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2 years ago

The higher the voltage across a bulb the b__________ it is.

Physics

2 answers:

podryga [215]2 years ago

8 0

Answer:

Brighter

Explanation:

Sonja [21]2 years ago

7 0

The higher the voltage across à bulbe the brighter it is.

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In a standard tensile test a steel rod of 22-mm diameter is subjected to a tension force of 75kN. Knowing that v = 0.30 and E =

VladimirAG [237]

Answer:

(a) Elongation of rod = 0.19732 mm

(b) Change in diameter = 0.00651 mm

Explanation:

Circular area at end of steel rod = pi * diameter^2 / 4

$Area = pi * (22 * 10^-^3)/4$

Area = 3.801 * 10^(-4) meter squared

Stress = force / area

Stress = 75000 / (3.801 * 10^-4)

Stress = 197316495 Pa OR 0.197 GPa

Modulus of elasticity = stress / strain

200 = 0.19732 / Strain

Strain = 0.0009866 (longitudinal)

(a) Strain = change in length / initial length

0.0009866 = Elongation of rod / 200

Elongation of rod = 0.19732 mm

(b) Poisson ratio = lateral strain / longitudinal strain

0.3 = lateral strain / 0.0009866

Lateral strain = 0.000296

Lateral strain = Change in diameter / original diameter

0.000296 = Change in diameter / 22

Change in diameter = 0.00651 mm

4 0

3 years ago

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One problem with weight training as a way to improve overall health is that the results of a weight-training program are not mea

Ede4ka [16]

One problem with weight training as a way to improve overall health is that the results of a weight-training program are not measurable.

B.False

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3 years ago

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A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the

zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

$E= \frac{kqr}{R^3}$

The potential at a distance can be represented as:

V(r) - V(0) = $-\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2$

V(r) - V(0) = $[\frac{qr^2}{8 \pi E_0R^3 }]$ ₀

V(r) = $-[\frac{qr^2}{8 \pi E_0R^3 }]$ ₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

= 0.56 × 10⁻² m

R = 1.29 cm

= 1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

$V(r)$ $= -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}$

$V(r)$ = -2.15 × 10⁻⁴ V

The difference between the radial distance and center can be expressed as:

V(r) - V(0) = $-\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2$

V(r) - V(0) = $[\frac{qr^2}{8 \pi E_0R^3 }]^R$

V(r) = $-\frac{qR^2}{8 \pi E_0R^3 }$

V(r) = $-\frac{q}{8 \pi E_0R }$

V(r) $= -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}$

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0

2 years ago

What is a good example of netwon 3rd law

satela [25.4K]

A fish pushes water backwards in order to move forward is a good example of Newton's 3rd Law.

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3 years ago

You are given an unknown solid substance of mass 500 g and you want to estimate its specific heat. unfortunately, you don’t have

Julli [10]

Answer:

Option C) 2,090 J/(mol K)

Explanation:

Data:

Volume in the beaker = 429 ml

temperature = 20° C

Density = 789 kg/m³

Equilibrium reading = 429

volume change = 29 ml

= 0.029 L

Energy change = mcΔT

U + PΔV

7 0

3 years ago