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Ugo [173]
3 years ago
11

Can we use a copper wire instead of eureka wire

Physics
1 answer:
natka813 [3]3 years ago
8 0
Yes you can amd it will actually work better than eureka wire copper wire holds more charge
You might be interested in
The wave property that is related to the height of a wave is the
DedPeter [7]

Answer:

Amplitude

Explanation:

The amplitude is maximum height a wave is measured from its rest position.

3 0
3 years ago
How many infrared photons of frequency 2.57 x 1013 Hz would need to be absorbed simultaneously by a tightly bound molecule to br
victus00 [196]

Answer:

94

Explanation:

f = 2.57 x 10^13 Hz

E = 10 eV = 10 x 1.6 x 10^-19 J = 1.6 x 10^-18 J

Energy of each photon = h f

Where, h is Plank's constant

Energy of each photon = 6.63 x 10^-34 x 2.57 x 10^13 = 1.7 x 10^-20 J

Number of photons = Total energy / energy of one photon

N = (1.6 x 10^-18) / (1.7 x 10^-20) = 94.11 = 94

6 0
2 years ago
Why do the constellations seem to move around in the sky?.
Gnoma [55]

Answer: As Earth spins on its axis, we, as Earth-bound observers, spin past this background of distant stars. As Earth spins, the stars appear to move across our night sky from east to west, for the same reason that our Sun appears to “rise” in the east and “set” in the west.

Explanation:

8 0
2 years ago
Sunlight reflects from a concave piece of broken glass, converging to a point 34 cm from the glass. what is the radius of curvat
wolverine [178]
The rays of light coming from the Sun are parallel to each other, so when they are reflected by the concave piece of glass (which acts as a concave mirror) they converge into the focus of the mirror, which is
f=34 cm
The radius of curvature of a concave mirror is twice its focal length, so in this case it is:
r=2f = 2 \cdot 34 cm=78 cm

6 0
3 years ago
If a cell phone is dropped from a very tall building, how far has the phone fallen after 2.7 seconds, neglecting air resistance?
Zielflug [23.3K]
The free fall of the phone is an uniformly accelerated motion toward the ground, with constant acceleration equal to
g=9.81 m/s^2

So, assuming the downward direction as positive direction of the motion, since the phone starts from rest the distance covered by the phone after a time t is given by
y(t) =  \frac{1}{2}gt^2
And if we substitute t=2.7 s, we find the distance covered:
y(t)=  \frac{1}{2}(9.81 m/s^2)(2.7 s)^2=35.8 m
6 0
2 years ago
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