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Harrizon [31]
2 years ago
13

the dimensions of a cuboidal block are 2.5 m * 2m * 1.2m its weight 900 n what is the minimum presseure exerted by the block on

the surface of its support.?​
Physics
2 answers:
aleksandr82 [10.1K]2 years ago
6 0

Answer:

your answer is here 180

Explanation:

900/2.5 * 2 = 180

daser333 [38]2 years ago
6 0

Answer:

The answer is 180 Pa.

Explanation:

<h3><u>Given</u>;</h3>

Here, weight = thrust = 900 N,

dimensions are 2.5 m × 2 m × 1.2 m

<h3><u>To </u><u>Find</u>;</h3>
  • Pressure exerted by the block on the surface of its support.
<h3><u>Formula</u>;</h3>
  • Pressure = Thrust ÷ Area

We know that, Pressure is inversely proportional to area. For the minimum pressure the area must be more.

So, if the block is resting with 2.5 m × 2 m surface on the support, the pressure will be minimum.

Pressure = 900 N ÷ 2.5 × 2.0 m² = 180 Pa

Thus, The pressure exerted by the block on the support is 180 Pa.

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3 years ago
A student is swimming south applying a force of 256 N. The water exerts a westward force of 104 N. If the student has a mass of
grigory [225]

Answer:

a=2.9\ m/sec^2

Explanation:

<u>Net Forces and Acceleration</u>

The second Newton's Law relates the net force F_r acting on an object of mass m with the acceleration a it gets. Both the net force and the acceleration are vector and have the same direction because they are proportional to each other.

\vec F_r=m\vec a

According to the information given in the question, two forces are acting on the swimming student: One of 256 N pointing to the south and other to the west of 104 N. Since those forces are not aligned, we must add them like vectors as shown in the figure below.

The magnitude of the resulting force F_r is computed as the hypotenuse of a right triangle

|F_r|=\sqrt{256^2+104^2}

|F_r|=276.32\ Nw

The acceleration can be obtained from the formula

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Note we are using only magnitudes here

\displaystyle a=\frac{F_r}{m}

\displaystyle a=\frac{276.32Nw}{95.3Kg}

\boxed{a=2.9\ m/sec^2}

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