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34kurt
3 years ago
14

Physics question, answer completely with work

Physics
1 answer:
Alenkasestr [34]3 years ago
7 0

The force is 2.0 N east

Explanation:

The impulse exerted by a force is defined as the product between the force itself and the time interval during which the force is applied. Mathematically, it is equal to the change in momentum experienced by the object on which the force is acting:

I=F\Delta t = \Delta p

Where

I is the impulse

F is the force

Delta t is the time interval during which the force is applied

\Delta p is the change in momentum

In this problem,

\Delta t = 3.0 s is the time interval

I=6.0 N\cdot s (east) is the impulse

Therefore, the magnitude of the force is

F=\frac{I}{\Delta t}=\frac{6.0}{3.0}=2.0 N

And the direction is the same as the impulse (east).

Learn more about impulse and change in momentum:

brainly.com/question/9484203

#LearnwithBrainly

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A projectile is launched horizontally from the top a 35.2m high cliff and lands a distance of 107.6m from the base of the cliff.
tankabanditka [31]

Answer:

v_o=40.14\ m/s

Explanation:

<u>Horizontal Launch </u>

It happens when an object is launched with an angle of zero respect to the horizontal reference. It's characteristics are:

  • The horizontal speed is constant and equal to the initial speed v_o
  • The vertical speed is zero at launch time, but increases as the object starts to fall
  • The height of the object gradually decreases until it hits the ground
  • The horizontal distance where the object lands is called the range

We have the following formulas

\displaystyle v_x=v_o

\displaystyle x=v_o.t

\displaystyle v_y=g.t

\displaystyle y=\frac{gt^2}{2}

Where v_o is the initial horizontal speed, v_y is the vertical speed, t is the time, g is the acceleration of gravity, x is the horizontal distance, and y is the height.

If we know the initial height of the object, we can compute the time it takes to hit the ground by using

\displaystyle y=\frac{gt^2}{2}

Rearranging and solving for t

\displaystyle 2y=gt^2

\displaystyle t^2=\frac{2\ y}{g}

\displaystyle t=\sqrt{\frac{2\ y}{g}}

We then replace this value in

\displaystyle x=v_o.t

To get

\displaystyle v_o=\frac{x}{t}

\displaystyle v_o=\frac{x}{\sqrt{\frac{2y}{g}}}

\displaystyle v_o=\sqrt{\frac{g}{2y}}.x

The initial speed depends on the initial height y=32.5 m, the range x=107.6 m and g=9.8 m/s^2. Computing v_o

\displaystyle v_o=\sqrt{\frac{9.8}{2(35.2)}}\ 107.6

The launch velocity is  

\boxed{v_o=40.14\ m/s}

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3 years ago
A. What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8cm from equilibrium and the
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Answer:

a) x = 8.8 cm * cos (9.52 rad/s * t)

b) x = 8.45 cm

Explanation:

This is a Simple Harmonic Motion, and most Simple Harmonic Motion equations start from the equilibrium point. In this question however, we are starting from the max displacement the equations, and thus, it ought to be different.

From the question, we are given that

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x = 8.8 cos (9.52 * 1.7)

x = 8.8 cos (16.184)

x = -8.45 cm

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