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Alex17521 [72]
2 years ago
10

FIRST PERSON WILL BE MARKED BRAINLIEST, THANKED, AND RATED A 5!

Physics
2 answers:
Whitepunk [10]2 years ago
7 0

Answer:

Strong force

Explanation:

The protons and neutrons are in an atom by strong force. It is approximately 137 times stronger than electromagnetic force and 10⁵ times stronger than weak force. Also, it is billion times stronger than gravitational force.

An atom consists of electrons, protons and neutrons. The negatively charged particles are electrons. The positively charged particles are protons. Neutrons have no charge.

Strong force is the strongest attractive force. Hence, the force shown in the diagram is strong force. So, the correct option is (C) " strong ".              

svetoff [14.1K]2 years ago
4 0
A is the answer to what your asking
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3 years ago
Fighter jet starting from airbase A flies 300 km east , then 350 km at 30° west of north and then 150km north to arrive finally
Tems11 [23]
A jet fighter flies from the airbase A 300 km East to the point M. Then 350 km at 30° West of North.
It means : at 60° North of West. So the distance from the final point to the line AM is :
350 · cos 60° = 350 · 0.866 = 303.1 km
Let`s assume that there is a line N on AM.
AN = 125 km and NM = 175 km.
And finally jet fighter flies 150 km North to arrive at airbase B.
NB = 303.1 + 150 = 453.1 km
Then we can use the Pythagorean theorem.
d ( AB ) = √(453.1² + 125²) = √(205,299.61 + 15,625) = 470 km
Also foe a direction: cos α = 125 / 470 = 0.266
α = cos^(-1) 0.266 = 74.6°
90° - 74.6° = 15.4°
Answer: The distance between the airbase A and B is 470 km.
Direction is : 15.4° East from the North. 
4 0
3 years ago
n outer space, a constant net force with a magnitude of 140 N is exerted on a 32.5 kg probe initially at rest. A) What accelerat
musickatia [10]

Answer:

a) 4.31 m/s²

b) 215.5 m

Explanation:

a) According to Newton's first law of motion

The net force applied to particular mass produced acceleration, a, according to

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m = 32.5 kg

a = ?

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a = 140/32.5 = 4.31 m/s²

b) Using the equations of motion, we can obtain the distance travelled by the object in t = 10 s

u = initial velocity of the probe = 0 m/s (since it was initially at rest)

a = 4.31 m/s²

t = 10 s

s = distance travelled = ?

s = ut + at²/2

s = 0 + (4.31×10²)/2 = 215.5 m

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Answer:

Average speed is total distance divided by total time.

v = d / t

5 0
2 years ago
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