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tester [92]
2 years ago
13

Which number can each term of the equation be multiplied by to eliminate the fractions before solving? 6 – x + = 6 minus StartFr

action 3 Over 4 EndFraction x plus StartFraction 1 Over 3 EndFraction equals StartFraction one-half EndFraction x plus 5.x + 5 2 3 6 12
Physics
2 answers:
Arte-miy333 [17]2 years ago
10 0

Answer:

12

Explanation:

i dont know hahaha but its 12 i cheat hahaha

zysi [14]2 years ago
5 0

Answer:

We need to multiply 12 to each term to eliminate fractions.

Explanation:

Given expression:

6-\frac{3}{4}x+\frac{1}{3}=\frac{1}{2}x+5

To eliminate the fraction we need to multiply each term by least common multiple of the denominators of the fraction.

The denominators in the above expressions are:

4, 3 and 2

The multiples of each can be listed below.

2⇒ 2,4,6,8,10,<u>12</u>,14,16.....

3⇒ 3,6,9,<u>12</u>,15,18

4⇒ 4,8,<u>12</u>.......

From the list of the multiples stated, we can see the least common multiple is 12.

So we will multiply each term by 12.

Multiplying 12 to both sides.

12(6-\frac{3}{4}x+\frac{1}{3})=12(\frac{1}{2}x+5)

Using distribution,

72-9x+4=6x+60

Thus we successfully eliminated the fractions.

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Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t
yarga [219]

Answer

The rate at which the magnetic field is changing is  [\frac{dB}{dt} ] =  0.000467 T/s

Explanation

From the question we are told that

   The electric field strength is E =  3.5mV/m =  3.5 *10^{-3} \ V/m

    The radius is  r =  1.5 \ m

The rate of change of the  magnetic  field  is mathematically represented as

        \frac{d \phi }{dt}  =  \int\limits^{} {E \cdot dl}

Where dl is change of a unit length

     \frac{d \phi}{dt}  =  A *  \frac{dB}{dt}

Where A is the area which is mathematically represented as

     A = \pi r^2

    So

    E \int\limits^{} {  dl} =  ( \pi r^2) (\frac{dB}{dt} )  

  E L  =  ( \pi r^2) (\frac{dB}{dt} )  

where L is the circumference of the circle which is mathematically represented as

     L = 2 \pi r

So

     E (2 \pi r ) =  (\pi r^2 ) [\frac{dB}{dt} ]

      E  =   \frac{r}{2}  [\frac{dB}{dt} ]

       [\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }

substituting values

      [\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }

      [\frac{dB}{dt} ] =  0.000467 T/s    

8 0
2 years ago
A boy drags a suitcase along the ground with a force of 100 N. If the frictional force opposing the motion of the suitcase is 50
stira [4]
Fortunately, 'force' is a vector.  So if you know the strength and direction
of each force, you can easily addum up and find the 'resultant' (net) force.

When we talk in vectors, one newton forward is the negative of
one newton backward.   Hold that thought, while I slog through
the complete solution of the problem.


            (100 N forward) plus (50 N backward)

        =  (100 N forward) minus (50 N forward)

        =           50 N forward .

That's it.
Is there any part of the solution that's not clear ?

4 0
2 years ago
Because the soles of your shoes have cleats, you can exert a forward force of 100 N even on slippery ice. A 10-kg picnic cooler
Brilliant_brown [7]

Answer:

you must throw 3 snowballs

Explanation:

We can solve this exercise using the concepts of conservation of the moment, let's define the system as formed by the refrigerator and all the snowballs. Let's write the moment

Initial. Before bumping that refrigerator

          p₀ = n m v₀

Where n is the snowball number

Final. When the refrigerator moves

         pf = (n m + M) v

The moment is preserved because the forces during the crash are internal

        n m v₀ = (n m + M) v

        n m (v₀ - v) = M v

        n = M/m    v/(vo-v)

Let's look for the initial velocity of the balls, suppose the person throws them with the maximum force if it slides in the snow (F = 100N), let's use the second law and Newton

          F = m a

          a = F / m

The distance the ball travels from zero speed to maximum speed is the extension of the arm (x = 1 m), let's look kinematically for the speed of the balls when leaving the arm

          v₁² = v₀² + 2 a x

          v₁² = 0+ 2 (100/1) 1

          v₁ = 14.14 m / s

This is the initial speed for the crash

         v₀ = v = 14.14 m / s

  Let's calculate

           n = M/m   v/ (v₀-v)

           n = 10/1   3 / (14.14 -3)

          n = 2.7 balls

you must throw 3 snowballs

7 0
2 years ago
A point charge Q = -400 nC and two unknown point charges, q1 and q2, are placed as shown. Point charge q1 is located 1.3 meters
nalin [4]
120 nC is the answer




Sorry if I’m wrong
6 0
2 years ago
Jenny puts a book on her desk. she lifts the book up with her finger, using a force of 0.5N .The cover is 10cm wide .
zepelin [54]

The turning moment on the cover of the book is 0.05 Nm.

Explanation:

Given:

Force applied (F) = 0.5 N

Distance covered (d) = 10 cm

Converting Distance covered from cm to meter we get (d)= 0.1 m

To find:

Turning Moment (M) on the cover of the book = ?

Formula to be used:                                    

Turning Moment (M) = F × d

                                  = 0.5 × 0.1

                                   = 0.05 Nm

Thus the turning moment on the cover of the book is found to be 0.05 Nm

6 0
2 years ago
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