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3 years ago

07. How do scientists use spectroscopy (the study of light frequencies released

1 answer:

8 0

Every element is able to be recognized individually in many different ways. A very easy and common way is using light absorption also known as spectroscopy. Every atom has electrons, and these electrons like to stay in their lowest-energy configuration. However, when photons collide with an electron it can increase it to a higher energy level.. This is absorption, and each element’s electrons absorb light at specific wavelengths related to the difference between energy levels in that atom. But the electrons want to return to their original levels, so they don’t hold onto the energy for long. When they emit the energy, they release photons with exactly the same wavelengths of light that were absorbed in the first place. An electron can release this light in any direction, so most of the light is emitted in directions away from our line of sight. Therefore, a dark line appears in the spectrum at that particular wavelength.

Because the wavelengths at which absorption lines occur are unique for each element, astronomers can measure the position of the lines to determine which elements are present in a target. The amount of light that is absorbed can also provide information about how much of each element is present.

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What kind of star gives rise to a type i supernova?

larisa86 [58]

a. It occures in binary systems where one of them is a whitedwarf

3 0

3 years ago

Which of the following is a TRUE statement?

pshichka [43]

Answer:

e. All of these statements are false.

Explanation:

As we know that heat transfer take place from high temperature to low temperature.

It is possible to convert all work into heat but it is not possible to convert all heat in to work some heat will be reject to the surrounding.

The first law of thermodynamics is the energy conservation law.

Second law of thermodynamics states that it is impossible to construct a device which convert all energy into work without rejecting the heat to the surrounding.

By using heat pump ,heat can transfer from cooler body to the hotter body.

Therefore all the answer is False.

5 0

3 years ago

A particle moves along the curve below. y = sqrt(1 + x^3) As it reaches the point (2, 3), the y-coordinate is increasing at a ra

blagie [28]

Answer:7 cm/s

Explanation:

Given

Particle move along curve

$y=\sqrt{1+x^3}$

As it reaches the (2,3) its y coordinate is increasing at 14 cm/s

Differentiating y w.r.t time

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3x^2}{2\sqrt{1+x^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

Now at (2,3)

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3\cdot 2^2}{2\sqrt{1+2^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$14=\frac{3\times 4}{2\times \sqrt{9}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$\frac{\mathrm{d} x}{\mathrm{d} t}=7 cm/s$

7 0

3 years ago

8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s

allsm [11]

Answer:

a) $\omega = 50\,\frac{rad}{s}$ , b) $\omega = 0\,\frac{rad}{s}$

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

$\tau = F \cdot r$

$\tau = m\cdot a \cdot r$

$\tau = m \cdot \alpha \cdot r^{2}$

Where $\alpha$ is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

$\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}$

$\alpha = -3\,\frac{rad}{s^{2}}$

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

$\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)$

$\omega = 50\,\frac{rad}{s}$

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

$t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}$

$t = 66.667\,s$

Since $t > 66.667\,s$ , then the angular velocity is equal to zero. Therefore:

$\omega = 0\,\frac{rad}{s}$

7 0

3 years ago

An egg is thrown upward with a velocity of 4.5 m/s. How long will it take to reach it's maximum height?

Ray Of Light [21]

Answer:

0.45 seconds

Explanation:

Letting the value of g = 10 m/s/s

final velocity (v) = 0 m/s (since the egg will come to rest at the maximum height)

initial velocity(u) = 4.5 m/s

acceleration = -10 m/s/s (since the gravity is acting against the egg)

time = t seconds

From the first equation of motion:

v = u + at

0 = 4.5 + (-10)t

t = -4.5 / -10

t = 0.45 seconds

3 0

3 years ago