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ArbitrLikvidat [17]
3 years ago
12

9. Santa brings a special baby a bouncy swing. The child bounces in a harness, with a spring constant k, and is suspended off of

the ground. a. If the spring stretches x1 = 0.27 m from equilibrium while supporting an 7.15-kg child, what is its spring constant, in newtons per meter?
Physics
1 answer:
Orlov [11]3 years ago
4 0

Answer:

259.52N

Explanation:

Force initiated on a spring that causes an extension is related by the expression below;

F= K×e

Where F is the Force

K is the spring constant

e is the extension caused by the spring.

By change of subject formula for K;

K = F/e

Now F is the same as weight and is given by mass× acceleration. In this case acceleration is g=9.8m/s2; this is because the child's mass is going to be under the influence of gravity as it swings up the harness}

Hence W =7.15×9.8=70.07N

Hence K = 70.07/0.27 =259.5185N/m

=259.52N/m to 2 decimal place.

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Answer:

Whenever you see the delta symbol infront of another, it always represents change. In other words:

Δu = u_f - u_i

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3 years ago
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(a) What magnitude point charge creates a 10,000 N/C electric field at a distance of 0.250 m? (b) How large is the field at 10.0
Sergio039 [100]
<h2>Answer:</h2>

(a) 6.95 x 10⁻⁸ C

(b) 6.25N/C

<h2>Explanation:</h2>

The electric field (E) on a point charge, Q, is given by;

E = k x Q / r²              ---------------(i)

Where;

k = constant = 8.99 x 10⁹ N m²/C²

r = distance of the charge from a reference point.

Given from the question;

E = 10000N/C

r = 0.250m

Substitute these values into equation(i) as follows;

10000 = 8.99 x 10⁹ x Q / (0.25)²

10000 = 8.99 x 10⁹ x Q / (0.0625)

10000 = 143.84 x 10⁹ x Q

Solve for Q;

Q = 10000/(143.84 x 10⁹)

Q = 0.00695 x 10⁻⁵C

Q = 6.95 x 10⁻⁸ C

The magnitude of the charge is 6.95 x 10⁻⁸ C

(b) To get how large the field (E) will be at r = 10.0m, substitute these values including Q = 6.95 x 10⁻⁸ C into equation (i) as follows;

E = k x Q / r²

E = 8.99 x 10⁹ x 6.95 x 10⁻⁸ / 10²

E = 8.99 x 10⁹ x 6.95 x 10⁻⁸ / 100

E = 6.25N/C

Therefore, at 10.0m, the electric field will be just 6.25N/C

3 0
3 years ago
If a 3-kg rock is thrown at a speed of 2 m/s in a gravity-free environment (presuming one could be found), then an unbalanced fo
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Answer:

a. False

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Newton's 1st Law of Motion states that an object will remain at rest unless acted upon by an unbalanced force, and an object will remain in motion unless acted upon by an unbalanced force.

Therefore, the unbalanced force of 6 N would not allow the rock to maintain its constant speed.

The answer to this question is A) False.

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