All categories

2 years ago

A uniform thin circular rubber band of mass M and spring constant k has an original radius R?

Physics

1 answer:

Gemiola [76]2 years ago

6 0

Answer: Yes

Explanation:

A ruber band can be described by his mas, M, the spring constant K, and his rest radius, wich is the radius of the circular rubber band when there is no force applied to it, so yes, a uniform thin circular rubber band of mass M and spring constant k has an original radius R

You might be interested in

Which type of listening response includes the use of head nods, facial expressions, and short utterances such as "uh-huh" that s

BARSIC [14]

This type of listening response is called back-channel signal. This allows the speaker to know that the listener is attentive or willing to engage a conversation between them. It is shown through short utterances, facial expressions, head nods and others.

4 0

2 years ago

The human eye can respond to as little as 10^−18 J of light energy. For a wavelength at the peak of visual sensitivity, 550 nm,

Bumek [7]

<span>The key equation is going to come from Mr Planck: E=h \nu

Where h is Plancks constant; and ν is the frequency. This equation gives you the energy per photon at a given frequency. Alas, you're given wavelength, but that's easy enough to convert to frequency given the following equation:

c= lambda / nu

where c is the speed of light; λ (lambda) is the wavelength; and ν is again frequency. As soon as you know the energy of a photon with a wavelength of 550nm, you should know how many photons you would require to accumulate 10^-18J. Be careful with your units.</span>

7 0

2 years ago

Read 2 more answers

If a current of 2.4 a is flowing in a cylindrical wire of diameter 2.0 mm, what is the average current density in this wire?

Gnom [1K]

The average current density in the wire is given by:

$J=\frac{I}{A}$

where I is the current intensity and A is the cross-sectional area of the wire.

The cross-sectional area of the wire is given by:

$A=\pi r^2$

where r is the radius of the wire. In this problem, $r=\frac{d}{2}=\frac{2.0 mm}{2}=1.0 mm=0.001 m$ , so the cross-sectional area is

$A=\pi (0.001 m)^2=3.14 \cdot 10^{-6} m^2$

and the average current density is

$J=\frac{I}{A}=\frac{2.4 A}{3.14 \cdot 10^{-6} m^2}=7.64 \cdot 10^5 A/m^2$

8 0

2 years ago

Read 2 more answers

A rigid vessel of 0.06 m3 volume contains an ideal gas , CV =2.5R, at 500K and 1 bar.a). if 15000J heat is transferred to the ga

andreev551 [17]

Answer:

Given that

V= 0.06 m³

Cv= 2.5 R= 5/2 R

T₁=500 K

P₁=1 bar

Heat addition = 15000 J

We know that heat addition at constant volume process ( rigid vessel ) given as

Q = n Cv ΔT

We know that

P V = n R T

n=PV/RT

n= (100 x 0.06)(500 x 8.314)

n=1.443 mol

Q = n Cv ΔT

15000 = 1.433 x 2.5 x 8.314 ( T₂-500)

T₂=1000.12 K

We know that at constant volume process

P₂/P₁=T₂/T₁

P₂/1 = 1000.21/500

P₂= 2 bar

Entropy change given as

$\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}$

Cp-Cv= R

Cp=7/2 R

Now by putting the values

$\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}$

$\Delta S=1.443\times 3.5\times 8.314\ln \dfrac{1000.21}{500}-1.443\times 8.314\ln \dfrac{2}{1}$

a)ΔS= 20.79 J/K

If the process is adiabatic it means that heat transfer is zero.

ΔS= 20.79 J/K

We know that

$\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}$

Process is adiabatic

$\Delta S_{surr}=0$

$\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}$

$\Delta S_{univ}= 20.79 +0$

$\Delta S_{univ}= 20.79$

3 0

2 years ago

Calculate the current through a 10.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V

Kipish [7]

Answer:

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

Explanation:

Given:

Length = l = 10 meter

$Radius = r = 0.321\ mm =0.321\times 10^{-3}\ meter$

$Resistivity=\rho=1.00\times 10^{-6}\ ohm\ meter$

V = 12 Volt

To Find:

Current, I =?

Solution:

Resistance for 0.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V battery given as

$R=\dfrac{\rho\times l}{A}$

Where,

R = Resistance

l = length

A = Area of cross section = πr²

$\rho=Resistivity=1.00\times 10^{-6}\ ohm\ meter$

Substituting the values we get

$R=\dfrac{1\times 10^{-6}\times 10}{3.14\times (0.321\times 10^{-3})^{2}}$

$R=\dfrac{1\times 10^{-5}}{3.23\times 10^{-7}}$

$R=\dfrac{1\times 10^{2}}{3.23}$

$R=30.95\ ohm$

Now by Ohm's Law,

$V= I\times R$

Substituting the values we get

$I=\dfrac{V}{R}=\dfrac{12}{30.95}=0.3876\ Ampere$

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

4 0

2 years ago