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AVprozaik [17]
3 years ago
9

How much work is done if a 15 newton suitcase is lifted 2 m ?​

Physics
1 answer:
nirvana33 [79]3 years ago
3 0

Answer:

196 J

Explanation:

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B. Each daughter cell receives its own copy of the parent cell's DNA.

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Cell division is the process in which a parent cell divides, giving rise to two or more daughter cells. It is an essential biological process in many organisms.

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The atmosphere of Neptune and Uranus have a blue color because of which gas?
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They are blue because of hydrogen helium and methane
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A rotating water pump works by taking water in at one side of a rotating wheel, and expelling it from the other side. If a pump
Xelga [282]

Answer:

Explanation:

initial angular velocity, ωo = 0 rad/s

angular acceleration, α = 30.5 rad/s²

time, t = 9 s

radius, r = 0.120 m

let the velocity is v after time 9 s.

Use first equation of motion for rotational motion

ω = ωo + αt

ω = 0 + 30.5 x 9

ω = 274.5 rad/s

v = rω

v = 0.120 x 274.5

v = 32.94 m/s

8 0
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Heat transfer from a hot surface to a cool surface in direct contact is called
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This is an example of conduction
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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}

7 0
3 years ago
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