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SCORPION-xisa [38]

3 years ago

8

A giant solar flare struck Earth in 1859, but it didn't seem to cause much damage. Why do people worry about one that large hitt

ing now? HELPPPPPP

1 answer:

Leya [2.2K]3 years ago

6 0

Answer:

electronics getting fried do to the electromagnetic surge. (the solar flare can act like an emp)

Explanation:

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Determine el valor de la potencia electrica que experimente un circuito cual se somete a 120 voltios emitidos por accion de las

Allisa [31]

Responder: 480 vatios

Explicación:

La potencia en un circuito eléctrico am puede expresarse como el producto si la corriente y el voltaje en el circuito.

Potencia = corriente × voltaje

Corriente dada = 4 amperios

Voltaje = 120 voltios

Potencia gastada en el circuito = 4 × 120

Potencia gastada en el circuito =

480 vatios

5 0

3 years ago

An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in

valkas [14]

Answer:

Volume of the sample: approximately $\rm 0.6422 \; L = 6.422 \times 10^{-4} \; m^{3}$ .

Average density of the sample: approximately $\rm 2.77\; g \cdot cm^{3} = 2.778 \times 10^{3}\; kg \cdot m^{3}$ .

Assumption:

$\rm g = 9.81\; N \cdot kg^{-1}$ .
$\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ .
Volume of the cord is negligible.

Explanation:

<h3>Total volume of the sample</h3>

The size of the buoyant force is equal to $\rm 17.50 - 11.20 = 6.30\; N$ .

That's also equal to the weight (weight, $m \cdot g$ ) of water that the object displaces. To find the mass of water displaced from its weight, divide weight with $g$ .

$\displaystyle m = \frac{m\cdot g}{g} = \rm \frac{6.30\; N}{9.81\; N \cdot kg^{-1}} \approx 0.642\; kg$ .

Assume that the density of water is $\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ . To the volume of water displaced from its mass, divide mass with density $\rho(\text{water})$ .

$\displaystyle V(\text{water displaced}) = \frac{m}{\rho} = \rm \frac{0.642\; kg}{1.000\times 10^{3}\; kg \cdot m^{-3}} \approx 6.42201 \times 10^{-4}\; m^{3}$ .

Assume that the volume of the cord is negligible. Since the sample is fully-immersed in water, its volume should be the same as the volume of water it displaces.

$V(\text{sample}) = V(\text{water displaced}) \approx \rm 6.422\times 10^{-4}\; m^{3}$ .

<h3>Average Density of the sample</h3>

Average density is equal to mass over volume.

To find the mass of the sample from its weight, divide with $g$ .

$\displaystyle m = \frac{m \cdot g}{g} = \rm \frac{17.50\; N}{9.81\; N \cdot kg^{-1}} \approx 1.78389 \; kg$ .

The volume of the sample is found in the previous part.

Divide mass with volume to find the average density.

$\displaystyle \rho(\text{sample, average}) = \frac{m}{V} = \rm \frac{1.78389\; kg}{6.42201 \times 10^{-4}\; m^{3}} \approx 2.778\; kg \cdot m^{-3}$ .

3 0

4 years ago

Identify the origins of breakdown when using a spectrum analyzer

Zanzabum

Four regions of the electromagnetic spectrum that astronomers use when observing objects in the space are the following enumerated answers.

1. First is Ultraviolet

2. Next is Infrared

3. Then the radio

4. Lastly the Visible lights.

These are the answers to the problem.

7 0

3 years ago

a steel block has a mass of 40g.it is in the form of a cube. each edge is 1.74cm long. calculate the density

Vinil7 [7]

Answer:

d ≈ 7,6 g/cm³

Explanation:

d = m/V = 40g/5,27cm³ ≈ 7,6 g/cm³

V = l³ = (1.74cm)³ ≈ 5,27 cm³

3 0

3 years ago

Two objects are 5 kg and 10 kg respectively, and they're 10 m apart. if the distance between them is increased to 20 m, what hap

Sati [7]

I think it's OPTION A, IT DECREASES.

6 0

3 years ago