Selina solutions

Selina solutions

Grade 6

Simple Machines | Numericals

Question 1

Q1) In a mechine an effort of 10 kgf is applied to lift a load of 100 kgf. What is its mechanical advantage?

Solution:

We are given that

Load = 100kgf

Effort = 10kg

Mechanical advantage = \frac{Load}{effort}=\frac{100kgf}{10kgf}=10

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subject-cta

Question 1

Q1) In a mechine an effort of 10 kgf is applied to lift a load of 100 kgf. What is its mechanical advantage?

Solution:

We are given that

Load = 100kgf

Effort = 10kg

Mechanical advantage = \frac{Load}{effort}=\frac{100kgf}{10kgf}=10

Still have questions? Our expert teachers can help you out

Book a free class now

Want to top your physics exam ?

Learn from an expert tutor.

Book a free class now
subject-cta
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