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4 years ago

10

Jeremy pulled on the spring scale below. how could he decrease the force needed?

1 answer:

rosijanka [135]4 years ago

5 0

Well since decrease me and down he could use his pulling force with a little less force.

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What component of a longitudinal sound wave is analogous to a trough of a transverse wave?

anzhelika [568]

Explanation:

There are two components of a longitudinal sound wave which are compression and rarefaction. Similarly, there are two components of the transverse wave, the crest, and trough.

The crest of a wave is defined as the part that has a maximum value of displacement while the trough is defined as the part which corresponds to minimum displacement.

While compression is that space where the particles are close together while the rarefaction is that space where the particles are far apart from each other.

So, the refraction or the rarefied part of a longitudinal sound wave is analogous to a trough of a transverse wave.

6 0

3 years ago

Malachi uses some glue to fix a broken plate. He puts the glue under a desk lamp but the glue doesn't dry. Then he puts the glue

Semmy [17]

Answer:

The sun gives off a type of light that carries energy, and the light from the desk lamp does not

Explanation:

I got it off a Quizlet hope this helps!!

4 0

3 years ago

Kettles heated on stoves used to be made of copper. was this a good choice?

zloy xaker [14]

Yes, this was a good choice

Hope that helped,

Stay safe happy & healthy!

5 0

3 years ago

Find the magnitude of the torque that acts on the molecule when it is immersed in a uniform electric field of 6.19×105 N/C with

Ivan

Answer:

$\tau=5.81\times 10^5p\ N-m$

Explanation:

We have,

Electric field, $E=6.19\times 10^5\ N/C$

The electric dipole vector at an angle of 69.9 degrees from the direction of the field.

The torque acting on a molecule is given by :

$\tau=p\times E\\\\\tau=pE\sin\theta$

p is electric dipole moment

$\tau=p\times 6.19\times 10^{5}\times \sin (69.9)\\\\\tau=5.81\times 10^5p\ N-m$

So, the magnitude of the torque acting on the molecule is $5.81\times 10^5p\ N-m$ .

7 0

4 years ago

A copper wire has a square cross section 2.0 mm on a side. The wire is 5.0 m long and carries a current of 2.0 A. The density of

kondor19780726 [428]

Answer:

30.22 hours

Explanation:

Given data:

A= l² = (2 x $10^{-3}$ )² = 4 x $10^{-6}$ m²

Length 'L' = 5m

current ' $I$ ' = 2 A

density of free electrons 'n'= 8.5 x $10^{28}$ /m³

Current Density 'J' = $I$ / A

J= 2/4 x $10^{-6}$

J= 5 x $10^{5}$ A/m²

We can determine the time required for an electron to travel the length of the wire by

T= L/ Vd

Where,

L is length and Vd is drift velocity.

Vd can be defined by J/ n|q|

where,

n is the charge-carrier number density

|q| is is the charge carried by each charge carrier =>1.6 x $10^{-19}$ C

T= L/ Vd

Therefore,

T= L . n|q| / J

T= (4 x 8.5 x $10^{28}$ x |1.6 x $10^{-19}$ |)/5 x $10^{5}$

T= 108800 seconds =>1813.33 minutes

Converting minute into hours:

T= 30.22 hours

Thus, time that is required for an electron to travel the length of the wire is 30.22 hours

4 0

3 years ago