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tia_tia [17]
3 years ago
9

Most problems addressed by the technological design process have only one solution true/false

Physics
2 answers:
ludmilkaskok [199]3 years ago
8 0
XD I think its false


Elanso [62]3 years ago
3 0
False for sure!!!!!!!!!
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A spring whose stiffness is 980 N/m has a relaxed length of 0. 50 m. If the length of the spring changes from 0. 25 m to 0. 81 m
Gelneren [198K]

151.9j

Explanation:

PE=1/2kx^2

PE=1/2(980)(.50)= 245j

PE=(1/2)(980)(.81)= 396.9j

396.9- 245= 151.9j

8 0
2 years ago
Ben wants to model how the motion of particles changes with temperature. He considers water in an iron pot. Ben draws a model to
Lilit [14]

Answer:

The particles should move faster and have more space between them.

Explanation:

As the molecules heat, they should start to vibrate with the energy. When they vibrate, the space between them increases.

7 0
4 years ago
A 0.717kg block is attached to a spring with spring constant 19.32N/m. While the block is sitting at rest, a student hits it wit
hjlf

Answer:

15.38 m.

Explanation:

The kinetic energy of the block is equal to potential energy of spring at maximum compression

1/ 2 m V² = 1 /2 K X²

m is mass of block , V is its velocity , K is spring constant and  X is maximum compression or its amplitude.

X = V\times\sqrt{\frac{m}{K} }

Putting the values

x = 79.856\times\sqrt{\frac{.717}{19.32} }

= 15.38 m.

7 0
3 years ago
Find the distance travelled when the initial angular velocity is 3 rad/s, the final angular velocity is 25 rad/s and the time of
Llana [10]

Answer:

Please find attached pdf

Explanation:

Download pdf
5 0
3 years ago
A 3.0 kg ball moving at 8 m/s to the right collides with a 1.0 kilogram ball at rest. After the collision, the 3
amid [387]

Let m₁ = 3.0 kg and v₁ = + 8 m/s (so right is positive), and m₂ = 1.0 kg and v₂ = 0. The total momentum of the two balls before and after collision is conserved, so

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

where v₁' = + 5 m/s and v₂' are the velocities of the two balls after colliding, so

(3.0 kg) (8 m/s) = (3.0 kg) (5 m/s) + (1.0 kg) v₂'

Solve for v₂' :

24 kg•m/s = 15 kg•m/s + (1.0 kg) v₂'

(1.0 kg) v₂' = 9 kg•m/s

v₂' = (9 kg•m/s) / (1.0 kg)

v₂' = + 9 m/s

which is to say, the second ball is given a speed of 9 m/s to the right after colliding with the first ball.

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3 years ago
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