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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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A 10N force pulls to the right and friction opposes 2N. If the object is 20kg,find the acceleraton.
zmey [24]

Force = mass * acceleration

10 N - 2 N = 20 kg * acceleration

8 N = 20 kg * acceleration

8 / 20 = acceleration

2/5 m/s^2 = acceleration

8 0
2 years ago
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Study the free body diagram above. Which scenario below can best be described with this free body diagram? A. a cup is at rest o
vekshin1

Answer: D

Explanation:

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2 years ago
A uniform piece of wire, 20 cm long, is bent in a right angle in the center to give it an L-shape. How far from the bend is the
zlopas [31]

Answer:

the center of mass is 7.07 cm apart from the bend

Explanation:

the centre of mass of a wire of length L is L/2 ( assuming uniform density). Then initially the x coordinate of the centre of mass is

x₁ = L/2 = 20 cm /2 = 10 cm

when the wire is bent in a right angle the coordinates of the new centre of mass will be

x₂ = L₂/2

y₂=  L₂/2

where L₂ is the length of the horizontal piece and vertical piece . Then L₂=L/2

x₂ = L₂/2 = L/4 = 20 cm/4 = 5 cm

y₂= L₂/2 = L/4 = 20 cm/4 = 5 cm

x₂=y₂=X

locating the bend in the origin (0,0) the distance to the centre of mass is

d = √(x₂²+y₂²) = √(2X²) = √2*X=√2*5cm = 7.07 cm

d = 7.07 cm

5 0
2 years ago
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A) The current theory of the structure of the Earth, called plate tectonics, tells us that the continents are in constant motion
suter [353]

A) The mass of the continent is 2.5\cdot 10^{21} kg

B) The kinetic energy is 2016 J

C) The speed of the jogger should be 7.1 m/s

Explanation:

A)

The mass of the continent can be calculated as

m = \rho V

where

\rho = 2800 kg/m^3 is its density

V is its volume

We have to calculate its volume. We know that the continent is represented as a slab of side 5900 km (so its surface is 5900 x 5900, assuming it is a square) and depth of 26 km, so its volume is:

V=(5900 km)^2 (26 km)=9.05\cdot 10^8 km^3 =9.05 \cdot 10^8 \cdot (10^9 m^3/k^3)=9.05\cdot 10^7 m^3

So, the mass of the continent is

m=\rho V = (2800)(9.05\cdot 10^{17})=2.5\cdot 10^{21} kg

B)

The kinetic energy of a body is given by

K=\frac{1}{2}mv^2

where

m is the mass of the body

v is its speed

For the continent, we have:

m=2.5\cdot 10^{21} kg is the mass

v=4 cm/year is the speed

We have to convert the speed into SI units. we have:

1 cm = 0.01 m

1 year = (365)(24)(60)(60) s = 3.15\cdot 10^7 s

So, the speed is

v=4 cm/year = 0.04 m/year \cdot \frac{1}{3.15\cdot 10^7}=1.27\cdot 10^{-9} m/s

Therefore, the kinetic energy is

K=\frac{1}{2}(2.5\cdot 10^{21} kg)(1.27\cdot 10^{-9} m/s)^2=2016 J

C)

Again, the kinetic energy of an object is

K=\frac{1}{2}mv^2

For the jogger in this problem, his mass is

m = 80 kg

And we want its kinetic energy to be equal to that of the continent, so

K = 2016 J

Re-arranging the equation for v, we find what speed the jogger needs to have this kinetic energy:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(2016)}{80}}=7.1 m/s

Learn more about kinetic energy here:

brainly.com/question/6536722

#LearnwithBrainly

8 0
2 years ago
How much gravitational potential energy does a 45.2 kg object have when it is 21.9m above the ground?
Blizzard [7]

Answer:

Explanation:

The formula for gravitational potential energy is

Ep = m · g · h   Assuming that the acceleration is g = 10m/s²

Ep = 45.4 · 10 · 21.9 = 9,942.6 J

God is with you!!!

6 0
2 years ago
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