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3 years ago

7

A spectroscope prism separates the white light from a star into a very wide spectrum. When widely spread, black lines appear in

the spectrum. Particular elements in a star create specific patterns in the lines.
What can you conclude about a spectroscope from this passage?

a
It allows scientists to determine a star's evolutionary stage.
b
It helps scientists measure a star's mass.
c
It helps scientists detect dark matter.
d
It allows scientists to determine which elements are present in a star.

2 answers:

topjm [15]3 years ago

5 0

Answer:

C. It helps scientists detect dark matter.

Explanation:

I got a 100% on my test. hope it help and I hope your having an awesome day :)

dolphi86 [110]3 years ago

5 0

Answer:

The answer is A)

I hope this helps.

Explanation:

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An insulated Thermos contains 140 cm3 of hot coffee at 85.0°C. You put in a 15.0 g ice cube at its melting point to cool the cof

Pavel [41]

Answer:

$T = 69^o C$

Explanation:

Here at thermal equilibrium we can say that thermal energy given by Hot coffee is equal to the thermal energy absorbed by ice cubes

So here we have

$Q_{ice} = Q_{coffee}$

now since ice cubes are added into coffee when it is at melting temperature

So here we can say that final temperature of coffee is T degree C

Now we have

$m_1L + m_1c_1\Delta T_1 = m_2c_2\Delta T_2$

here we have

$m_1 = 15 gram$

L = 333 kJ/kg = 333 J/g[/tex]

$c_1 = c_2 = 4186 J/kg C = 4.186 J/g C$

$\Delta T_1 = T - 0$

$\Delta T_2 = 85 - T$

now we have

$15(333) + 15(4.186)(T - 0) = 140(4.186)(85 - T)$

$4995 + 62.79T = 49813.4 - 586.04T$

$648.83 T = 44818.4$

$T = 69^o C$

6 0

3 years ago

The ideal mechanical advantage of an inclined plane is 3.5, and its efficiency is 0.6. What is the mechanical advantage of the i

Nadya [2.5K]

Mechanical advantage = ideal mechanical advantage x efficiency = 3.5 x 0.6 = 2.1
The mechanical advantage of the inclined plane is 2.1

7 0

3 years ago

A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this

wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

$r=r_o\times A^{\frac{1}{3}}$

$r_o=1.25 \times 10^{-15} m$ = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

$r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}$

$r=4.6656\times 10^{-15} m=4.6656 fm$

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) $F=k\times \frac{q_1q_2}{a^2}$

k= $9\times 10^9 N m^2/C^2$ = Coulombs constant

$q_1,q_2$ = charges kept at distance 'a' from each other

F = electrostatic force between charges

$q_1=+1.602\times 10^{-19} C$

$q_2=+1.602\times 10^{-19} C$

Force of repulsion between two protons on opposite sides of the diameter

$a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m$

$F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}$

$F=2.6527 N$

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

6 0

3 years ago

The uniform rods AB and BC weigh 24 ky and kg, respectively,and the small wheel at C is of negligible weight. If the wheel ismov

victus00 [196]

The velocity of pin B after rod AB has rotated through 90* is vb = 3.2549 m/s.

<h3>What is Potential and Kinetic energy?</h3>

Potential energy is the energy that is stored in any item or system as a result of its location or component arrangement. The environment outside of the object or system, such as air or height, has no impact on it. In contrast, kinetic energy refers to the energy of moving particles inside a system or an item.

mass of rod, mab = 2.4kg

mass of rod, mbc = 4kg

conservation of energy

$T_{1} + V_{1} = T_{2} + V_{2}$

$h_{ab} = h_{bc} = 0.18m$

potential energy at position 1,

$V1 = m_{ab} gh_{ab} + m_{bc} gh_{bc}$

V1 = 2.5 * 9.81 * 0.18 + 4 * 9.81 * 0.18

V1 = 11.30112

kinetic energy T1 at position 1 is zero

potential energy at position 2 is zero

K.E at position 2,

$T_{2} = \frac{1}{2} l_{ab} w^{2}_{ab} + \frac{1}{2} m_{bc} v^{2}_{G} + \frac{1}{2} lw^{2}_{bc}$

$l_{ab} =\frac{m_{ab} l^{2}_{ab} }{3}$

= 1/3 *4 * (0.36)²

=0.10368kg m²

$l =\frac{m_{bc} l^{2}_{bc} }{12}$

= 1/12 *4 * (0.6)²

=0.12kg m²

on putting the values in above equation we get,

T₂ = 1.0667vb²

0 + 11.30112 = 1.0667vb² + 0

vb = 3.2549 m/s

to learn more about Kinetic and potential energy go to - brainly.com/question/18963960

#SPJ4

5 0

1 year ago

When an automobile moves with constant speed down a highway, most of the power developed by the engine is used to compensate for

sveta [45]

Answer:

F = -4567.40 N

Explanation:

Given that,

The power developed by the engine, P = 196 hp

1 hp = 746 W

196 hp = 146157 W

Speed of the car, v = 32 m/s

Let F is the total friction force acting on the car. The product of force and velocity is called the power developed by the engine. It is given by :

$P=-F\times v$

$F=\dfrac{-P}{v}$

$F=\dfrac{-146157\ W}{32\ m/s}$

F = -4567.40 N

So, the total frictional force acting on the car is 4567.40 N. Hence, this is the required solution.

7 0

3 years ago