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oksian1 [2.3K]
2 years ago
10

NEED HELP PLEASE!!!

Physics
1 answer:
kenny6666 [7]2 years ago
3 0

Answer:

The gravitational potential energy the barbells have at the maximum height, is P.E. = 3680·h J = 1.10 × 10⁴ units

Where;

h = The maximum height to which he lifts the barbells

Explanation:

The given parameters are;

The  amount of work done by the championship lifter in lifting the weight, W = 1.10 × 10⁴ units

The weight of the barbells lifted by the championship lifter, N = 3680 N

The gravitational potential energy, P.E., the barbells had at their maximum height of lift is given as follows;

P.E. = m × g × h

Where;

m = The mass of the barbells;

g = The acceleration due to gravity = 9.8 m/s²

h = The maximum height to which the barbells are lifted by the championship weight lifter

m × g = The weight of the barbells = 3680 N

∴ P.E. = 3680 N × h = 3680·h J

By the conservation of energy principle, work done by the championship weight lifter = The maximum gravitational potential energy gained by the barbell = The gravitational potential energy at the maximum height, P.E.

∴ The gravitational potential energy the barbells have at the maximum height, P.E. = 3680·h J = W = 1.10 × 10⁴ units

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Answer:

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Answer:

The magnitude of the net electric field is:

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Explanation:

The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).

On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).

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Let's find first E1 and E2.

The electric field equation is given by:

|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}

Where:

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|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}

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And E2 will be:

|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}

|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}

|E_{2}|=40.39\: N/C

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E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}

E_{net}=\sqrt{80.84^{2}+40.39^{2}}

E_{net}=90.37\: N/c

I hope it helps you!

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