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N76 [4]
2 years ago
8

How close would you have to bring 1 C of positive chargeand 1 C of negative charge for them to exert forces of 1 N onone another

?
Physics
1 answer:
Harman [31]2 years ago
4 0

Answer:

94,800 m

Explanation:

F = kq1 q2/r^2

1 = 9 x 10^9 x 1 / r^2

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.
eimsori [14]

Explanation:

d =  \frac{1}{2} a {t}^{2}  + vt \\ d = 75 \\ t = 5 \\ v = 0 \\  \\ 75 =  \frac{1}{2}  \times 25a + 0 \\ a = 6 \frac{m}{ {s}^{2} }  \\

7 0
3 years ago
What occurs as a ray of light passes from<br> al inilo water?
iVinArrow [24]

Answer:

this may be wrong but I am not sure

3 0
3 years ago
An empty semi-truck sits on the scale of a weighing station. It is then loaded with 100 irritated pigeons. These two events then
mina [271]

Answer:

The reading in the scale is going to be the same, or if it experiences some change it would be minimum. The reason is because the truck will act as a big closed cage, therefore, when the pigeons fly, the air they move with their wings in order to keep flying,  exerts the same force on the closed cage, that if they were standing on the ground.

If the truck however, allows the air flow, the weight might change, because under this scenario, the air flowing could represent less force exerted on the balance.

Explanation:

4 0
3 years ago
Um ônibus percorre a distância de 480 km, entre Santos e Curitiba, com velocidade escalar média de 80 km/h. De Curitiba a Floria
Mariulka [41]

Answer:

10 h

Explanation:

velocidade é a taxa de variação da distância no tempo. é a razão entre a distância e o tempo

de Santos e Curitiba:

distância (d) de 480 km, velocidade (s) de 80 km/h

s=\frac{d}{t}\\ t=\frac{d}{s} =\frac{480}{80}=6h

de Curitiba e Florianópolis:

distância (d) de 300 km, velocidade (s) de 75 km/h

s=\frac{d}{t}\\ t=\frac{d}{s} =\frac{300}{75}=4h

tempo médio de ônibus entre Santos e Florianópolis = 6h + 4h = 10h

7 0
3 years ago
Read 2 more answers
Calculate the average speed of a complete round trip in which the outgoing 300 km is covered at 93 km/h , followed by a 1.0-h lu
madreJ [45]

Answer : 63 km/h

Explanation :

Outgoing distance = 300 km.

Outgoing speed = 93 km/h.

Break time = 1 h

Return distance = 300 km

Return speed = 56 km/h

Outgoing time  = outgoing distance / outgoing speed

Outgoing time   = 300 km / 93 km/h = 3.225806451612 h

Return time  = return distance / return speed

Return time = 300 km / 56 km/h = 5.357142867142 h

Total distance =. Outgoing distance + return distance travelled

Total distance  = 300 km + 300 km = 600 km

Total time = outgoing time + break time + return time

Total time = 3.225806451612 h + 1 h+ 5. 357142867142 h = 9.582949308754 h

Average speed = total distance / total time

Average speed = 600 km / 9.582949308754 h

Average speed = 62.61120462042 km/h = 63 km/h

3 0
3 years ago
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