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3 years ago

How do I figure out 7.68+3.18÷12=

Physics

1 answer:

Alinara [238K]3 years ago

7 0

Answer:

7.945

Explanation:

do the division first

3.18/12=0.265

then addition

7.68+0.265=

7.945

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Say you want to make a sling by swinging a mass M of 1.9 kg in a horizontal circle of radius 0.042 m, using a string of length 0

padilas [110]

Answer: T= 715 N

Explanation:

The only external force (neglecting gravity) acting on the swinging mass, is the centripetal force, which. in this case, is represented by the tension in the string, so we can say:

T = mv² / r

At the moment that the mass be released, it wil continue moving in a straight line at the same tangential speed that it had just an instant before, which is the same speed included in the centripetal force expression.

So the kinetic energy will be the following:

K = 1/2 m v² = 15. 0 J

Solving for v², and replacing in the expression for T:

T = 1.9 Kg (3.97)² m²/s² / 0.042 m = 715 N

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3 years ago

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Vladimir79 [104]

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3 years ago

A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp

Jobisdone [24]

Answer:

The tension is $T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by hinge $Fx= \frac{11}{4\sqrt{3} } Mg$

Explanation:

From the question we are told that

The mass of the beam is $m_b =M$

The length of the beam is $l = L$

The hanging mass is $m_h = 3M$

The length of the hannging mass is $l_h = \frac{3}{4} l$

The angle the cable makes with the wall is $\theta = 60^o$

The free body diagram of this setup is shown on the first uploaded image

The force $F_x \ \ and \ \ F_y$ are the forces experienced by the beam due to the hinges

Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

$\sum F =0$

Now about the x-axis the moment is

$F_x -T cos \theta = 0$

=> $F_x = Tcos \theta$

Substituting values

$F_x =T cos (60)$

$F_x= \frac{T}{2} ---(1)$

Now about the y-axis the moment is

$F_y + Tsin \theta = M *g + 3M *g ----(2)$

Now the torque on the system is zero because their is no rotation

So the torque above point 0 is

$M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0$

$\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0$

$\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}$

$T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }$

$T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by the hinge is

$F_x= \frac{T}{2} ---(1)$

Now substituting for T

$F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}$

$Fx= \frac{11}{4\sqrt{3} } Mg$

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3 years ago

51. A balloon starts rising from the ground, vertically upwards, uniformly at the rate of 1 m s-1. At the end of 4 seconds, a bo

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Answer:

Explanation:

when a bOdy rises into the air,the time it takes to reach a particular height is the same as the time it will take the body to fall from that height to the ground.

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1 year ago

A 2.00 kg object is moving in a circular path with a radius of 5.00 cm. The object starts from rest and with constant angular ac

Rom4ik [11]

Answer:

0.800 m/s²

Explanation:

First, calculate the angular acceleration:

ω = αt + ω₀

6.00 rad/s = α (3.00 s) + 0 rad/s

α = 2.00 rad/s²

Now calculate the angular velocity at t = 2.00 s:

ω = αt + ω₀

ω = (2.00 rad/s²) (2.00 s) + 0 rad/s

ω = 4.00 rad/s

Calculate the linear velocity:

v = ωr

v = (4.00 rad/s) (0.0500 m)

v = 0.200 m/s

Finally, calculate the centripetal acceleration:

a = v² / r

a = (0.200 m/s)² / (0.0500 m)

a = 0.800 m/s²

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3 years ago

How do I figure out 7.68+3.18÷12=​

How do I figure out 7.68+3.18÷12=