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3 years ago

14

It takes you 8.3 min to walk with an average velocity of 1.6 m/s to the north from the bus stop to the museum entrance. How far

did you walk? Answer in units of m.

1 answer:

Anarel [89]3 years ago

3 0

solution:

1.6 m/s = 96 m/min (in other words, 1.6 m/s x 60 s/min)

96 m/min x 8.3 min = 796.8 m

$s=ut +\frac{1}{2}at^2\\there is no accleration mentioned so,\\s= uv\\8.3\times60=498(s)\\510\times1.6=816(m)$

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A jet airliner moving initially at 693 mph (with respect to the ground) to the east moves into a region where the wind is blowin

Nesterboy [21]

Answer:

The new speed of the aircraft with respect to the ground is 1414.3 mph.

Explanation:

Given that,

Angle = 37°

Velocity of jet airliner = 693 mph

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We know that,

The new velocity of the aircraft with respect to the ground

$v=v_{a}+v_{w}$

We need to calculate the new speed of the aircraft with respect to the ground

Using formula for velocity

$v=\sqrt{(v_{a})^2+(v_{w})^2+2v_{a}v_{w}\cos\theta}$

Put the value into the formula

$v=\sqrt{(693)^2+(798)^2+2\times693\times798\cos37}$

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Hence, The new speed of the aircraft with respect to the ground is 1414.3 mph.

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3 years ago

A 1.25 kg block is attached to a spring with spring constant 17.0 N/m . While the block is sitting at rest, a student hits it wi

Free_Kalibri [48]

Answer:

The amplitude of the subsequent oscillations is 13.3 cm

Explanation:

Given;

mass of the block, m = 1.25 kg

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speed of the block, v = 49 cm/s = 0.49 m/s

To determine the amplitude of the oscillation.

Apply the principle of conservation of energy;

maximum kinetic energy of the stone when hit = maximum potential energy of spring when displaced

$K.E_{max} = U_{max}\\\\\frac{1}{2} mv^2 = \frac{1}{2} kA^2\\\\where;\\\\A \ is \ the \ maximum \ displacement = amplitude \\\\mv^2 = kA^2\\\\A^2 = \frac{mv^2}{k} \\\\A = \sqrt{\frac{mv^2}{k}} \\\\A = \sqrt{\frac{1.25\ \times \ 0.49^2}{17}} \\\\A = 0.133 \ m\\\\A = 13.3 \ cm$

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Answer:

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Explanation:

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3 years ago