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3 years ago

Kūna veikia jėga kurios momentas 0,4 N × m petys - 5 cm. Koks šios jėgos dydis?

Physics

1 answer:

umka21 [38]3 years ago

8 0

Answer:

F = 8 N

Explanation:

The question says "The body is subjected to a force with a moment of 0.4 N × m shoulder - 5 cm. What is the magnitude of this force?

Given that,

Moment/Torque, $\tau=0.4\ N-m$

Distance moved, d = 5 cm = 0.05 m

We need to find the magnitude of this force. We know that, the torque acting on an object is given by :

$\tau=Fd\\\\F=\dfrac{\tau}{d}\\\\F=\dfrac{0.4\ N-m}{0.05\ m}\\\\F=8\ N$

So, the magnitude of force is equal to 8 N.

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3 years ago

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A fireboat is to fight fires at coastal areas by drawing seawater with a density of 1030 kg/m3 through a 10-cm-diameter pipe at

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Answer:

50.93 m/s

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Explanation:

From the question, the nozzle exit diameter = 5 cm, Radius= diameter/2= 5cm/2= 2.5cm. we can convert it to metre for unit consistency= (2.5×0.01)=

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A= π ×0.025^2

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mgh = 1/2mv²

We can make "h" subject of the formula, which is the height of required head of water

h = (1/2mv²)/mg

h= v² / 2g

h = 50.93² / (2 ×9.81)

h = 132.21m

From the question;

The total irreversible head loss of the system = 3 m,

the given position of nozzle = 3 m

the total head the pump needed=(The total irreversible head loss of the system + the position of the nozzle + required head of water )

=(3 + 3 + 132.21m)

=138.21m

mass of water pumped in a seconds can be calculated since we know that mass is a product of volume and density

Volume= 0.1m³

Density of sea water=1030 kg/m

(0.1 m^3× 1030)

= 103kg

We can calculate the Potential enegry, which is = mgh

= (103 ×9.81 × 138.21)

= 139651.5 Watts

= 139.65kW

To determine required shaft power input to the pump and the water discharge velocity

Energy= efficiency × power

But we are given efficiency of 70 percent, then

139651.5 Watts = 0.7P

=199502.18 Watts

P=199.5 kW

Therefore, the required shaft power input to the pump and the water discharge velocity is 199.5 kW

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2 years ago

Density of water is 1000 kg/m^3. What will be the volume of 35000 kg water?

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3 years ago

If a 20 N object has been lifted 5 meters above the ground, how much gravitational potential energy does it have?

Andreyy89

The gravitational potential energy of the object is 100 J.

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Explanation:

I hope this helps :)

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3 years ago

Kūna veikia jėga kurios momentas 0,4 N × m petys - 5 cm. Koks šios jėgos dydis?​

Kūna veikia jėga kurios momentas 0,4 N × m petys - 5 cm. Koks šios jėgos dydis?