(PLEASE HELP ITS DUE SOON ILL MARK BRAINLIEST AND 5 STARS & PLEASE SHOW WORK!!)(And the answer is not 44 I already tried that and it doesn’t start with 4 either)

Answer:

B.  Energy increases with decreasing wavelength and increasing frequency.

Explanation:

You draw 3 circles around the stations with the size of the circle equal to the distance from the earthquake. Then you simply find where the edge circles all overlap.

Answer:

31.29 m/sec

Explanation:

We have given density of substance

We have convert this into

We know that 1 lb = 0.4535 kg. so 0.14 lb = 0.14×0.4535 = 0.06349 kg

We know that 1 kg = 1000 g ( 1000 gram )

So 0.06349 kg = 63.49 gram

And we know that 1 gram = 1000 milligram

So 63.49 gram

We know that

So

So =\frac{63.49\times 10^3}{0.2249\times 10^{-5}}=276.74\times 10^8lb/m^3[/tex]

In second part we have to convert 70 mi/hr to m/sec

We know that 1 mi = 1609.34 meter

So 70 mi = 70×1609.34 = 112653.8 meter

1 hour = 3600 sec

So 70 mi/hr

Answer: i think it is B

Explanation:

Answer:

Explanation:

Pressure on the hydraulic system is expressed as;

Pressure = Force/Area

Given

Force on the fluid = 400N

Area = 0.001m²

Pressure in the fluid = 400/0.001

Pressure in the fluid = 400,000N/m²

1N/m² = 0.001kPa

400,000N/m² = x

x = 400,000 × 0.001

x = 400kPa

Hence the pressure in kPa is 400kPa

b) Using the formula;

Pa = Pb

Fa/Aa = Fb/Ab

Pa = Fb/Ab

Fb = PaAb

Fb = 400,000(0.2)

Fb = 80,000N

Hence the magnitude of the force exerted on the load bearing piston by the hydraulic fluid is 80,000N

Final answer:

In a hydraulic system, a force exerted creates a pressure that is transmitted equally throughout. The induced pressure in the hydraulic system is 400 kPa. By applying the same pressure across the larger piston (0.2 m2), a force of 80,000 N is generated on the load-bearing piston.

Explanation:

The subject of this question is the physics topic of hydraulic systems, specifically how forces and pressure interactions. Use Pascal's principle which states that a change in pressure at any point in an enclosed fluid under equilibrium will be transmitted equally to all parts of the fluid.

a. To find the pressure in kPa induced by the applied pressure, use the formula P=F/A, where P is pressure, F is force and A is area. With the force F = 400 N and area A = 0.001 m2, the induced pressure is P = 400 N / 0.001 m2 = 400,000 Pa or 400 kPa.

b. The force exerted on the load-bearing piston by the hydraulic fluid is calculated by rearranging the formula to F=PxA. So, F = 400kPa x 0.2m2 = 80,000 N. Therefore, the force exerted on the load-bearing piston by the hydraulic fluid is 80,000N.

Learn more about Hydraulic Systems here:

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