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andrew-mc [135]
3 years ago
6

Suponga que usted pone en contacto 2 cuerpos, uno con carga +6 y otro con carga -8. ¿Con qué carga queda cada uno al final?

Physics
1 answer:
soldi70 [24.7K]3 years ago
4 0
What is the question?
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1.020 please help!!<br><br> Mark brainliest
Viktor [21]

Answer:

0.0035 mg

Explanation:

1 mcg = 1e-3 mg

》35 mcg = 35×10^-3 mg

= 0.0035 mg

5 0
3 years ago
Please help me I will give you 20 points !!
nadezda [96]

Answer:

D. The primary source, because it was written by the researcher

Explanation:

One should learn to trust the primary source more because it is the real work of the researcher.

A primary work defines the work of research from his or her experimental findings.

  • The primary source presents experimental data and other subordinates ones to reach a conclusion as seen from the view of the researcher.
  • A secondary source implies someone else documenting their own opinion about the primary experimental set up.
  • This can get dicey in the sense that results can be twisted to serve other ulterior motives.

8 0
3 years ago
A 40kg skier starts at the top of a 12 meter high slope at the bottom she is traveling g 10m/a how much energy does she lose to
nikdorinn [45]
Energy at top = U = mgh = 40 * 9.8 * 12 = 4704 J

Energy at bottom = 1/2 mv² = 1/2 * 40 * 10² = 4000 / 2 = 2000 J

Energy Lost = Final - Initial = 4704 - 2000 = 2704 J

In short, Your Answer would be 2704 Joules

Hope this helps!
3 0
4 years ago
Question 1 (1 point)
MatroZZZ [7]

Answer:

The work done by the frictional force is 600J.

Explanation:

The work W done by the frictional force is

W= Fd.

Now, F = 60N and d =10m; therefore,

W= (60N)(10m)

\boxed{W = 600J.}

Hence, the work done by friction is 660J.

7 0
3 years ago
A rock is dropped from a 110-m-high cliff. How long does it take to fall (a) the first 55.0 m and (b) the second 55.0 m?
Genrish500 [490]

Answer:

a) t = 3.35[s]; b) t = 1.386[s]

Explanation:

We can solve this problem by dividing it into two parts, for the first 55 [m] and then the second part with the remaining 55 [m].

We will take the initial velocity as zero, as the problem does not mention that the Rock was thrown at initial velocity.

And using kinematics equations:

v_{f}^{2}= v_{o}^{2}+2*g*y\\where:\\v_{o}=0\\g=gravity = 9.81[m/s^2]\\y=55 [m]\\v_{f}^{2}=0+2*9.81*55\\v_{f}=\sqrt{2*9.81*55} \\v_{f}=32.85[m/s]

Now we can calculate the time:

v_{f}=v_{o}+g*t\\t=\frac{v_{f}-v_{o}}{g}\\ t=\frac{32.85-0}{9.81}\\ t=3.35[s]

Now we can calculate the second time, but using as a initial velocity 32.85[m/s].

The final velocity will be:

v_{f}^{2}= v_{o}^{2}+2*g*y\\v_{f}=\sqrt{v_{o}^{2}+2*g*y} \\v_{f}=\sqrt{32.85^{2}+2*9.81*55 } \\v_{f}=46.45[m/s]

Now we can calculate the second time:

t=\frac{46.45-32.85}{9.81} \\t= 1.386[s]

Note: The reason the second time is shorter even though it is the same distance is that the acceleration of gravity increases the speed of the rock more and more as it falls.

7 0
3 years ago
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