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3 years ago

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The plane is flying with a velocity of 220m/s when it encounters a 45 m/s wind blowing same direction as the planes motion calul

ate the new velocity of the plane relative to the ground

1 answer:

Inga [223]3 years ago

5 0

Answer:

I have no idea

Explanation:

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A hoop and a solid disc are relased from rest

Murljashka [212]

Answer:

1) The hoop and a solid disc rolling without slipping down an incline plane.

Their final velocities are proportional to their moment of inertia.

The condition for moment of inertia: v = ωR

We will use conservation of energy.

<u>For the hoop:</u>

$K_1 + U_1 = K_2 + U_2\\0 + m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}I\omega_h^2 + 0$

They are released from rest, so their initial kinetic energy is zero. And when they reach the bottom, their final potential energy is also zero.

The moment of inertia of a hoop is

$I_h = m_hR^2$

Let's continue with the energy equations:

$m_h gh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}(m_hR^2)(\frac{v_h^2}{R^2})\\m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}m_hv_h^2\\m_hgh = m_hv_h^2\\v_h = \sqrt{gh}$

Similarly <u>for the solid disk</u> with a moment of inertia of (1/2)mR^2:

$K_1 + U_1 = K_2 + U_2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}I_d\omega_d^2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}(\frac{1}{2}m_dR^2)(\frac{v_d^2}{R^2})\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{4}m_dv_d^2\\m_dgh = \frac{3}{4}m_dv_d^2\\v_d = \sqrt{\frac{4gh}{3}}$

Comparing the final velocities, we can conclude that the solid disk reaches the bottom first.

2) The angular acceleration of the pebble is equal to the angular acceleration of the tire, since they stuck together. We can deduce the angular acceleration of the tire from the linear acceleration of the bicycle.

The kinematics equations states that

$v = v_0 + at\\4.47 = 0 + 2a\\a = 2.235 ~m/s^2$

where a is the linear acceleration.

The relation with the angular and linear acceleration is

$a = \alpha R$

where R is the radius of the tire. Since it is not given in the question, we will leave it as R.

The angular acceleration of the small pebble is

$\alpha = 2.235/R ~m/s^2$

4 0

4 years ago

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A hockey goalie is standing on ice. Another player fires a puck (m = 0.170 kg) at the goalie with a velocity of +41.0 m/s. (a) I

viktelen [127]

Answer:

2991.42 N

Explanation:

For this problem, we'll use the equations: momentum= mass x velocity and impulse = change in momentum, and impulse=force x time.

initial momentum; p1 = 0.17 x 41 = 6.97 kg.m/s

final momentum; p2 = 0, because final velocity is 0 m/s

Thus,

impulse = p1 - p2= 6.97 - 0 = 6.97 kg.m/s

Finally, impulse= Force x time,

Thus, Force = Impulse/time

Force= 6.97/ (2.33 x 10^(-3)) = 2991.42 N

4 0

4 years ago

Plzzzzz help me swear will mark u brainiest

photoshop1234 [79]

Answer:

B

Explanation:

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3 years ago

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Please help i don't understand. will mark brainliest

soldier1979 [14.2K]

Answer:

marblebrainiest plz\c cmarble

Explanation:

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3 years ago

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A vertical spring with stiffness k originally is at rest with no mass attached. Then, a mass M is attached, and the spring rocks

Alina [70]

Answer:

The position of the spring in terms of g, m & k is $x = \frac{m g}{k}$

Explanation:

Stiffness of the spring = k

Mass = m

When a mass m is attached with the spring then spring stretched. in that case the force exerted on the spring is equal to weight of the mass attached.

⇒ Force exerted on the spring F = k x

⇒ m g = k x

⇒ $x = \frac{m g}{k}$

This is the position of the spring in terms of g, m & k.

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4 years ago