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Lady_Fox [76]
2 years ago
12

Please help me on this!

Physics
1 answer:
aalyn [17]2 years ago
5 0

Answer:

erosion

Explanation:

grows into the eirth each year. the atmosphere and how the worldchanges every year

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There is photons in low energy waves
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A cylinder is given a push and then rolls up an inclined plane. If the origin is the starting point, sketch the position, veloci
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A 2 kg mass is free falling in the negative Y direction when a 10 N force is exerted in the minus X direction. What is the accel
lara31 [8.8K]

Answer:

The mass's acceleration is 5 m/s^2 in the minus X direction and 9,8 m/s^2 in the minus Y direction.

Explanation:

By applying the second Newton's law in the X and Y direction we found that in the minus X direction an external force of 10 N is exerted, while in the minus Y direction the gravity acceleration is acting:

X-direction balance force: -10 [N] = m.ax

Y-direction balance force: -m*9,8 \frac{m}{s^2} = m.ay

Where ax and ay are the components of the respective acceleration and m is the mass. By solving for each acceleration:

ax=(-10 [N]) / m

ay=-m*9,8\frac{m}{s^2} / m

Note that for the second equation above the mass is cancelled and, the Y direction acceleration is minus the gravity acceleration:

ay=-9,8\frac{m}{s^2}

For the x component aceleration we must replace the Newton unit:

N =\frac{kg.m}{s^2}

ax= -10 \frac{kg.m}{s^2} / (2 kg)

ax= - 5 \frac{m}{s^2}

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3 years ago
What happens to jetstream’s as they get closer to the equator
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6 0
3 years ago
Read 2 more answers
A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in m
ArbitrLikvidat [17]

Answer:

<h2>206.67N</h2>

Explanation:

The sum of force along both components x and y is expressed as;

\sum Fx = ma_x  \ and \ \sum Fy = ma_y

The magnitude of the net force which is also known as the resultant will be expressed as R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

a_x = \frac{d^2 x }{dt^2}

a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4  )\\\\a_x = \frac{d}{dt}(12t  )\\\\a_x = 12m/s^{2}

Similarly,

a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2}   )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2

\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N

R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

7 0
3 years ago
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