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3 years ago

To protect her new two-wheeler, Iroda Bike

Physics

1 answer:

goldfiish [28.3K]3 years ago

6 0

Answer:

The length of chain she is allowed is 1.169 ft

Explanation:

The given parameters are;

The linear density of the chain = 0.83 lb/ft

The weight limit of the chain she wants = 1.4 lb

The formula for linear density = Weight/length

Therefore, in order to keep the chain below 1.4 lb, we have;

Linear density = Weight/length

Therefore;

The maximum length she wants = Weight/(Linear density)

Which gives;

The maximum length she wants = 1.4 lb/(0.83 lb/ft) =1.169 ft

Therefore;

The length of chain she is allowed = 1.169 ft.

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g Consider a (12.5 A) cm long metal bar moving horizontally across a vertical magnetic field at a speed of (2.40 B) m/s. The mag

Misha Larkins [42]

Answer:

Explanation:

emf due to movement of a rod of length L in a perpendicular to magnetic field B with velocity v is given as

emf = BLv

Putting in the given values ,

E = 2.45 x 12.5 x 10⁻² x 2.4

= .735 V

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2 years ago

A particle undergoes a constant acceleration of 3.90 m/s2. After a certain amount of time, its velocity is 12.2 m/s. (Where appl

Law Incorporation [45]

Answer:

$s_{f} = 14.312 m$

Explanation:

Since the particle is experimenting a constant acceleration, the displacement can be found by using this formula:

$v_{f}^{2} = v_{o}^{2} + 2 \cdot a \cdot (s_{f} - s_{o})$

Since $s_{o} = 0 m$ , the equation is simplified to this form:

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Then, the displacement is now isolated:

$s_{f} = \frac{v_{f}^{2}-v_{o}^{2}}{2 \cdot a}$

Terms are replaced herein:

$s_{f} = \frac{(12.2 \frac{m}{s})^{2}-(6.1 \frac{m}{s})^{2}}{2 \cdot (3.90 \frac{m}{s^{2}}) } \\s_{f} = 14.312 m$

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3 years ago

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Hello there,
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Hope this helps :))

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3 years ago

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