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Eddi Din [679]
3 years ago
5

An object of mass m1​= 5kg placed on a frictionless, horizontal table is connected to a string that passes over a pulley and the

n is fastened to a hanging object of mass m2​= 8kg as shown in figure.
Calculate,
(a) the acceleration (a) of the objects
(b) the tension (T) in the string.

(a) The acceleration of the objects (in m/s2) is

Answer for part 1
(b) The tension in the string (in Newtons) is

Answer for part 2​
Physics
1 answer:
horrorfan [7]3 years ago
5 0

Answer:

a. 6.04 m/s^2 (3 sig. fig.) b.  T = 30.2 N

Explanation:

a) Assume that the acceleration due to gravity is 9.81 m/s^2 downwards

8x9.81 = (5+8)a

a = 6.04 m/s^2 (3 sig. fig.)

b) Using the results from (a)

   T = 5(6.04)

   T = 30.2 N

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Which of the following describes the relationship between the weight of fluid
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2 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}

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3 years ago
Galileo's discovery shows that, to an observer on Earth, Venus has phases just like Earth's moon. How did this observation provi
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<u>Galileo's discovery shows that, to an observer on Earth, Venus has phases just like Earth's moon. This observation provide support for the heliocentric model of the solar system as follows:</u>

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TEST
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Yes yes sir yes I’ll dm is the day I get that I wanna go home and get my
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