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Anna11 [10]
3 years ago
13

**HELP ASAP PLEASE!**

Physics
2 answers:
guapka [62]3 years ago
8 0
I think, according to Neil DeGrasse Tyson's Cosmos, that it would be either A. or B. That should eliminate the answers that I am pretty sure don't make sense. Though I am pretty sure it should be A...... 

P.S. You should watch Cosmos! It's pretty good!
Setler79 [48]3 years ago
7 0
They use a spectrograph so the answer would be B.
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What type of friction is using chalk in the summer to draw on the ground in Copley square?
Lera25 [3.4K]
The second option rolling friction
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4 years ago
A lens with f = +11cm is paired with a lens with f = −25cm. What is the focal length of the combination?
Zepler [3.9K]

Answer:

19.642 cm

Explanation:

f₁ = Focal length of first lens = 11 cm

f₂ = Focal length of second lens = -25 cm

Combined focal length formula

\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}\\\Rightarrow \frac{1}{f}=\frac{1}{11}+\frac{1}{-25}\\\Rightarrow \frac{1}{f}=\frac{14}{275}\\\Rightarrow f=\frac{275}{14}\\\Rightarrow f=19.642\ cm

Combined focal length is 19.642 cm

5 0
3 years ago
. What is the velocity of a free-<br> falling object after 5 seconds?<br> (Use 10 m/s2 for gravity.)
Viefleur [7K]

Answer:

vf = 50 m/s

Explanation:

The equation for this kinematic problem is:

vf = vi + at

We are given:

a = 10m/s^2

vi = 0m/s

t = 5 sec

vf = ?

Solve for final velocity:

vf = 0 + 10(5)

vf = 50 m/s

8 0
3 years ago
Dave is moving 3 m/s when he crashes his bike into a wall, which stops him in 0.6 seconds. If Dave and his bike have a mass of 9
Arisa [49]
Force = change of momentum / time taken
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7 0
3 years ago
The top of a tower much like the leaning bell tower at Pisa, Italy, moves toward the south at an average rate of 1.4 mm/y. The t
Gemiola [76]

Answer:

\omega=7.16*10^{-13}\frac{rad}{s}

Explanation:

The angular speed is given by:

\omega=\frac{v}{r}

Here v is the linear speed and r is the radius of the circular motion. The height of the tower is equal to the radius of the circular motion of the top of the tower, since is rotating about its base. We need to convert the given linear speed to \frac{m}{s}:

1.4\frac{mm}{y}*\frac{10^{-3}m}{1mm}*\frac{1y}{3.154*10^7s}=4.44*10^{-11}\frac{m}{s}

Now, we calculate the angular speed:

\omega=\frac{4.44*10^{-11}\frac{m}{s}}{62m}\\\omega=7.16*10^{-13}\frac{rad}{s}

8 0
4 years ago
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