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dem82 [27]
3 years ago
9

A load of 800 N is lifted using a block and tackle having 5 pulleys. If the applied effort is 200 N, calculate

Physics
2 answers:
ser-zykov [4K]3 years ago
8 0

Explanation:

Load=800N

Effort=200N

1. Mechanical Advantage = LOAD/EFFORT

= 800N/200N

= 4

2 Velocity Ratio = no. Of pulleys =5

3. Efficiency = Mechanical advantage / velocity ratio × 100%

= (4/5)×100%

=80%

4. output work= load×load distance

= 800N × 5m

= 4 × 1000J

5. Efficiency = (output work/input work) ×100%

Or, 80% = (4000J/input work) ×100%

Or, 80%/100% = 4000J/inputwork

Or, 4/5 = 4000J/inputwork

Or, input work =4000J × 5/4

Input work = 5×1000J

I hope it helped! ;-)

never [62]3 years ago
4 0

Answer:

I got the same answers.

Explanation:

I agree with the first person respone.

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Find the density of seawater at a depth where the pressure is 500 atm if the density at the surface is 1100 kg/m^3 . Seawater ha
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The density of seawater at a depth where the pressure is 500 atm is 1124kg/m^3

Explanation:

The relationship between bulk modulus and pressure is the following:

B=\rho_0 \frac{\Delta p}{\Delta \rho}

where

B is the bulk modulus

\rho_0 is the density at surface

\Delta p is the variation of pressure

\Delta \rho is the variation of density

In this problem, we have:

B=2.3\cdot 10^9 N/m^2 is the bulk modulus

\rho_0 =1100 kg/m^3

\Delta p = p-p_0 = 500 atm - 1 atm = 499 atm = 5.05\cdot 10^7 Pa is the change in pressure with respect to the surface (the pressure at the surface is 1 atm)

Therefore, we can find the density of the water where the pressure is 500 atm as follows:

\rho = \rho_0 + \Delta \rho = \rho_0+\frac{\rho_0 \Delta p}{B}=\rho_0 (1+\frac{\Delta p}{B})=(1100)(1+\frac{5.05\cdot 10^7}{2.3\cdot 10^9})=1124kg/m^3

Learn more about pressure in a fluid:

brainly.com/question/9805263

#LearnwithBrainly

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