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LuckyWell [14K]
3 years ago
13

Why do astronomers use spectroscopes to analyze light from distant objects?

Physics
1 answer:
RoseWind [281]3 years ago
7 0
Spectroscopy — the use of light from a distant object to work out the object is made of — could be the single-most powerful tool astronomers use, says Professor Fred Watson from the Australian Astronomical Observatory. ... "It lets you see the chemicals being absorbed or emitted by the light source.
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What is the name of a variable that you change in an experiment
stealth61 [152]
Independent variable
3 0
3 years ago
Read 2 more answers
Which of the following is true about the speed of light?
Kazeer [188]

Answer:

B. In a vacuum, it is a constant value that does not depend on the observer.

6 0
2 years ago
Final exam answers for physical science apex?
Tju [1.3M]
First table:

1. For height, we usually use cm. If it is a must to use the conversion to solve for number 1, we can first convert the ft into m then to cm and then convert the in to cm and then add the values.

5ftx\frac{0.305m}{1ft}  = 1.525mx\frac{100cm}{1m} =152.5cm

then we convert the remaining 2in into cm and add it up:
2inx\frac{2.54cm}{1in}=5.08cm

152.5cm + 5.08cm = 157.58cm or 157.6cm

Another way to do this is to convert the ft to in and add it up to the remaining 2 in then convert to cm. The answer would be more or less the same. 

2. For weight, we will convert lbs to kg. 

105lbsx\frac{1kg}{2.2lbs} =  \frac{105kg}{2.2} = 47.7kg

3. For distance we will convert mi to km. 

1.1mix\frac{1.61km}{1mi} = 1.8km

4. The unit mph is miles per hour and we will want to convert that into kph or kilometers per hour. You will use the same method that you did above to get it. We do not need to convert hours anymore, so it will stay as is.

65 \frac{mi}{hr}x\frac{1.61km}{1mi} = 104.7\frac{km}{hr}

5. Now for the next one it is very straightforward because 1V = 1V no matter where you go. So if the given is 220V then that means that the converted value is also 220V. 

6. Now to convert Fahrenheit to Celsius, then all you have to do is use the formula given above and fill in what you know. 

The given is  70°F

°C = \frac{5}{9} (F - 32)
     = \frac{5}{9} (70 - 32)
     = \frac{5}{9} (38)
     = \frac{190}{9}
     = 21.1°C

The second table is very easy, all you need to do is convert the values to scientific notation or the other way around. All you need to do is move the decimal till you have a coefficient that is between 1 to 10 and then count how many times you moved the decimal. That will be the number or the exponent of 10. 

1. $150,000 = $1.5 x 10^5

If you moved the decimal to the left, the exponent will be positive. 

Now the other way around. To convert scientific notation to standard form, just move the decimal the number of places indicated by the exponent of 10. If it is positive, you will move to the right and if it is negative, you will move to the left. 

2. 2.59x10^6 = 2,590,000m

Now notice that the next value is less than 1. This means you will be moving the the decimal to the right to make it have a coefficient of 1 to 10. You need to move it 3 times to the right so that means the exponent will be negative. 

3. 0.005in = 5.0 x 10^(-3)in

Now do not forget to put in your units. In math or even in science the unit is EXTREMELY IMPORTANT. 

Hope this helped you and good luck!
4 0
3 years ago
Blood pressure is usually measured by wrapping a closed air-filled jacket equipped with a pressure gage around the upper arm of
Sever21 [200]

Answer:

a) High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

Explanation:

a) <em>High and low pressures in kilopascals</em>:

101.325 kPa equals 760 mm Hg, then, we can obtain the values by a single conversion:

p_{high} = 120\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}

p_{high} = 15.999\,kPa

p_{low} = 80\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}

p_{low} = 10.666\,kPa

High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) <em>High and low pressures in pounds per square inch</em>:

14.696 psi equals 760 mm Hg, then, we can obtain the values by a single conversion:

p_{high} = 120\,mm\,Hg\times \frac{14.696\,psi}{760\,mm\,Hg}

p_{high} = 2.320\,psi

p_{low} = 80\,mm\,Hg\times\frac{14.696\,psi}{760\,mm\,Hg}

p_{low} = 1.547\,psi

High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) <em>High and low pressures in meter water column in meters water column</em>:

We can calculate the equivalent water column of a mercury column by the following relation:

\frac{h_{w}}{h_{Hg}} = \frac{\rho_{Hg}}{\rho_{w}}

h_{w} = \frac{\rho_{Hg}}{\rho_{w}}\times h_{Hg} (Eq. 1)

Where:

\rho_{w}, \rho_{Hg} - Densities of water and mercury, measured in kilograms per cubic meter.

h_{w}, h_{Hg} - Heights of water and mercury columns, measured in meters.

If we know that \rho_{w} = 1000\,\frac{kg}{m^{3}}, \rho_{Hg} = 13600\,\frac{kg}{m^{3}}, h_{Hg, high} = 0.120\,m and h_{Hg, low} = 0.080\,m, then we get that:

h_{w, high} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.120\,m

h_{w, high} = 1.632\,m

h_{w, low} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.080\,m

h_{w, low} = 1.088\,m

High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

4 0
3 years ago
People have proposed driving motors with the earth's magnetic field. This is possible in principle, but the small field means th
NISA [10]

Answer:

Explanation:

Torque on a loop in a magnetic field

Maximum torque = M B

M is magnetic moment  of loop and B is magnetic field

M = area x current

= 3.14 x .10² x i

Maximum torque = M B

1 x 10⁻³ = 3.14 x .10² x i x .65 x 10⁻⁴

i = 490 A

Current = 490 A.

3 0
3 years ago
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