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Fantom [35]
3 years ago
13

What are some of the physical properties of stars?

Physics
1 answer:
Arisa [49]3 years ago
8 0
Distance, luminosity, brightness, radius, chemical composition and temperature
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What problems can you imagine arising from a nation of mixed nationalities and faiths? Provide real world examples, if you can.
JulsSmile [24]
Many different problem such as racial slurring and racial profiling
4 0
3 years ago
A vertical spring gun is used to launch balls into the air. If the spring is compressed by 4.9 cm, the ball of mass 5.5 g is lau
AleksandrR [38]

We know, by conservation of energy :

\dfrac{kx^2}{2}=mgh

Therefore,

\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}

Putting given values, we get :

\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}\\\\\dfrac{4.9^2}{x_2^2}=\dfrac{50.2}{2\times 50.2}\\\\x_2^2=2\times 4.9^2\\\\x_2 = 4.9\times \sqrt{2}\\\\x_2=6.93\ cm

Therefore, the spring be compressed to 6.93 cm to send the ball twice as high.

Hence, this is the required solution.

6 0
2 years ago
A rectangular block of steel measures 4cm*2cm*1.5cm*. its mass is 93.6g
nata0808 [166]

Answer:

b

Explanation:

5 0
2 years ago
4
taurus [48]

Answer:

31,360J

Explanation:

Gravitation potential energy (gpe) is calculated from the formula mgh.

That implies, gpe = mgh

Therefore substituting the values of m and h as given in the question, knowing in mine that the acceleration due to gravity( g) is 9.8 N/kg, will give 31,360J

Never forget to put your SI units, because even if your answer is numerical correct, it will be incorrect because it represents no physical quantity.

5 0
3 years ago
Un cubo de madera de densidad 0.780 g/cm³ mide 11.2 cm en un lado. Cuando se coloca en agua, ¿qué altura del bloque flotará sobr
Stolb23 [73]

Answer:

2.464 cm above the water surface

Explanation:

Recall that for the cube to float, means that the volume of water displaced weights the same as the weight of the block.

We calculate the weight of the block multiplying its density (0.78 gr/cm^3) times its volume (11.2^3  cm^3):

weight of the block = 0.78 * 11.2^3  gr

Now the displaced water will have a volume equal to the base of the cube (11.2 cm^2) times the part of the cube (x) that is under water. Recall as well that the density of water is 1 gr/cm^3.

So the weight of the volume of water displaced is:

weight of water = 1 * 11.2^2 * x

we make both weight expressions equal each other for the floating requirement:

0.78 * 11.2^3 = 11.2^2 * x

then x = 0.78 * 11.2 cm = 8.736 cm

This "x" is the portion of the cube under water. Then to estimate what is left of the cube above water, we subtract it from the cube's height (11.2 cm) as follows:

11.2 cm - 8.736 cm = 2.464 cm

6 0
3 years ago
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