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klemol [59]
2 years ago
9

In which mechanical test is a specimen deformed with a gradually increasing load that is applied uniaxially along the long axis

of the specimen that contracts the specimen, i.e. the specimen is squeezed?
Physics
1 answer:
IRINA_888 [86]2 years ago
7 0

Answer:

Compression Test

Explanation:

The Specimen is undergoing a compression test. It is similar to tensile test with the difference that the force is compressive and applied along the direction of stress. Both Tensile and compression tests are performed on Universal Testing machine. Compression test is done to determine the product's reaction when it is compressed, squashed and crushed.

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A spring scale exerts a net force of 8.5 N on an object. What is the object's mass if it has an acceleration of
Amiraneli [1.4K]

Answer:

mass of the object is 2.18 kg

Explanation:

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Force (F) = 8.5 N = 8.5 kg.m/s^{2}

acceleration (a) = 3.9 m/s^{2}

Mass (m) = ?

We know that the newton's second law of motion gives the relation between mass of ab object. force acted upon and the amount the object is accelerated. It is expressed in the form of an equation:

F = ma

mass, m = F/a

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4 0
3 years ago
A ball is hit with a paddle, causing it to travel straight upward. It takes 2.90 s for the ball to reach its maximum height afte
zmey [24]

Answer:

A. 28.42 m/s

B. 41.21 m

Explanation:

From the question given above, the following data were obtained:

Time (t) to reach the maximum height = 2.90 s

Initial velocity (u) =?

Maximum height (h) =?

A. Determination of the initial velocity of the ball.

Time (t) to reach the maximum height = 2.90 s

Final velocity (v) = 0 m/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =?

v = u – gt (since the ball is going against gravity)

0 = u – (9.8 × 2.9)

0 = u – 28.42

Collect like terms

0 + 28.42 = u

u = 28.42 m/s

Thus, the initial velocity of the ball is 28.42 m/s

B. Determination of the maximum height reached by the ball.

Final velocity (v) = 0 m/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) = 28.42 m/s

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 28.42² – (2 × 9.8 × h)

0 = 807.6964 – 19.6h

Collect like terms

0 – 807.6964 = – 19.6h

– 807.6964 = – 19.6h

Divide both side by – 19.6

h = – 807.6964 / – 19.6

h = 41.21 m

Thus, the maximum height reached by the ball is 41.21 m

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