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Ilya [14]
3 years ago
13

Do energy drinks actually provide energy?

Physics
2 answers:
denis-greek [22]3 years ago
6 0

Answer:

Most energy drinks contain large amounts of caffeine, which can provide a temporary energy boost. Some energy drinks contain sugar and other substances. The boost is short-lived, however, and may be accompanied by other problems.

Explanation:

Phoenix [80]3 years ago
4 0

Answer:

kinda

Explanation:

Most energy drinks contain large amounts of caffeine, which can provide a temporary energy boost. Some energy drinks contain sugar and other substances. The boost is short-lived, however, and may be accompanied by other problems.

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An object moves in the eastward direction at constant speed. A net force directed northward acts on the object for 5.0 s. At the
Pie

Answer: D

The final velocity of the object will be directed north of east

Explanation:

8 0
4 years ago
Read 2 more answers
A typical atomic polarizability is 1 × 10-40 (C·m)/(N/C). If the q in p = qs is equal to the proton charge e, what charge separa
Alenkasestr [34]

Answer:

s = 6.25 10⁻²² m

Explanation:

Polarizability is the separation of electric charges in a structure, in the case of the atom it is the result of the separation of positive charges in the nucleus and the electrons in their orbits, macroscopically it is approximated by

                p = q s

               s = p / q

let's calculate

              s = 1 10⁻⁴⁰ / 1.6 10⁻¹⁹

              s = 0.625 10⁻²¹ m

              s = 6.25 10⁻²² m

We see that the result is much smaller than the size of the atom, therefore this simplistic model cannot be taken to an atomic scale.

6 0
3 years ago
A 5-cm-high peg is placed in front of a concave mirror with a radius of curvature of 20 cm.
Andrei [34K]

Answer:

Explanation:

Using the magnification formula.

Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)

M = v/u = H1/H2

v/u = H1/H2...1

3) Given the radius of curvature of the concave lens R = 20cm

Focal length F = R/2

f = 20/2

f = 10cm

Object distance u = 5cm

Object height H2= 5cm

To get the image distance v, we will use the mirror formula

1/f = 1/u+1/v

1/v = 1/10-1/5

1/v = (1-2)/10

1/v =-1/10

v = -10cm

Using the magnification formula

(10)/5 = H1/5

10 = H1

H1 = 10cm

Image height of the peg is 10cm

4) If u = 15cm

1/v = 1/f-1/u

1/v = 1/10-1/15

1/v = 3-2/30

1/v = 1/30

v = 30cm

30/15 = H1/5

15H1 = 150

H1/= 10cm

5) if u = 20cm

1/v = 1/f-1/u

1/v = 1/10-1/20

1/v = 2-1/20

1/v = 1/20

v = 20cm

20/20 = H1/5

20H1 = 100

H1 = 5cm

6) If u = 30cm

1/v = 1/f-1/u

1/v = 1/10-1/30

1/v = 3-1/30

1/v = 2/30

v = 30/2 cm

v =>15cm

15/30 = Hi/5

30H1 = 75

H1 = 75/30

H1 = 2.5cm

4 0
3 years ago
A jogger runs at a speed of 3 m/s. How far does he run in 120 seconds?<br>​
Licemer1 [7]

Answer: *360 mph*

Explanation:

I am pretty sure that it is 360 mph

3 times 120 = 360

8 0
3 years ago
A proton moves with a speed of 1.00 x 106 m/s perpendicular to a magnetic field, B. As a result, the proton moves in circle of r
Softa [21]

Answer:

0.0109 m ≈ 10.9 mm

Explanation:

proton speed = 1 * 10^6 m/s

radius in which the proton moves = 20 m

<u>determine the radius of the circle in which an electron would move </u>

we will apply the formula for calculating the centripetal force for both proton and electron ( Lorentz force formula)

For proton :

Mp*V^2 / rp  = qp *VB   ∴  rp = Mp*V / qP*B    ---------- ( 1 )

For electron:

re = Me*V/ qE * B -------- ( 2 )

Next: take the ratio of equations 1 and 2

re / rp = Me / Mp                                 ( note: qE = qP = 1.6 * 10^-19 C )

∴ re ( radius of the electron orbit )

= ( Me / Mp ) rp

= ( 9.1 * 10^-31 / 1.67 * 10^-27 ) 20

= ( 5.45 * 10^-4 ) * 20

= 0.0109 m ≈ 10.9 mm

5 0
3 years ago
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