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Yuliya22 [10]
3 years ago
6

Which statement describes the relationship of resistance and current

Physics
1 answer:
slavikrds [6]3 years ago
6 0

This is picture describe as

More current mean less resistance

Less resistance mean more current

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To which category does galaxy #1 belong? Why does it belong in this category?
vodka [1.7K]

Answer:

This galaxy belongs to the elliptical galaxy category. This is because it does not have spiral arms.

5 0
3 years ago
An undiscovered planet, many light-years from Earth, has one moon, which has a nearly circular periodic orbit. If the distance f
EleoNora [17]

Answer:

364days

Explanation:

Pls see attached file

Explanation:

8 0
3 years ago
Which characteristic of a sound wave is directly related to the loudness of the sound?
aniked [119]
The amplitude of a sound wave is directly related to the loudness of the sound. 
5 0
3 years ago
Read 2 more answers
(b) If you decrease the length of the pendulum by 25%, how does the new period TN compare to the old period T?
malfutka [58]

Answer:

The new period will be reduced by 50%

Explanation:

The period of pendulum is given by;

T= 2\pi\sqrt{\frac{L}{g} }\\\\\frac{T}{2\pi} = \sqrt{\frac{L}{g} }\\\\(\frac{T}{2\pi} )^2 = {\frac{L}{g}}\\\\\frac{T^2}{4\pi ^2} = {\frac{L}{g}}\\\\T^2(\frac{g}{4\pi ^2}) = L\\\\ \frac{g}{4\pi ^2}= \frac{L}{T^2}\\\\\frac{L_1}{T_1^2} = \frac{L_2}{T_2^2}

When the length is decreased by 25%, the new length L₂ is given by;

L₂ = 25/100(L₁)

L₂ = 0.25L₁

\frac{L_1}{T_1^2} = \frac{L_2}{T_2^2}\\\\T_2^2 = \frac{T_1^2L_2}{L_1} \\\\T_N^2 = \frac{T^2(0.25L_1)}{L_1}\\\\ T_N^2 =0.25T^2\\\\T_N = \sqrt{0.25T^2}}\\\\T_N = 0.5 T

Thus, the new period will be reduced by 50%

8 0
3 years ago
A 1.00-kg mass is attatched to a string 1.0m long and completes a horizontal circle in 0.25s. What is the centripetal accelerati
liraira [26]

-- The string is 1 m long.  That's the radius of the circle that the mass is
traveling in.  The circumference of the circle is  (π) x (2R) = 2π meters .

-- The speed of the mass is (2π meters) / (0.25 sec) = 8π m/s .

-- Centripetal acceleration is  V²/R = (8π m/s)² / (1 m) = 64π^2 m/s²

-- Force = (mass) x (acceleration) = (1kg) x (64π^2 m/s²) =

                                                         64π^2 kg-m/s² = 64π^2 N = about <span>631.7 N .

</span>
That's it.  It takes roughly a 142-pound pull on the string to keep
1 kilogram revolving at a 1-meter radius 4 times a second !<span> 
</span>
If you eased up on the string, the kilogram could keep revolving
in the same circle, but not as fast.

You also need to be very careful with this experiment, and use a string
that can hold up to a couple hundred pounds of tension without snapping. 
If you've got that thing spinning at 4 times per second and the string breaks,
you've suddenly got a wild kilogram flying away from the circle in a straight
line, at 8π meters per second ... about 56 miles per hour !  This could definitely
be hazardous to the health of anybody who's been watching you and wondering
what you're doing.


3 0
3 years ago
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