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Brilliant_brown [7]
3 years ago
6

PLEASE HELP!!! I have so much work left to do. PLEASE!!!

Physics
1 answer:
Neko [114]3 years ago
6 0

Answer:

the answer is C.

Explanation:

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What is the volts for a cord that has a current of 18 and resistance of 2 ohms
Dvinal [7]

Answer:

9

Explanation:

i think not too sure but yea

5 0
2 years ago
A bug flying horizontally at 0.65 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, th
irina [24]

The angular momentum is defined as,

L=I\omega

Acording to this text we know for conservation of angular momentum that

L_i=L_f

Where L_iis initial momentum

L_f is the final momentum

How there is a difference between the stick mass and the bug mass, we define that

Mass of the bug= m

Mass of the stick=10m

At the point 0 we have that,

L_i=mvl

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

vector from the point of reference (O), and ve is the velocity

At the end with the collition we have

L_f=(I_b+I_s)\omega

Substituting

L_f=(ml^2+\frac{10ml^2}{3})\omega

L_f=\frac{13}{10}ml^2w

m(0.65)l=\frac{13}{10}ml^2 \omega

\omega=\frac{1}{2l}

Applying conservative energy equation we have

\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'

\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)

Replacing the values and solving

l=\frac{13}{0.54g}

Substituting

l=\frac{13}{0.54(9.8)}

l=2.45cm

7 0
3 years ago
Hans Full is pulling on a rope to drag his backpack to school across the ice. He pulls upwards and rightwards with a force of 22
natka813 [3]

Answer:

2420 J

Explanation:

From the question given above, the following data were obtained:

Force (F) = 22.9 N

Angle (θ) = 35°

Distance (d) = 129 m

Workdone (Wd) =?

The work done can be obtained by using the following formula:

Wd = Fd × Cos θ

Wd = 22.9 × 129 × Cos 35

Wd = 22.9 × 129 × 0.8192

Wd ≈ 2420 J

Thus, the workdone is 2420 J.

3 0
3 years ago
A space station, in the form of a wheel 119 m in diameter, rotates to provide an "artificial gravity" of 2.20 m/s2 for persons w
Rina8888 [55]
<span>Well, since it's in the shape of a wheel and the person walks around the edge of it, they must have a centripetal acceleration. Since a=v^2/r you can solve for "v" using 2.20 as your "a" and 59.5 as your "r" (r=half of the diameter).
</span> a=v^2/r 
 v=(a*r)^(1/2)=((2.20)*(59.5))^(1/2)=<span> <span>11.44 m/s.
</span></span><span> After you get "v," plugged that into T=2 pi r/ v. This will give you the 1rev per sec.
</span> T=2 pi r/ v= T=(2)*(pi)*(59.5)/(11.44)= <span> <span>32.68 rev/s
</span></span> Use dimensional analysis to get rev per min (1rev / # sec) times (60 sec/min). 
 (32.68 rev/s)(60 s/min)=<span> <span>1960.74 rev/min
</span></span>
5 0
3 years ago
What kind of energy does a flying bullet have?
Musya8 [376]
It mainly travels by kinetic energy
3 0
3 years ago
Read 2 more answers
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