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kakasveta [241]
3 years ago
14

A cat pushes a porcelain statue off a bookshelf with a speed of 0.5 m/s and it smashed on the floor 0.85 sec later.

Physics
1 answer:
Bad White [126]3 years ago
5 0

Answer:

167?

Explanation:

i added both

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Explain the benefits of understanding sound waves properties in real life.
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The vibrations from sound waves cause our ears to send signals to our brains to create sound. The speed of sound waves will determine the sound's pitch, or how high or low something sounds. Sound waves are important because they allow us to hear important messages and emergency signals to protect ourselves.

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An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twi
shusha [124]

Answer:

Rectangular path

Solution:

As per the question:

Length, a = 4 km

Height, h = 2 km

In order to minimize the cost let us denote the side of the square bottom be 'a'

Thus the area of the bottom of the square, A = a^{2}

Let the height of the bin be 'h'

Therefore the total area, A_{t} = 4ah

The cost is:

C = 2sh

Volume of the box, V = a^{2}h = 4^{2}\times 2 = 128            (1)

Total cost, C_{t} = 2a^{2} + 2ah            (2)

From eqn (1):

h = \frac{128}{a^{2}}

Using the above value in eqn (1):

C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}

C(a) = 2a^{2} + \frac{256}{a}

Differentiating the above eqn w.r.t 'a':

C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}

For the required solution equating the above eqn to zero:

\frac{4a^{3} - 256}{a^{2}} = 0

\frac{4a^{3} - 256}{a^{2}} = 0

a = 4

Also

h = \frac{128}{4^{2}} = 8

The path in order to minimize the cost must be a rectangle.

8 0
3 years ago
hydraulic lift is to be used to lift a 2100-kg weight by putting a weight of 25 kg on a piston with a diameter of 10 cm. Determi
trasher [3.6K]

Answer:

840 cm

Explanation:

Note: A hydraulic press operate based on pascal's principle.

From pascal's principle

W₁/d₁ = W₂/d₂...................... Equation 1

Where W₁ and W₂ are the first and second weight, and d₁ and d₂ are the  first and second diameter of the piston.

make d₁ the subject of the equation

d₁ = W₁×d₂/W₂................ Equation 2

Given: W₁ = 2100 kg, W₂ = 25 kg, d₂ = 10 cm = 0.1 m.

Substitute these values into equation 2

d₁ = 2100(0.1)/25

d₁ = 8.4 m

d₁ = 840 cm

3 0
3 years ago
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