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2 years ago

Which of the following are electromagnetic waves?

Physics

2 answers:

olchik [2.2K]2 years ago

5 0

Answer:

Microwaves

Radiowaves.

Lana71 [14]2 years ago

3 0

Answer:

radio waves

microwaves

infrared (IR)

visible light

ultraviolet light (UV)

X-rays

gamma rays

Explanation:

these are the electromagnetic spectrum. they are arranged in decreasing wavelength and increasing energy and frequency.

You might be interested in

You have a 100 ohm resistor. How

sp2606 [1]

Answer:

R2 = 300 Ohms

Explanation:

Let the two resistors be R1 and R2 respectively.

RT is the total equivalent resistance.

Given the following data;

R1 = 100 Ohms

RT = 75 Ohms

To find R2;

Mathematically, the total equivalent resistance of resistors connected in parallel is given by the formula;

$RT = \frac {R1*R2}{R1 + R2}$

Substituting into the formula, we have;

$75 = \frac {100*R2}{100 + R2}$

Cross-multiplying, we have;

75 * (100 + R2) = 100R2

7500 + 75R2 = 100R2

7500 = 100R2 - 75R2

7500 = 25R2

R2 = 7500/25

R2 = 300 Ohms

4 0

2 years ago

Three resistors of resistances, R1=10Ω, R2=5Ω, R3=20Ω are connected in a circuit in such way that same amount of current flows t

Nezavi [6.7K]

Answer

Explanation:

As the three resistors are connected in series, the expression to be used for the

calculation of RT equivalent resistance

is:

RT = R1 + R2 + R3

We replace the data of the statement in the previous expression and it remains:

5 10 15 RT + R1 + R2 + R3 + +

We perform the mathematical operations that lead us to the result we are looking for:

RT - 30Ω

5 0

2 years ago

An object weighs 63.8 N in air. When it is suspended from a force scale and completely immersed in water the scale reads 16.8 N.

I am Lyosha [343]

Answer:

The density of this object is approximately $1.36\; {\rm kg \cdot L^{-1}}$ .

The density of the oil in this question is approximately $0.600\; {\rm kg \cdot L^{-1}}$ .

(Assumption: the gravitational field strength is $g =9.806\; {\rm N \cdot kg^{-1}}$ )

Explanation:

When the gravitational field strength is $g$ , the weight $(\text{weight})$ of an object of mass $m$ would be $m\, g$ .

Conversely, if the weight of an object is $(\text{weight})$ in a gravitational field of strength $g$ , the mass $m$ of that object would be $m = (\text{weight}) / g$ .

Assuming that $g =9.806\; {\rm N \cdot kg^{-1}}$ . The mass of this $63.8\; {\rm N}$ -object would be:

$\begin{aligned} \text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{63.8\; {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 6.506\; {\rm kg}\end{aligned}$ .

When an object is immersed in a liquid, the buoyancy force on that object would be equal to the weight of the liquid that was displaced. For instance, since the object in this question was fully immersed in water, the volume of water displaced would be equal to the volume of this object.

When this object was suspended in water, the buoyancy force on this object was $(63.8\; {\rm N} - 16.8\; {\rm N}) = 47.0\; {\rm N}$ . Hence, the weight of water that this object displaced would be $47.0 \; {\rm N}$ .

The mass of water displaced would be:

$\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{47.0\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 4.793\; {\rm kg}\end{aligned}$ .

The volume of that much water (which this object had displaced) would be:

$\begin{aligned}\text{volume} &= \frac{\text{mass}}{\text{density}} \\ &\approx \frac{4.793\; {\rm kg}}{1.00\; {\rm kg \cdot L^{-1}}} \\ &\approx 4.793\; {\rm L}\end{aligned}$ .

Since this object was fully immersed in water, the volume of this object would be equal to the volume of water displaced. Hence, the volume of this object is approximately $4.793\; {\rm L}$ .

The mass of this object is $6.50\; {\rm kg}$ . Hence, the density of this object would be:

$\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{6.506\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 1.36\; {\rm kg \cdot L^{-1}} \end{aligned}$ .

(Rounded to $\text{$3$ sig. fig.}$ )

Similarly, since this object was fully immersed in oil, the volume of oil displaced would be equal to the volume of this object: approximately $4.793\; {\rm L}$ .

The weight of oil displaced would be equal to the magnitude of the buoyancy force: $63.8\; {\rm N} - 35.6\; {\rm N} = 28.2\; {\rm N}$ .

The mass of that much oil would be:

$\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{28.2\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 2.876\; {\rm kg}\end{aligned}$ .

Hence, the density of the oil in this question would be:

$\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{2.876\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 0.600\; {\rm kg \cdot L^{-1}} \end{aligned}$ .