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Levart [38]
3 years ago
9

How do you solve this system of equations. x+y−z = 0 5.08x − 9.96y = 5.31 9.96y − 14.72z = −10.37?

Physics
1 answer:
liraira [26]3 years ago
5 0
Just know this:
x + y < or equal to z
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Read the sentence.
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Answer:

assembled

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Consider a plywood square mounted on an axis that is perpendicular to the plane of the square and passes through the center of t
belka [17]

Answer:

T= 8.061N*m

Explanation:

The first thing to do is assume that the force is tangential to the square, so the torque is calculated as:

T = Fr

where F is the force, r the radius.

if we need the maximum torque we need the maximum radius, it means tha the radius is going to be the edge of the square.

Then, r is the distance between the edge and the center, so using the pythagorean theorem, r i equal to:

r = \sqrt{(0.38m)^2+(0.38m)^2}

r = 0.5374m

Finally, replacing the value of r and F, we get that the maximun torque is:

T = 15N(0.5374m)

T= 8.061N*m

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3 years ago
Answer the following question: “Do the particles in a gas ever slow down and stop? Include information about the three states of
jarptica [38.1K]

Answer:

I hope it is no too late

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hmmm,

In a gas, for example, the molecules are traveling in random directions at a variety of speeds - some are fast and some are slow. ... If more energy is put into the system, the average speed of the molecules will increase and more thermal energy or heat will be produced.

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A car starts from rest and after 7 seconds it is moving at 42 m/s. What is the car’s average acceleration? A. 0.17 m/s2 B. 1.67
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Acceleration =velocity /time
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A 25 kg block is held against a compressed spring and then the spring is allowed to decompress giving the block a velocity. The
Alex787 [66]

Answer:

h=18.05 cm

Explanation:

Given that

m= 25 kg

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x=26.4 cm

θ= 19.5 ∘

When the block just leave the spring then the speed of block = v m/s

From energy conservation

\dfrac{1}{2}Kx^2=\dfrac{1}{2}mv^2

Kx^2=mv^2

v=\sqrt{\dfrac{kx^2}{m}}

By putting the values

v=\sqrt{\dfrac{kx^2}{m}}

v=\sqrt{\dfrac{1300\times 0.264^2}{25}}

v=1.9 m/s

When block reach at the maximum height(h) position then the final speed of the block will be zero.

We know that

V_f^2=V_i^2-2gh

By putting the values

0^2=1.9^2-2\times 10\times h

h=0.1805 m

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