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Yuki888 [10]
3 years ago
8

Which form of

Physics
1 answer:
Soloha48 [4]3 years ago
3 0
I think it’s C b/c it works for me
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A truck heading east has an initial velocity of 6 m/s. It accelerates at 2 m/s2 for 12 seconds. What distance does the truck tra
gavmur [86]
I believe it would be 144 m!
7 0
3 years ago
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A scene in a movie has a stuntman falling through a floor onto a bed in the room below. The plan is to have the actor fall on hi
IgorC [24]

Answer:

m=17.79Kg

Explanation:

In this process energy must be conserved. On the initial stage, there will be only gravitational potential energy, while on the final stage there will be only elastic potential energy, so they will be equal. We write this as:

U_g=U_e

Which is the same as:

mgh=\frac{k \Delta x^2}{2}

So we can obtain our mass from there, and for our values:

m=\frac{k \Delta x^2}{2gh}=\frac{(65144 N/m)(0.1333m)^2}{2(9.8m/s^2)(3.32m)}=17.79Kg

4 0
3 years ago
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}

7 0
3 years ago
O Which two elements will be the most alike and why?
Pavel [41]
C. Because they are in the same group, so they have the same number of valence (outer) electrons
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An object has an emissivity of 0.725. At temperature T, it radicates heat at a rate of 10 W. What is its rate of heat radiation
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Radiative power scales with the fourth power of temperature, so 160 W.
3 0
3 years ago
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