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3 years ago

As an airplane is flying 20.0 degrees north of east at a speed of 750 km/h. How fast is it moving to the north

Physics

2 answers:

dedylja [7]3 years ago

7 0

<span>in this case velocity of airoplane will be 750*sin20'

</span>

Gwar [14]3 years ago

3 0

Answer:

$v_y=71.25\ m/s$

Explanation:

Given that,

Speed of an airplane, v = 750 km/hr = 208.33 m/s

An airplane is flying 20.0 degrees north of east, $\theta=20^{\circ}$

To find,

Velocity in north direction.

Solution,

Let $v_y$ is the speed of the airplane to the north. The y component of any motion is given by :

$v_y=v\ sin\theta$

$v_y=208.33\ m/s\ sin(20)$

$v_y=71.25\ m/s$

So, the airplane is moving with a speed of 71.25 m/s to the north direction.

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A 0.520-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of

Veseljchak [2.6K]

Answer:

Explanation:

Given that,

Mass attached to spring

M = 0.52kg

Force constant K = 8N/m

Amplitude A = 11.6 cm

a. Maximum speed?

Angular velocity is calculated using

w = √k/m

w = √8/0.52

w = √15.385

w = 3.922rad/s

Then, the relation ship between angular velocity and linear velocity is given as

v = - w•A

v = - 3.922 × 11.6

v = - 45.5 cm/s

Then, the maximum velocity is

vmax = |v|= 45.5cm/s

b. Acceleration a?

Acceleration can be determine using the formula

a = -w²• A

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Magnitude of the acceleration is 178.46cm/s²

c. Speed when the object is at 9.6cm from equilibrium position?

Generally,

The position of the object at equilibrium is

x(t) = A•Cos(wt)

x(t) = 11.6 Cos (3.922t)

Then, when x(t) = 9.6cm

9.6 = 11.6 Cos(3.92t)

Cos(3.922t) = 9.6/11.6

Cos(3.922t) = 0.8276

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Note: the angle is in radiant

3.922t = 0.596

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t = 0.152 second

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v(t) = -45.5Sin(3.922t)

Now, when t =0.152

v(t) = -45.5 Sin(3.922×0.152)

v(t) = -45.5Sin(0.596)

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