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azamat
3 years ago
9

Name the plate boundary responsible for this. Please help!!

Physics
2 answers:
zheka24 [161]3 years ago
8 0

Hello,

Contintial crust


Brainliest!?!


~Rendorforestmusic

Korolek [52]3 years ago
5 0
Continental crust plate name
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An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic
malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

\omega_f = \omega_0 + \alpha t

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\omega_f =Final Angular Velocity

\omega_0 =Initial Angular velocity

\alpha = Angular acceleration

t = time

The relation between the tangential acceleration is given as,

a = \alpha r

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s

\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s

t = 11s

Replacing the previous equation with our values we have,

\omega_f = \omega_0 + \alpha t

9.8436 = 6.5973 + \alpha (11)

\alpha = \frac{9.8436- 6.5973}{11}

\alpha = 0.295rad/s^2

The tangential velocity then would be,

a = \alpha r

a = (0.295)(0.2)

a = 0.059m/s^2

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

\omega_f^2=\omega_0^2+2\alpha\theta

Replacing with our values and re-arrange to find \theta,

\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}

\theta = \frac{9.8436^2-6.5973^2}{2*0.295}

\theta = 90.461rad

That is equal in revolution to

\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev

The linear displacement of the system is,

x = \theta*(2\pi*r)

x = 14.397*(2\pi*\frac{0.25}{2})

x = 11.3m

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3 years ago
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2 years ago
A golf ball is struck with a five iron on level ground. it lands 100.0 m away 4.60 s later. what was the magnitude and direction
finlep [7]

consider the motion in x-direction

v_{ox} = initial velocity in x-direction = ?

X = horizontal distance traveled = 100 m

a_{x} = acceleration along x-direction = 0 m/s²

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Using the equation

X = v_{ox} t + (0.5) a_{x} t²

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consider the motion along y-direction

v_{oy} = initial velocity in y-direction = ?

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Using the equation

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0 = v_{oy} (4.60) + (0.5) (- 9.8) (4.60)²

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v_{o} = sqrt((v_{ox})² + (v_{oy})²)

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direction: θ = tan⁻¹(22.54/21.7) = 46.12 deg

6 0
3 years ago
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