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liq [111]
3 years ago
14

If a material, such as concrete, can withstand a large applied weight but gives away (ruptures) quickly when a crack forms, then

it is considered _____.
Physics
2 answers:
MrMuchimi3 years ago
6 0
The correct answer is brittle:)
mart [117]3 years ago
5 0
Brittle, weak, unstable.
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An arrow is shot into a falling apple. What type of collision is this an example of?
Ad libitum [116K]

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The answer is B. Inelastic.

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To determine a waves' frequency, you must know the
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The correct answer is<span> number of oscillations in a given period of time

This is measured in what is called the Hertz measurement and the period of time is usually taken to be per second.</span>
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Free 25 points cause why the hell not
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I love nice people. Doing anything interesting for Thanksgiving?

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While traveling along a highway a driver slows from 34 m/s to 17 m/s in 6 seconds. What is the automobiles acceleration?​
xxTIMURxx [149]

Answer:

-2.83 m/s²

Explanation:

  • Initial velocity (u) = 34 m/s
  • Final velocity (v) = 17 m/s
  • Time taken (t) = 6 seconds

❖ Acceleration is defined as the rate of change in velocity with time.

→ a = (v - u)/t

  • v denotes final velocity
  • a denotes acceleration
  • u denotes initial velocity
  • t denotes time

→ a = (17 - 34)/6 m/s²

→ a = -17/6 m/s²

<h3>→ Acceleration = -2.83 m/s²</h3>

(Minus sign implies that the velocity is decreasing.)

5 0
3 years ago
Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin
likoan [24]

Answer:

13.6 N

Explanation:

Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.

Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

So

Mg - T = Ma    (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0     and T - mgcos57 - F = m(0) = 0

Mg - T = 0    (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

Mg - mgcos57 - F = 0

So, Mg - mgcos57 = F  

F = Mg - mgcos57

F = (M - mcos57)g

Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

We find F, thus

F = (2.94 kg - 2.85 kgcos57)9.8 m/s²

F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

6 0
3 years ago
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