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N76 [4]
3 years ago
15

A student creates a model by placing raisins in bread dough and allowing the dough to rise for several hours. The student sketch

es the model before and after the dough rises.
The student claims that the model accurately describes the motion of the galaxies in the universe. What evidence from the model supports the student’s claim?


As the dough rises, the size of the raisins increases, resembling the expansion of the galaxies.
As the dough rises, the size of the raisins increases, resembling the expansion of the galaxies.
As the dough rises, the size of the raisins increases, resembling the expansion of the galaxies.

As the dough rises, the distance between the raisins decreases, indicating that the galaxies will collide.
As the dough rises, the distance between the raisins decreases, indicating that the galaxies will collide.

As the dough rises, the size of the raisins decreases, resembling the emission of light from stars.
As the dough rises, the size of the raisins decreases, resembling the emission of light from stars.

As the dough rises, the distance between the raisins increases, indicating that the galaxies are moving apart.
Physics
1 answer:
lbvjy [14]3 years ago
3 0

As the dough rises, the distance between the raisins increases, indicating that the galaxies are moving apart. D

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A model rocket is launched straight upward with an initial speed of 52.0 m/s. It accelerates with a constant upward acceleration
JulsSmile [24]

Explanation:

(a) After the engines stop, the rocket reaches a maximum height at which it will stop and begin to descend in free fall due to gravity.

(b) We must separate the motion into two parts, when the rocket's engines is on  and when the rocket's engines is off.

First we must find the rocket speed when the engines stop:

v_f^2=v_0^2+2ay_1\\v_f^2=(52\frac{m}{s})^2+2(1\frac{m}{s^2})(160m)\\v_f^2=3024\frac{m^2}{s^2}\\v_f=\sqrt{3024\frac{m^2}{s^2}}=54.99\frac{m}{s}

This final speed is the initial speed in the second part of the motion, when engines stop until reach its maximun height. Therefore, in this part the final speed its zero and the value of g its negative, since decelerates the rocket:

v_f^2=v_0^2+2gy_{2}\\y_{2}=\frac{v_f^2-v_0^2}{2g}\\y_{2}=\frac{0^2-(54.99\frac{m}{s})^2}{2(-9.8\frac{m}{s^2})}=154.28m

So, the maximum height reached by the rocket is:

h=y_1+y_2\\h=160m+154.28m=314.28m

(c) In the first part we have:

v_f=v_0+at_1\\t_1=\frac{v_f-v_0}{a}\\t_1=\frac{54.99\frac{m}{s}-52\frac{m}{s}}{1\frac{m}{s^2}}\\t_1=2.99s

And in the second part:

t_2=\frac{v_f-v_0}{g}\\t_2=\frac{0-54.99\frac{m}{s}}{-9.8\frac{m}{s^2}}\\t_2=5.61s

So,  the time it takes to reach the maximum height is:

t_3=t_1+t_2\\t_3=2.99s+5.61s=8.60s

(d) We already know the time between the liftoff and the maximum height, we must find the rocket's time between the maximum height and the ground, therefore, is a free fall motion:

v_f^2=v_0^2+2ay\\v_f^2=0^2+2(9.8\frac{m}{s^2})(314.28m)\\v_f=\sqrt{6159.888\frac{m^2}{s^2}}=78.48\frac{m}{s}

t_4=\frac{v_f-v_0}{g}\\t_4=\frac{78.48\frac{m}{s}-0}{9.8\frac{m}{s^2}}\\t_4=8.01s

So, the total time is:

t=t_3+t_4\\t=8.60s+8.01s\\t=16.61s

7 0
3 years ago
Our Sun emits most of its radiation at a wavelength of 550 nm. If a star were 3.50 times hotter than our Sun, it would emit most
Wittaler [7]

The wavelength of the radiation emitted by the star is 183 nm.

Explanation:

As per Wien's displacement law, the product of emitted wavelength and temperature of the star will be equal to 2.898 × 10⁻³ mK.

T*wavelength = 2.898 * 10^{-3}

So if the wavelength of the emitted radiation by Sun is given as 550 nm, then the temperature of the Sun will be

Temperature of Sun = \frac{2.898*10^{-3} }{wavelength}

Temperature of Sun = \frac{2.898*10^{-3} }{550 * 10^{-9} } = 5.27 * 10^{3} K

Then if the temperature of star is said to be 3.5 times hotter than Sun, then the temperature of Star = 3.5×5.27×10³ = 15.81×10³ K.

With this temperature, the wavelength of the emitted radiation can be found as follows:

Wavelength = \frac{2.898 * 10^{-3} }{15.81 * 10^{3} } =183 nm

So, the wavelength of the radiation emitted by the star is 183 nm.

8 0
3 years ago
A meter stick with a mass of 0.167 kg is pivoted about one end so it can rotate without friction about a horizontal axis. The me
SSSSS [86.1K]

Answer:

Part a)

change in potential energy is given as

\Delta U = 0.82 J

Part B)

angular speed of the rod is given as

\omega = 5.42 rad/s

Part c)

Linear speed of the end of the rod is given as

v = 5.42 m/s

Part d)

when a particle falls from rest to distance d = 1 m

v = 4.42 m/s

Explanation:

Part A)

As we know that the gravitational potential energy change is given as

\Delta U = mgH

\Delta U = 0.167(9.81)(0.5)

\Delta U = 0.82 J

Part B)

As we know that change in gravitational energy is equal to gain in kinetic energy

so we have

\Delta U = \frac{1}{2}I\omega^2

0.82 = \frac{1}{2}(\frac{mL^2}{3})\omega^2

\omega^2 = \frac{6 \times 0.82}{0.167 (1)^2}

\omega = 5.42 rad/s

Part c)

Linear speed of the end of the rod is given as

v = L\omega

v = 5.42 m/s

Part d)

when a particle falls from rest to distance d = 1 m

so we will have

v = \sqrt{2gL}

v = \sqrt{2(9.81)(1)}

v = 4.42 m/s

4 0
3 years ago
Select the best answer for the question. 6. The first step in using time more efficiently is A. studying how successful people u
SIZIF [17.4K]
I would say C.
Developing a plan is important because it will motivate you and tell you when to get things done. It is the most efficient

Hope this helped :)
6 0
3 years ago
Read 2 more answers
How fast can the 140 a current through a 0.200 h inductor be shut off if the induced emf cannot exceed 80.0 v?
Vesna [10]
Recall that to compute for the emf of a circuit given current and inductance, we must recall that 

emf = - M \frac{\Delta I }{\Delta t}

where I is the current (A), M is the mutual inductance (h), and t is the time (ms). Since the current must not exceed 80.0 V, we have

80.0 \geq 0.200(\frac{140}{t})
t \geq \frac{28.0}{80}
t \geq 0.35

From this, we see that it must take at least 0.35 ms so it doesn't exceed 80 V.
Answer: 0.35 ms

7 0
3 years ago
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