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3 years ago

40 POINTS!!! PLEASE HELPP!!!

Physics

2 answers:

NeX [460]3 years ago

7 0

Answer:

Explanation:

if it was B it would say from one to another to another

djyliett [7]3 years ago

7 0

Answer A

Single transformation

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Nuclide X has a higher rate of decay than nuclide Y. Based on this information, which of the following statements must be true?

marshall27 [118]

Answer: Answer down below.

Explanation:

8 0

2 years ago

At constant volume, the heat of combustion of a particular compound is − 3550.0 kJ / mol. When 1.075 g of this compound ( molar

swat32

Answer:

$C=1,25\cdot 10^{5} kJ/^{\circ}C$

Explanation:

First of all let's define the specific molar heat capacity.

$C = \frac{-Q}{n\cdot \Delta T}$ (1)

Where:

Q is the released heat by the system

n is the number of moles

ΔT is the difference of temperature of the system

Now, we can find n with the molar mass (M) the mass of the compound (m).

$n=\frac{m}{M}=6.95\cdot 10^{-3} moles$

Using (1) we have:

$C=\frac{-3550}{6.95\cdot 10^{-3} 4.073}$

$C=1,25\cdot 10^{5} kJ/^{\circ}C$

I hope it helps!

6 0

3 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:

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

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

7 0

2 years ago

If the volume on your TV is low, turning the volume up one click of the remote control will make the TV seem louder than if the

marusya05 [52]

From what we know, we can confirm that this ratio (turning up the volume by one click relative to the TV's overall volume) can be quantified as the Weber fraction.

<h3>What is the Weber fraction?</h3>

This fraction describes the ratio needed for change to a stimulus in which the change is just barely noticeable. This question is a prime example in that it seeks to find out just how low of a difference is needed in TV volume in order for the difference to be noticeable.

Therefore, we can confirm that this ratio (turning up the volume by one click relative to the TV's overall volume) can be quantified as the Weber fraction.

To learn more about Weber visit:

brainly.com/question/5004433?referrer=searchResults

7 0

2 years ago

According to the Big Bang Theory, the density and temperature of the Universe is ___________ now than in the past.

34kurt

Answer:

According to the Big Bang Theory, the density and temperature of the Universe is lower now than in the past.

Hope this answer correct :)

5 0

3 years ago