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Vanyuwa [196]
3 years ago
15

Two forces act on a 8.50-kg object. One of the forces is 14.0 N. If the object accelerates at 3.50 m/s2 , what is the greatest p

ossible magnitude of the other force
Physics
1 answer:
astra-53 [7]3 years ago
7 0

Answer:

The magnitude of the other force is 43.75 N.

Explanation:

Given that,

Mass of the object, m = 8.5 kg

Force 1, F_1=14\ N

Acceleration of the object, a=3.5\ m/s^2

To find,

The greatest possible magnitude of the other force.

Solution,

Let F_2 is the magnitude of other force that is acting on the object. For the greatest force, the two forces must be act in opposite direction such that :

F_2-F_1=ma

F_2=ma+F_1

F_2=14+8.5\times 3.5

F_2=43.75\ N

So, the magnitude of the other force is 43.75 N.

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A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent
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The wavelengths of the constituent travelling waves CANNOT be 400 cm.

The given parameters:

  • <em>Length of the string, L = 100 cm</em>

<em />

The wavelengths of the constituent travelling waves is calculated as follows;

L = \frac{n \lambda}{2} \\\\n\lambda = 2L\\\\\lambda = \frac{2L}{n}

for first mode: n = 1

\lambda = \frac{2\times 100 \ cm}{1} \\\\\lambda = 200 \ cm

for second mode: n = 2

\lambda = \frac{2L}{2} = L = 100 \ cm

For the third mode: n = 3

\lambda = \frac{2L}{3} \\\\\lambda = \frac{2 \times 100}{3} = 67 \ cm

For fourth mode: n = 4

\lambda = \frac{2L}{4} \\\\\lambda = \frac{2 \times 100}{4} = 50  \ cm

Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.

The complete question is below:

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:

A. 400 cm

B. 200 cm

C. 100 cm

D. 67 cm

E. 50 cm

Learn more about wavelengths of travelling waves here: brainly.com/question/19249186

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