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pochemuha
2 years ago
11

One of the great triumphs of spectroscopy was when astronomers identified a new element in the sun (one that was only later foun

d on earth). today, this element is called?
Physics
1 answer:
Nezavi [6.7K]2 years ago
6 0

Answer:

When Helium is identified by astronomers is one of the great triumphs of spectroscopy.

Explanation:

Janssen managed to do this great triumphs on August 18, 1868 . Janssen was the first person to introduce the helium, an element that never seen before on Earth, in the solar spectrum. At that time, he didn’t know that what he’d seen—he just think that it was something new. In the mid of 1800, the spectroscope instrument is introduced in astronomy.

Later on we heard that all helium in Universe has been created by the fusion of hydrogen nuclei.

To know more about Spectroscopy:

brainly.com/question/14677550

#SPJ4

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A cruise ship made a trip to Guam and back. The trip there took 12 hours and the trip
Alecsey [184]

Answer:

The average speed is 8.0 km/h

Explanation:

We use the data for the return journey to calculate the distance travelled using the constant velocity equation:

s  =  v  t  =6*4=24km

 

Note I didn't change any units so the answer comes out in kilometres.

Now use the distance and time taken to travel to Guam to find the average speed:

v=st=24/3=8.0km/h

6 0
3 years ago
What are the density, specific gravity and mass of the air in a room whose dimensions are 4 m * 6 m * 8 m at 100 kPa and 25 C.
Volgvan

Answer:

Density = 1.1839 kg/m³

Mass = 227.3088 kg

Specific Gravity = 0.00118746 kg/m³

Explanation:

Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³

Now, from tables, density of air at 25°C is 1.1839 kg/m³

Now formula for density is;

ρ = mass(m)/volume(v)

Plugging in the relevant values to give;

1.1839 = m/192

m = 227.3088 kg

Formula for specific gravity of air is;

S.G_air = density of air/density of water

From tables, density of water at 25°C is 997 kg/m³

S.G_air = 1.1839/997 = 0.00118746 kg/m³

4 0
3 years ago
A 10-cm-long spring is attached to theceiling. When a 2.0 kg mass is hung from it,the spring stretches to a length of 15 cm.a.Wh
alekssr [168]

(a) 392 N/m

Hook's law states that:

F=k\Delta x (1)

where

F is the force exerted on the spring

k is the spring constant

\Delta x is the stretching/compression of the spring

In this problem:

- The force exerted on the spring is equal to the weight of the block attached to the spring:

F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N

- The stretching of the spring is

\Delta x=15 cm-10 cm=5 cm=0.05 m

Solving eq.(1) for k, we find the spring constant:

k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m

(b) 17.5 cm

If a block of m = 3.0 kg is attached to the spring, the new force applied is

F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N

And so, the stretch of the spring is

\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm

And since the initial lenght of the spring is

x_0 = 10 cm

The final length will be

x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm

8 0
3 years ago
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A beam of electrons with of wavelength of 7.5 x 10-6 m is incident on a pair of narrow rectangular slits separated by 0.75 mm. T
ale4655 [162]
Just try your hardest
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3 years ago
A mars prototype carrying scientific instruments has a mass of 1060kg. a net force of 52000 n is applied to this roverat a test
Nitella [24]

F = 52000 N

m = 1060 kg

a= F/m = 52000 N/1060 kg = 49.0566 m/s^2

8 0
4 years ago
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