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nekit [7.7K]
3 years ago
10

A force of 50n is applied to an object at an angle of 50 degrees measured relative to the horizontal. Determine the horizontal a

nd vertical components of the applied force?
Physics
1 answer:
kobusy [5.1K]3 years ago
3 0

Answer:

Fx = 32.14 [N]

Fy  = 38.3 [N]

Explanation:

To solve this problem we must decompose the force vector, for this we will use the angle of 50 degrees measured from the horizontal component.

F = 50 [N]

Fx = 50*cos(50) = 32.14 [N]

Fy = 50*sin(50) = 38.3 [N]

We can verify this result using the Pythagorean theorem.

F = \sqrt{(32.14)^{2}+ (38.3)^{2}} \\F = 50 [N]

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Answer:

The applied torque is 3.84 N-m.      

Explanation:

Given that,

Moment of inertia of the wheel is 2\ kg-m^2

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Final angular speed is 25 rad/s

Time, t = 13 s

The relation between moment of inertia and torque is given by :

\tau=I\alpha \\\\\tau=I\times \dfrac{\omega_f}{t}\\\\\tau=2\times \dfrac{25}{13}\\\\\tau=+3.84\ N-m

So, the applied torque is 3.84 N-m.

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b

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A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
Natasha2012 [34]

Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

5 0
3 years ago
If there is a break at any point in a series circuit the current will
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Answer:

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hope it works out

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3 years ago
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