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3 years ago

Microwaves have a higher frequency than radio waves, so why is it that they don't travel faster?

1 answer:

4 0

Answer: Energy requirement or consumption also increases as frequency goes higher. Hence, those low-frequency to mid-frequency waves are commonly referred to as radio waves and essentially, they have longer wavelengths. On the other hand, microwaves have higher frequencies and shorter wavelengths.

Explanation: therefore that's why they don't travel faster.

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What does the 2 mean in the formula 5Mg3(PO4)2?

stira [4]

D. There are two phosphate ions in a molecule of magnesium phosphate

3 0

3 years ago

Read 2 more answers

Three metal fishing weights, each with a mass of 1.00x102 g and at a temperature of 100.0°C, are placed in 1.00x102 g of water a

worty [1.4K]

Answer:

Approximately $0.253\; {\rm J \cdot g^{-1} \cdot K^{-1}}$ assuming no heat exchange between the mixture and the surroundings.

Explanation:

Consider an object of specific heat capacity $c$ and mass $m$ . Increasing the temperature of this object by $\Delta T$ would require $Q = c\, m \, \Delta T$ .

Look up the specific heat of water: $c(\text{water}) = 4.182\; {\rm J \cdot g^{-1} \cdot K^{-1}}$ .

It is given that the mass of the water in this mixture is $m(\text{water}) = 1.00 \times 10^{2}\; {\rm g}$ .

Temperature change of the water: $\Delta T(\text{water}) = (45 - 35)\; {\rm K} = 10\; {\rm K}$ .

Thus, the water in this mixture would have absorbed :

$\begin{aligned}Q &= c\, m\, \Delta T \\ &= 4.182\; {\rm J \cdot g^{-1}\cdot K^{-1}} \\ &\quad \times 1.00 \times 10^{2}\; {\rm g} \times 10\; {\rm K} \\ &= 4.182 \times 10^{3}\; {\rm J}\end{aligned}$ .

Thus, the energy that water absorbed was: $Q(\text{water}) = 4.182 \times 10^{3}\; {\rm J}$ .

Assuming that there was no heat exchange between the mixture and its surroundings. The energy that the water in this mixture absorbed, $Q(\text{water})$ , would be the opposite of the energy that the metal in this mixture released.

Thus: $Q(\text{metal}) = -Q(\text{water}) = -4.182 \times 10^{3}\; {\rm J}$ (negative because the metal in this mixture released energy rather than absorbing energy.)

Mass of the metal in this mixture: $m(\text{metal}) = 3 \times 1.00 \times 10^{2}\; {\rm g} = 3.00 \times 10^{2}\; {\rm g}$ .

Temperature change of the metal in this mixture: $\Delta T(\text{metal}) = (100 - 45)\; {\rm K} = 55\; {\rm K}$ .

Rearrange the equation $Q = c\, m \, \Delta T$ to obtain an expression for the specific heat capacity: $c = Q / (m\, \Delta T)$ . The (average) specific heat capacity of the metal pieces in this mixture would be:

$\begin{aligned}c &= \frac{Q}{m\, \Delta T} \\ &= \frac{-4.182 \times 10^{3}\; {\rm J}}{3.00 \times 10^{2}\; {\rm g} \times (-55\; {\rm K})} \\ &\approx 0.253\; {\rm J \cdot g^{-1} \cdot K^{-1}}\end{aligned}$ .

6 0

3 years ago

What is the formula for displacement

pogonyaev

Displacement is usually given to you as it is, but you can also get displacement through velocity by Δd= Δv*t, where <span>Δv is the change in velocity and t is the change in time.

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4 0

3 years ago

A student stands with both feet on some scales in order to measure his weight.

Neko [114]

Answer:

500 N

Explanation:

Because even if he lift one foot his weight will be same as the pressure applied on the scale will be the same and will not change it is not that the scale measures each foot separately

3 0

3 years ago

Can work be done on a system if there is no motion?

Minchanka [31]

Answer:

D) No, because of the way work is defined

Explanation:

The work done on an object is given by:

$W=Fd cos \theta$

where

F is the force applied on the object

d is the displacement of the object

$\theta$ is the angle between the direction of the force and the displacement

From the formula, we observe that if there is no motion involved, then the displacement of the object is zero:

d = 0

As a result, the product in the formula is also zero, therefore the work done will be zero as well.

5 0

3 years ago