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2 years ago

How big is the image produced by the periscope compared to the size of the object

Physics

1 answer:

ipn [44]2 years ago

8 0

As long as both mirrors are set at 45% and the same size then you see the same as is reflected in the upper mirror 

Put a lens in the middle of the tube 

? 

We use mirrors when we drive cars ect 

Normally they are set across from a concealed entrance or one that is hard to see both ways like the inside of a hairpin bend. Sometimes only to help in one direction. 

Sonar which is sound waves that are sent out at a set rate then reflected by objects. The longer the gap between the two the further away it is, They still use periscopes to target boats though. 

The periscope can only reflect what is outside so if you could see it because there is enough light then Yes. If you could not see it because it is dark then No unless you get into Info-Red light or Image Intensifying systems as well

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the gas in a balloon has P=100000 pa and v=0.0279 m^3. if the pressure increases to 120000 pa at constant temperature, what is t

mash [69]

Answer:

New volume of the baloon is 0.02325m^3

Explanation:

To answer this question we need to know the ideal gas law, which says:

p•V = n•R•T

p is pressure, V is volume, n is amount of substance (in moles), R is constant value and T is temperature.

Since it's stated that n and T are constant, and we know that R is a constant too, that means that p•V = constant value. Basically, that means that p1•V1 (pressure and volume before the pressure increase) equals to p2•V2 (pressure and volume after the pressure increase).

That means that:

100000 Pa • 0.0279 m^3 = 120000 Pa • V2. Next, V2= 100000•0.0279/120000. So, V2=0.02325m^3.

6 0

2 years ago

Read 2 more answers

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.}$ )

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1 year ago

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2 years ago

Amy swims 500 m (0.5 Km) in 6 minutes (0.1 hour). What was her speed in Kilometers per hour? A. 8.3 Km/hr B. 5 Km/hr C. 50 Km/hr

kondor19780726 [428]

Speed = Distance ÷ Time so divide .5 km by .1h. .5 km÷.1h=5 km/h, so the answer is B. 5km/h.

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2 years ago

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Austin buys a new moped he travel 3 km south and then 4 km east and then 3km north how far does he need to go to get back to whe

Ira Lisetskai [31]

The first thing to do is to define the origin of the coordinate system as the point at which the moped journey begins.
Then, you must write the position vector:
r = -3j + 4i + 3j
Rewriting
r = 4i
To go back to where you started, you must go
d = -4i
That is to say, must travel a distance of 4Km to the west.
Answer
West, 4km.

4 0

2 years ago