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dalvyx [7]
4 years ago
6

Using software, a simulation of a javelin being thrown both on Earth and on the moon is created. In both cases, the velocity and

angle of projection are kept the same. On Earth, the javelin covers a distance y. What distance would it cover on the moon? (The moon’s gravity is about 1/6th that of Earth.) A) 6y B) 6/y C) y/6 D) (y + 6)/2
Physics
2 answers:
strojnjashka [21]4 years ago
7 0
The answer would be A: 6y
Great explanation below.  Keep studying ole Chap
SSSSS [86.1K]4 years ago
6 0
Since the gravity at the moon is around 6 times weaker, we have that the downwards pull is around 6 times smaller. In equations, we have that g of the moon=g of Earth/6 and since the weight is F=m*g, the force is 1/6th of the force on earth. Hence, the distance is going to be around 6 times larger than the distance traversed on Earth and thus the correct relation is A=6y.
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3 years ago
What is the electrical force between q1 and q2? Recall that k = 8.99 × 109 N•meters squared over Coulombs squared.. 4.3 × 10 N 3
Rudiy27

Answer:

Explanation:

Incomplete question but for understanding.

We want to find the electrical force between two charges, then you can use the coulombs law which states that the force of attraction or repulsion between two charges is directly proportional to the product of the two charges and inversely proportional to the square of their distance apart,

So,

F = kq1•q2 / r²

Where k is a constant and it is given as

K = 8.99 × 10^9 Nm²/C²

q1 and q2 are the charges and in this question it is not given, so the question is incomplete. Let assume that,

q1 = - 1.609 × 10^-19 C electron

q2 = 1.609 × 10^-19 C proton

Since unlike charges attract, then it is force of attraction

Also, r is the distance apart and it is not given, let assume the distance between the two charges is 2 × 10^-5m

Then,

F = kq1•q2 / r²

F = 8.99 × 10^9 × 1.609 × 10^-19 × 1.609 × 10^-19 / (2 × 10^-5)²

F = 5.82 × 10^-19 N

7 0
4 years ago
Read 2 more answers
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