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4 years ago

Define the difference between the rigid body problem and a single particle problem.

1 answer:

3 0

. In single particle problem whole mass is concentrated at a single point so it has a single displacement, single velocity and single acceleration. while, in rigid body mass is distributed

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What is the minimum kinetic energy in electron volts that an electron must have to be able to ionize a hydrogen atom (that is, r

marissa [1.9K]

13.6/4^2 - (-13.6/3^2) = -13.6/16 + 13.6/9 = .66 eV<span>
</span>

8 0

3 years ago

A heavy piece of hanging sculpture is suspended by a90-cm-long, 5.0 g steel wire. When the wind

Yuliya22 [10]

As we know that fundamental frequency is given as

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

here we know that

$\mu = \frac{m}{l}$

here we have

m = mass of wire = 5 g

l = length of wire = 90 cm

$\mu = \frac{0.005}{0.90} kg/m$

$\mu = 5.56 \times 10^{-3} kg/m$

from above formula now

$80 = \frac{1}{2(0.90)}\sqrt{\frac{T}{5.56\times 10^{-3}}}$

$144 = \sqrt{180 T}$

$T = 115.2 N$

now we know that tension is due to weight of the sculpture so we will have

$Mg = 115.2 N$

$M = 11.76 kg$

so its mass will be 11.76 kg

6 0

4 years ago

What is the approximate weight of the air inside the tire in English Engineering Units (tire outside diameter = 49", rim diamete

Mademuasel [1]

Answer:

1.265 Pounds

Explanation:

Data provided:

Tire outside diameter = 49"

Rim diameter = 22"

Tire width = 19"

Now,

1" = 0.0254 m

thus,

Tire outside radius, r₁ = 49"/2 = 24.5" = 24.5 × 0.0254 = 0.6223 m

Rim radius, r₂ = 22" / 2 = 11" = 0.2794 m

Tire width, d = 19" = 0.4826 m

Now,

Volume of the tire = π ( r₁² - r₂² ) × d

on substituting the values, we get

Volume of air in the tire = π ( 0.6223² - 0.2794² ) × 0.4826 = 0.46877 m³

Also,

Density of air = 1.225 kg/m³

thus,

weight of the air in the tire = Density of air × Volume air in the tire

weight of the air in the tire = 1.225 × 0.46877 = 0.5742 kg

also,

1 kg = 2.204 pounds

Hence,

0.5742 kg = 0.5742 × 2.204 = 1.265 Pounds

3 0

3 years ago

Explain how the gravitational potential energy of an object can be changed.

inna [77]

Answer:

Change in Potential energy, pe = Final potential energy - Initial potential energy

Explanation:

∆p.e = Final p.e - Initial p.e

From the equation ∆p.e = (mgh) final - (mgh) initial

4 0

3 years ago

What profession is most likely to make use of amperes and candelas? A. nurse practitioner B. marine biologist C. electrical engi

ASHA 777 [7]

Ampere is the unit for the current that passes through the circuit. Further, the word "candela" is the unit for the luminous intensity. The professional that would make use of these units are the electrical engineer. Thus, the answer is letter C.

5 0

3 years ago

Read 2 more answers