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Luba_88 [7]
3 years ago
10

Convert 26.4 mi to km

Physics
1 answer:
expeople1 [14]3 years ago
4 0

Answer:

42.48668

Explanation:

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Electroencephalography (EEG), in which electrodes placed on the scalp pick up electrical signals underneath, and magnetoencephal
Alexxx [7]
Does electrical curents flows of charges across the membrane and the depolarization continues also due to charges that reject each other and the flow continues .so there appears a magnetic field as well
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One of the main differences between the intaglio and the relief printing processes is that with intaglio the ink ________ the su
TEA [102]
The answer to this question is "LIES BELOW THE SURFACE" happens or occurs. When one of the main differences between the two which is the Intaglio and the other one is the relief printing processes is that with the Intaglio the ink LIES BELOW the surface of the printing plate.
5 0
3 years ago
Ben hit a 0.25kg nail into a wood with a 5.2kg hammer. If the hammer moves witht the speed of 52 m/s and two fifth of its kineti
Art [367]

Answer:

Explanation:

Mass of nails is 0.25kg

Mass of hammer 5.2kg

Speed of hammer is =52m/s

Then, Ben kinetic energy is given as

K.E= ½mv²

K.E= ½×5.2×52²

K.E= 7030.4J

Given that, two-fifth of kinetic energy is converted to internal energy

Internal energy (I.E) = 2/5 × K.E

Internal energy (I.E) = 2/5 × 7030.4

I.E=2812.16J.

Energy increase is total Kinetic energy - the internal energy

∆Et= K.E-I.E

∆Et= 7030.4 - 2812.16

∆Et= 4218.24J

7 0
3 years ago
A hockey puck has a coefficient of kinetic friction of μk = .35. If the puck feels a normal force (FN) of 5 N, what is the frict
alina1380 [7]

Answer:

The frictional force is  F_f =  1.75 \  N

Explanation:

From the question we are told that

     The coefficient of kinetic force is  μk = 0.35

     The normal force felt by the puck is  F_N  =  5 \  N

Generally the frictional force that acts on the puck is mathematically represented as

          F_f =  \mu_k  *  F_N

=>       F_f =  0.35  *  5

=>       F_f =  1.75 \  N

3 0
3 years ago
If 0.035pC of charge is transferred via the movement of Al3+ ions, how's many of these must be transferred in total? Please add
mr Goodwill [35]

Each Al^+^3 ion contains three extra protons. Hence, the extra charge on each  Al^+^3 = 3 \times 1.6 \times 10^-^1^9 C

Total charge = 0.035 pC

Total charge (Q) = 0.035 \times 10^-^1^2 C

Let the number of Al^+^3 ions be n.

According to question:

n \times 3 \times 1.6 \times 10^-^1^9 =0.035 \times 10^-^1^2

n = \frac{0.035 \times 10^-^1^2}{3 \times 1.6 \times 10^-^1^9}

n = 7.29167 \times 10^4

n = 72917

Hence, the total number of ions needed to be transferred is 72917

3 0
3 years ago
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