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1 year ago

A box has the dimensions of 50 cm × 30 cm × 15 cm, weighs 150 N, and is to be

1 answer:

8 0

The surface can only withstand a pressure Face A (50cm × 30cm) and Face C (50cm × 15cm) as the surface can only withstand a pressure of 0.25 N/cm³.

<h3>What is pressure?</h3>

The physical force used to apply pressure to an object is defined as such. Per square inch of an object, a force is applied perpendicularly to its surface. For pressure, the fundamental formula is F/A. (Force per unit area). The Pascal is the unit of pressure (Pa).

The four different types of pressure are absolute, atmospheric, differential, and gauge pressure. Have you ever noticed that when you use a straw to drink something, the air actually gets suked out? In reality, you're applying "Pressure" as you drink the beverage.

A box has the dimensions of 50 cm × 30 cm × 15 cm

Let each face be A, B and C

The weight of the box = 150 N

Formula for pressure is

P = F/A

To find out which face of the box can withstand a pressure of 0.25 N/cm

we need find the area of each face and find its pleasure

Face A = 50 cm × 30 cm

Area A = l × b

= 50 × 30

= 1500 cm²

Pressure A = 150/1500

= 0.1 N/cm³

0.25 > 0.1

The surface can definitively withstand the pressure of Face A

Face B = 30 cm × 15 cm

Area A = l × b

= 30 × 15

= 450 cm²

Pressure A = 150/450

= 0.3 N/cm³

0.25 < 0.3

The surface could not withstand the pressure of Face B

Face C = 50 cm × 15 cm

Area A = l × b

= 50 × 15

= 750 cm²

Pressure A = 150/750

= 0.2 N/cm³

0.25 > 0.2

The surface can definitively withstand the pressure of Face C

Thus, The surface can only withstand a pressure Face A (50cm × 30cm) and Face C (50cm × 15cm) as the surface can only withstand a pressure of 0.25 N/cm³.

Learn more about Pressure

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A ball of mass m, attached to the end of a horizontal cord, is rotated in a circle of radius r on a frictionless horizontal surf

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Given

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3 years ago

A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected

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Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

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$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta$

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$5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45$

$15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}$

$15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}$ ....(I)

Along y -axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta$

Put the value into the formula

$0+0=5\times v_{1}\sin30-3\times v_{2}\sin45$

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From equation (I) and (II)

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Put the value of v₁ in equation (I)

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Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

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Answer:

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