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gladu [14]
2 years ago
14

If you place 1 C of positive charge on Earth and 1 C of negative charge on the moon, 384,500 km away, how much force would the p

ositive charge on Earth experience?
Physics
1 answer:
ivolga24 [154]2 years ago
4 0

Answer:

6.1 x 10^-8 newtons

Explanation:

F = 8.98 *109 *1*1/3845000002

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Calculate the energy released by the electron-capture decay of 5727Co. Consider only the energy of the nuclei (ignore the energy
erma4kov [3.2K]

Answer:

Explanation:

⁵⁷Co₂₇  + e⁻¹  =  ²⁷Fe₂₆

mass defect = 56.936296 + .00055 - 56.935399

= .001447 u

equivalent energy

= 931.5 x .001447 MeV

= 1.3479 MeV .

= 1.35 MeV

energy of gamma ray photons = .14  + .017

= .157 MeV .

Rest of the energy goes to neutrino .

energy going to neutrino .

= 1.35 - .157

= 1.193 MeV.

5 0
3 years ago
2. How can we make the percent error value as minimal as possible?
andriy [413]

Answer:

Percent error can be reduced by improving both your accuracy and precision.

8 0
3 years ago
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A major-league pitcher can throw a ball in excess of 40.1 m/s. If a ball is thrown horizontally at this speed, how much will it
mote1985 [20]

Answer:

The ball will drop 0.881 m by the time it reaches the catcher.

Explanation:

The position of the ball at time "t" is described by the position vector "r":

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

Where:

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

When the ball reaches the catcher, the position vector will be "r final" (see attached figure).

The x-component of the vector "r final", "rx final", will be 17.0 m. We have to find the y-component.

Using the equation of the x-component of the position vector, we can calculate the time it takes the ball to reach the catcher (notice that the frame of reference is located at the throwing point so that x0 and y0 = 0):

x = x0 + v0x · t

17.0 m = 0 m + 40.1 m/s · t

t = 17.0 m/ 40. 1 m/s = 0.424 s

With this time, we can calculate the y-component of the vector "r final", the drop of the ball:

y = y0 + v0y · t + 1/2 · g · t²

Initially, there is no vertical velocity, then, v0y = 0.

y = 1/2 · g · t²

y = -1/2 · 9.8 m/s² · (0.424 s)²

y = -0.881 m

The ball will drop 0.881 m by the time it reaches the catcher.

8 0
3 years ago
A spacecraft is traveling in interplanetary space at a constant velocity. What is the estimated distance traveled by the spacecr
vekshin1

i would say that the correct answer would be cc


3 0
3 years ago
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BRAINLIEST AND MAY DOUBLE POINTS!!!
raketka [301]

Given that:

 Energy of bulb  (Work ) = 30 J,

 Time (t) = 3 sec

The power consumption =  ?

 We know that, Power can be defined as rate of doing work

              Power (P) = Work(Energy supplied) ÷ time

                               = 30 ÷ 3

                              = 10 Watts

<em> The power consumption is 10 W.</em>


3 0
3 years ago
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