p}=10615(10)^{-9}m=10615nm  This is the peak wavelength for radiation from ice at 273 K, and corresponds to the<u> infrared.</u>

8 0
3 years ago
When a submarine dives to a depth of 5.0 × 102 m, how much pressure, in pa, must its hull be able to withstand? how many times l
nordsb [41]
<span>Depth = 5.0 Ă— 10^2 m
 Density of sea water = 1.025 x 10^3
 Pd = Po + pgh
Atmospheric pressure is standard Patm = 1.01325 x 10^5 Pa
  Since the normal pressure is retained in the hull, no need to bother about Po Pd = pgh = 1.025 x 10^3 x 9.8 x 5.0 x 10^2 = 50.225 x 10^5
 So now Pd / Patm = 50.225 x 10^5 / 1.01325 x 10^5 = 49.56
 So it is 49.56 times larger.</span>
5 0
3 years ago
Read 2 more answers
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