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3 years ago

Fire could be considered a living thing

1 answer:

5 0

anything that contains one cell of life is considered a living thing, a fire however does not contain any cells so it is not considered a living thing/organism

hope this helped you :)

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1. Water flows through a hole in the bottom of a large, open tank with a speed of 8 m/s. Determine the depth of water in the tan

lora16 [44]

Answer:

3.26m

Explanation:

See attached file

5 0

3 years ago

A solid ball of radius rb has a uniform charge density rho.

Oksana_A [137]

Note: question B is incomplete.

Complete Question

A solid ball of radius rb has a uniform charge density ρ.

a. What is the magnitude of the electric field E(r) at a distance r>rb from the center of the ball? Express your answer in terms of ρ, rb, r, and ϵ0.

b. What is the magnitude of the electric field E(r) at a distance r<rb from the center of the ball? Express your answer in terms of ρ, r, rb, and ϵ0.

c. Let E(r) represent the electric field due to the charged ball throughout all of space. Which of the following statements about the electric field are true?

1. E(0) = 0.

2. E(rb) = 0

3. lim E(r) = 0.

4. The maximum electric field occurs when r = 0.

5. The maximum electric field occurs when r = rb.

6. The maximum electric field occurs as r to infinity.

Answer:

a) the magnitude of E(r)= ρr³/3 ε₀r²

b) the magnitude at distance r from the centre E(r)= ρr/3 ε ₀ ( if r<rb)

c) statements 1(E(0) = 0), 3(E(0) = 0) and 5(The maximum electric field occurs when r = rb.) are true

Explanation:

given

charge density = ρ , ε

Volume of sphere , V = (⁴/₃)πr³

a) charge density = charge/volume

ρ = q ÷ V

make charge the subject of the formula

∴q = ρ × V= ρ× (⁴/₃)πr³

where r³ = rb³(at distance rb³)

recall

E= q/4πε₀r²

E= ρ × (⁴/₃)πrb³/4πε₀r²

∴E(r)= ρrb³/3 ε ₀r²

(b) The Gaussian surface is inside the ball, therefore, surface only encloses a portion of ball's charge .

The net charge enclosed by the Gaussian surface is different to the of net charge enclosed in (a)

Recall

E= q/4πε₀r²

V= (⁴/₃)πr³

E= ρ × (⁴/₃)πr³/4πε₀r²

∴E(r)= ρr/3 ε₀

limr-----∝

E(r)= 0

The maximum electric field occurs when r=rb.

4 0

4 years ago

¿ qué es un motor de explosión?

ArbitrLikvidat [17]

Answer:

es un motor de combustión interna con encendido por chispa.

5 0

3 years ago

The pulley system below uses a gasoline engine to raise a drill head up through a smooth drill pipe. The engine provides a const

polet [3.4K]

NB: The diagram of the pulley system is not shown but the information provided is sufficient to answer the question

Answer:

Power = 2702.56 W

Explanation:

Let the power consumed be P

Energy expended = E = mgh

height, h = 5 m

E = 80 * 9.8 * 5

E = 3920 J

$Power = \frac{Energy}{time}$

To calculate the time, t

From F = ma

F = 900 N

900 = 80 a

a = 900/80

a = 11.25 m/s²

From the equation of motion, $s = ut + 0.5at^{2}$

The drill head starts from rest, u = 0 m/s

$5 = 0 * t + (0.5*11.25*t^{2} )\\5 = 5.625t^{2}\\t^{2} = 5/5.625\\t^{2} = 0.889\\t = 0.943 s$

Power, P = E/t

P = 3920/0.0.943

P = 4157.79 W

But Efficiency, E = 0.65

P = 0.65 * 4157.79

Power = 2702.56 W

6 0

3 years ago

a toy propeller fan with a moment of inertia of .034 kg x m^2 has a net torque of .11Nxm applied to it. what angular acceleratio

Harman [31]

Answer:

The angular acceleration is $\alpha = 3.235 \ rad/s ^2$

Explanation:

From the question we are told that

The moment of inertia is $I = 0.034\ kg \cdot m^2$

The net torque is $\tau = 0.11\ N \cdot m$

Generally the net torque is mathematically represented as

$\tau = I * \alpha$

Where $\alpha$ is the angular acceleration so

$\alpha = \frac{\tau }{I}$

substituting values

$\alpha = \frac{0.1 1}{ 0.034}$

$\alpha = 3.235 \ rad/s ^2$

6 0

3 years ago