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2 years ago

A 22.0 ohm and 75.0 ohm resistor are in parallel, connected to a 5.00 v battery. how much current flows out of the battery

Physics

1 answer:

Julli [10]2 years ago

5 0

Answer:

294 mAmps

Explanation:

The two resistors are equal to 22 * 75 / ( 22+ 75) = 17.01 ohm

V = IR

V/R = I

5 / 17.01 = .294 A

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g When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 3.2 10-15 m before

AnnyKZ [126]

Answer:

Time, $t=1.07\times 10^{-23}\ s$

Explanation:

Given that,

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel $3.2\times 10^{-15}\ m$ before interacting.

Let t is the time interval required for the strong interaction to occur. It will move with the speed of light. So,

$t=\dfrac{d}{c}\\\\t=\dfrac{3.2\times 10^{-15}}{3\times 10^8}\\\\t=1.07\times 10^{-23}\ s$

So, the time interval is $1.07\times 10^{-23}\ s$

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3 years ago

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Answer:

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3 years ago

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The latent heat of fusion for Aluminium is 3.97 x 105. How much energy would be required to melt 0.75 kg of it?

RoseWind [281]

Answer:

$E = 2.9775\times10^5 J$

Explanation:

Given: The latent heat of fusion for Aluminum is $L = 3.97\times10^5 J/Kg$

mass to be malted m = 0.75 Kg

Energy require to melt E = mL

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$E = 2.9775\times10^5 J$

5 0

3 years ago

Calculate the work against gravity required to build the right circular cone of height 4 m and base of radius 1.2 m out of a lig

Nana76 [90]

Answer:

Work done = 35467.278 J

Explanation:

Given:

Height of the cone = 4m

radius (r) of the cone = 1.2m

Density of the cone = 600kg/m³

Acceleration due to gravity, g = 9.8 m/s²

Now,

The total mass of the cone (m) = Density of the cone × volume of the cone

Volume of the cone = $\frac{1}{3}\pi r^2 h$

thus,

volume of the cone = $\frac{1}{3}\pi 1.2^2\times 4$ = 6.03 m³

therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg

The center of mass for the cone lies at the $\frac{1}{4}$ times the total height

thus,

center of mass lies at, h' = $\frac{1}{4}\times4=1m$

Now, the work gone (W) against gravity is given as:

W = mgh'

W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J

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3 years ago

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lord [1]

Answer:

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