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2 years ago

Which of the following would decrease the resistance in a wire?

Physics

1 answer:

padilas [110]2 years ago

6 0

<u>Increase the thickness of the wire</u> would decrease the resistance in a wire

Explanation:

Thicker wires have a larger cross-section that increases the surface area with which electrons can flow unimpeded. The thicker the wire, therefore, the lower the resistance.

Thin wires have very high resistance the reason the thin tungsten in a bulb glows because it is heated from the high resistance of many electrons trying to pass through a very small cross-section.

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Which benefit outweighs the risks in the technological design of headphones?

Dmitriy789 [7]

Answer:

The Answer is A They can damage hearing.

Explanation:

I took the Test

7 0

2 years ago

A 3.00 x 10^2-W electric immersion heater is

andre [41]

Answer

t = 367.77 s = 6.13 min

Explanation:

According to the law of conservation of energy:

$Heat\ Supplied\ By \ Heater = Heat\ Absorbed\ by\ Glass + Heat\ Absorbed\ by\ Water\\Pt = m_gC_g\Delta T_g + m_wC_w\Delta T_w\\$

where,

P = Electric Power of Heater = 300 W

t = time required = ?

m_g = mass of glass = 300 g = 0.3 kg

m_w = mass of water = 250 g = 0.25 kg

C_g = speicific heat of glass = 840 J/kg.°C

C_w = specific heatof water = 4184 J/kg.°C

ΔT_g = ΔT_w = Change in Temperature of Glass and water = 100°C - 15°C

ΔT_g = ΔT_w = 85°C

Therefore,

$(300\ W)(t) = (0.3\ kg)(840\ J/kg.^oC)(85^oC)+(0.25\ kg)(4184\ J/kg.^oC)(85^oC)\\$

8 0

2 years ago

What are the structures shown below ??

forsale [732]

The body system on the chart

8 0

3 years ago

If a ball is thrown straight up into the air with an initial velocity of 65 ft/s, its height in feet after t seconds is given by

fgiga [73]

Answer:

a) $v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s$

$v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s$

$v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s$

$v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s$

b) $v=65-32(2)=1ft/s$

Explanation:

From the exercise we got the ball's equation of position:

$y=65t-16t^{2}$

a) To find the average velocity at the given time we need to use the following formula:

$v=\frac{y_{2}-y_{1} }{t_{2}-t_{1} }$

Being said that, we need to find the ball's position at t=2, t=2.5, t=2.1, t=2.01, t=2.001

$y_{t=2}=65(2)-16(2)^{2} =66ft$

$y_{t=2.5}=65(2.5)-16(2.5)^{2} =62.5ft$

$v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s$

$y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft$

$v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s$

$y_{t=2.01}=65(2.01)-16(2.01)^{2} =66.0084ft$

$v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s$

$y_{t=2.001}=65(2.001)-16(2.001)^{2} =66.001ft$

$v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s$

b) To find the instantaneous velocity we need to derivate the equation

$v=\frac{df}{dt}=65-32t$

$v=65-32(2)=1ft/s$

7 0

2 years ago

The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 3 months. The line

Sladkaya [172]

Answer:

Explanation:

given

T = 3months = 7.9 × 10⁶s

orbital speed = 88 × 10³m/s

V= 2πr÷T

∴ r = (V×T) ÷ 2π

r = (88km × 7.9 × 10⁶s) ÷ 2π

r = 1.10 × 10⁸km

using kepler's 3rd law

mass of both stars = (seperation diatance)³/(orbital speed)²

M₁ + M₂ = (2r)³/( $\frac{1}{4}$ year)²

= (1.06 × 10²⁵)/(6.2×10¹³)

1.71×10¹²kg

since M₁ = M₂ =1.71×10¹²kg ÷ 2

M₁ = M₂ = 8.55×10¹¹kg

6 0

2 years ago