Question

яку силу треба прикласти щоб підняти під водою Pв=1000км/м3,камін масою 30кг об'єм якого 0,012м3?(g=10H/m)(1задача)Який тиск створює вістря увеха на думку,якщо площа війра увеха 0,6мм2,а сила ,з якою він діє на думку =30H?(поможіть срочно ,дуже треба)​

Answer 1

Ответ: 180Н; 50МПа

Объяснение:

Дано:

m30кг

V=0,012м³

ρ=1000кг/м³

g=10Н/кг

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ΔF-?

ΔF=Fт-Fa=mg-ρgV=g(m-ρV)=10*(30-1000*0,012)=180Н

Дано:

S=0,6мм²=0,0000006м²

F=30Н

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Р-?

P=F/S=30/0,0000006=50000000Па=50МПа

Answer 2

Problem 1: Finding the force required to lift a stone underwater

To find the force required to lift a stone underwater, we can use Archimedes' principle, which states that the buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.

Given: - Density of water, ρv = 1000 kg/m^3 - Mass of the stone, m = 30 kg - Volume of the stone, V = 0.012 m^3 - Acceleration due to gravity, g = 10 m/s^2

To calculate the force required to lift the stone underwater, we need to find the weight of the water displaced by the stone. The weight of an object is given by the formula:

Weight = mass × acceleration due to gravity

The weight of the water displaced by the stone is equal to the weight of the stone itself. Therefore, the force required to lift the stone underwater is:

Force = Weight = mass × acceleration due to gravity

Substituting the given values:

Force = 30 kg × 10 m/s^2 = 300 N

Therefore, the force required to lift the stone underwater is 300 Newtons.

Problem 2: Calculating the pressure exerted by a needle

To calculate the pressure exerted by a needle, we can use the formula:

Pressure = Force / Area

Given: - Area of the needle, A = 0.6 mm^2 = 0.6 × 10^(-6) m^2 - Force exerted by the needle, F = 30 N

Substituting the given values into the formula:

Pressure = 30 N / 0.6 × 10^(-6) m^2

Simplifying the expression:

Pressure = 5 × 10^7 N/m^2

Therefore, the pressure exerted by the needle is 5 × 10^7 Newtons per square meter (or Pascal).

Please let me know if there's anything else I can help you with!