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melomori [17]
3 years ago
8

The gravity between me (68 kg) and my laptop (0.91 kg)

Physics
1 answer:
zepelin [54]3 years ago
4 0
If you mean gravitational force, then it is GMm/r^2, which is G(68)(.91)/ (the distance between you and the laptop), where G is the universal gravitational constant
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Why is the term minority group no longer effective
Cloud [144]
The term minority group is no longer effective because these groups now make up significant percentages of the total population
8 0
3 years ago
Which phrase describes the thermal energy of a sample of matter?
fiasKO [112]

Answer:

where are the phrase's?

5 0
3 years ago
In still​ water, a boat averages 18 18 miles per hour. it takes the same amount of time to travel 16 miles 16 miles ​downstream,
Vladimir79 [104]
<span>The current is 6 miles per hour.
   Let's create a few equations:
 Traveling with the current:
 (18 + c)*t = 16

   Traveling against the current:
 (18 - c)*t = 8

   Let's multiply the 2nd equation by 2
 (18 - c)*t*2 = 16

   Now subtract the 1st equation from the equation we just doubled.
 (18 - c)*t*2 = 16
 (18 + c)*t = 16

   (18 - c)*t*2 - (18 + c)*t = 0
 Divide both sides by t
 (18 - c)*2 - (18 + c) = 0

   Now solve for c
 (18 - c)*2 - (18 + c) = 0
 36 - 2c - 18 - c = 0
 36 - 2c - 18 - c = 0
 18 - 3c = 0
 18 = 3c
  6 = c

   So the current is 6 mph.
   Let's verify that.
 (18 + 6)*t = 16
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4 0
3 years ago
If the distance between two charges is doubled, by what factor is the magnitude of the electric force changed? F_e final/F_e, in
motikmotik

To solve this problem we will apply the concepts related to Coulomb's law for which the Electrostatic Force is defined as,

F_{initial} = \frac{kq_1q_2}{r^2}

Here,

k = Coulomb's constant

q_{1,2} = Charge at each object

r = Distance between them

As the distance is doubled so,

F_{final} = \frac{kq_1q_2}{( 2r )^2}

F_{final} = \frac{ kq_1q_2}{ 4r^2}

F_{final} = \frac{1}{4} \frac{ kq_1q_2}{r^2}

F_{final} = \frac{1}{4} F_{initial}

\frac{F_{final}}{ F_{initial}} = \frac{1}{4}

Therefore the factor is 1/4

6 0
3 years ago
Consider electrons of kinetic energy 6.0 eV and 600 keV. For each electron, find the de Broglie wavelength, particle speed, phas
irinina [24]

Answer:

For 6.0 eV

0.5 nm, 1.45*10^6 m/s, 6.17*10^10 m/s, 1.45*10^6 m/s

For 600 eV

1.26*10^-3 nm, 2.66*10^8 m/s, 3.37*10^8 m/s, 2.66*10^8 m/s

Explanation:

See attachment for calculation

5 0
3 years ago
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