A stars mass determines ?

A wire is hung between two towers and has a length of 208 m. A current of 154 A exists in the wire, and the potential difference

NARA [144]

Answer:

Mass = 4152kg

Explanation:

Given

L = 208m

I = 154A

V = 0.245V

Density = 3610 kg/m3

ρ = 4.23 x 10-8Ω·m = resistivity of wire

Resistance R = ρL/ A

R = voltage / current = V/I = 0.245/154 = 1.59×10-³ohms

1.59×10-³ = 4.23 x 10-⁸×208/A

Rearranging,

A = 4.23 x 10-⁸×208/1.59×10-³

A = 5.53×10-³m²

Mass = density × volume

Volume = L×A = 208×5.53×10-³m³= 1.15m³

Mass = 3610×1.15 = 4152kg

5 0

3 years ago

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Express the surface area of a cube as a function of its volume v.

vredina [299]

The surface area of a cube is:

S = 6l², making l the subject:
l = √(S/6)

While the volume is:

V = l³

If we substitute l to get:

V = (√(S/6))³

√(S/6) = ³√V

S/6 = (³√V)²

S = 6(³√V)² ⇒ 6V^(2/3)

5 0

4 years ago

A person 1.8m tall stands 0.75m from a reflecting globe in a garden.

Maru [420]

Answer:

1. The image of the person is 1.41 m, virtual and formed at the back of the surface of the globe.

2. The person's image is 3.38 m tall.

Explanation:

From the given question, object distance, u = 0.75 m, object height = 1.8 m, radius of curvature of the reflecting globe, r = 8 cm = 0.08 m.

f = $\frac{r}{2}$ = $\frac{0.08}{2}$ = 0.04 m

1. The image distance, v, can be determined by applying mirror formula:

$\frac{1}{f}$ = $\frac{1}{u}$ + $\frac{1}{v}$

$\frac{1}{0.04}$ = $\frac{1}{0.75}$ + $\frac{1}{v}$

$\frac{4}{100}$ - $\frac{75}{100}$ = $\frac{1}{v}$

$\frac{1}{v}$ = $\frac{4 - 75}{100}$

= - $\frac{71}{100}$

⇒ v = - $\frac{100}{71}$

= - 1.41 m

The image of the person is 1.41 m, virtual and formed at the back of the surface of the globe.

2. $\frac{image distance}{object distance}$ = $\frac{image height}{object height}$

$\frac{1.41}{0.75}$ = $\frac{v}{1.8}$

v = $\frac{2.538}{0.75}$

= 3.384

v = 3.38 m

The person's image is 3.38 m tall.

6 0

3 years ago

3. Una cuerda de guitarra tiene 60 cm de longitud y una masa de 0.05 kg de masa. Si se tensiona mediante una fuerza de 20 N. La

jok3333 [9.3K]

Answer:

f1 = 12.90 Hz

Explanation:

To calculate the first harmonic frequency you use the following formula for n = 1:

$f_n=\frac{n}{2L}\sqrt{\frac{T}{M/L}}$

$f_1=\frac{1}{2L}\sqrt{\frac{T}{M/L}}$ ( 1 )

It is necessary that the unist are in meters, then you have:

L: length of the string = 60cm = 0.6m

M: mass of the string = 0.05kg

T: tension on the string = 20 N

you replace the values of L, M and T in the expression (1) for getting f1:

$f_1=\frac{1}{2(0.6m)}\sqrt{\frac{20N}{0.05kg/0.6m}}=12.90\ Hz$

Hence, the first harmonic has a frequency of 12.90 Hz

4 0

4 years ago

A charge of 8.4*10^-4 moves at an angle of 35 degrees to a magnetic field that has a field strentgh of 6.7

12345 [234]

The force on a charged particle in a magnetic field is given by

the speed of the charged particle = 10842 m/s.

Explanation:

F= q V B sinθ

F=force=3.5 x 10⁻²N

q= charge= 8.4 x 10⁻⁴ C

B= magnetic field= 6.7 x 10⁻³ T

θ=35⁰

Thus the velocity is given by V= $\frac{F}{q B sin35}$

V=(3.5 x 10⁻²)/[(8.4 x 10⁻⁴)(6.7 x 10⁻³)(sin35)]

V=10842 m/s

3 0

3 years ago