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hichkok12 [17]
3 years ago
6

Two balloons having the same charge of 1.2 × 10-6 coulombs each are kept 5.0 × 10-1 meters apart. What is the magnitude of the r

epulsive force between them? (k = 9.0 × 109 newton·meter2/coulombs2)
1.2 × 10-2 newtons

3.2 × 10-3 newtons

6.5 × 10-3 newtons

3.0 × 10-4 newtons

5.2 × 10-2 newtons
Physics
1 answer:
ra1l [238]3 years ago
4 0
Steps: 1)Data: q1=charge of first balloon=1.2 × 10^-6 q2=charge of second balloon=1.2 × 10^-6 Condition:q1=q2 then for convenience,we say q instead of q1 and q2. Radius=r=5*10^-1 Radius is the distance between balloons! Magnitude of repulsive force=F=? 2)Solution: F=kq1q2/r^2 As q1=q2=q So F=kqq/r^2 =kq^2/r^2 Putting values: F=((9*10^9)(1.2 × 10^-6)^2)/(5*10^-1)^2 Using calculator: F=0.05184 N
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Answer:

0.625 c

Explanation:

Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.

In the context,

The relative speed of body 2 with respect to body 1 can be expressed as :

$u'=\frac{u-v}{1-\frac{uv}{c^2}}$

Speed of rocket 1 with respect to rocket 2 :

$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$

$u' = \frac{0.7 c}{1.12}$

u'=0.625 c

Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c

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Why can’t we see black holes? Question 11 options: A.Because they don't want to be seen. B.Because the sun is too far away from
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We can't see black holes because D) no light can get out

Explanation:

Black holes are the result of the gravitational collapse of a supermassive star.

The life of a supermassive star ends with a huge explosion, called supernova, that leaves behing a super-dense core called black hole.

Black holes are the most dense objects of the universe, having a huge mass in a super small size. For this reason, the gravitational force exerted by a black hole in its proximity is so strong that even light is not able to escape from the gravitational field. For this reason, light from a black hole is not able to reach us, and so we are not able to see these objects.

The "edge" of space beyond which light remains "trapped" inside the black hole is called event horizon, and no object can escape this region of space.

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3 years ago
A constant force of 3.2 N to the right acts on a 18.2 kg mass for 0.82 s. (a) Find the final velocity of the mass if it is initi
zheka24 [161]

Explanation:

The given data is as follows.

      F = 3.2 N,      m = 18.2 kg,

      t = 0.82 sec

(a)  Formula for impulse is as follows.

          I = Ft = \Delta P

        Ft = m(v_{f} - v_{i})

or,    v_{f} = \frac{Ft}{m} + v_{i}

Putting the given values into the above formula as follows.

      v_{f} = \frac{Ft}{m} + v_{i}

              = \frac{3.2 \times 0.82}{18.2} + 0

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Therefore, final velocity of the mass if it is initially at rest is 0.144 m/s.

(b)  When velocity is 1.85 m/s to the left then, final velocity of the mass will be calculated as follows.

           Ft = m(v_{f} - v_{i})

or,      v_{f} = \frac{Ft}{m} + v_{i}

                  = \frac{3.2 \times 0.82 sec}{18.2} - 1.85

                  = -1.705 m/s

Hence, we can conclude that the final velocity of the mass if it is initially moving along the x-axis with a velocity of 1.85 m/s to the left is 1.705 m/s towards the left.

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