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ololo11 [35]
2 years ago
9

a load of 400 Newton is lifted by a first class lever in which the load is at the distance of 20 cm and the effort is at the dis

tance of 60 cm from the fulcrum if 150 Newton effort is required to lift the load what is its efficiency ?​
Physics
1 answer:
Len [333]2 years ago
7 0

Answer:

  1. solution,
  2. Given
  3. load =400N
  4. ld=0.2m
  5. ed=0.6m
  6. effort =150N

Explanation:

efficiency =output work/input work ×100%

l×ld/e×ed×100%

400×0.2/150×0.6×100%

80/90×100%

88.89%ans

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ryzh [129]

Answer:

0.375 m/s north & 0.375 m/s east

Explanation:

8 0
3 years ago
In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a fur
Margaret [11]

Answer:

D = -4/7 = - 0.57

C = 17/7 = 2.43

Explanation:

We have the following two equations:

3C + 4D = 5\ --------------- eqn (1)\\2C + 5D = 2\ --------------- eqn (2)

First, we isolate C from equation (2):

2C + 5D = 2\\2C = 2 - 5D\\C = \frac{2 - 5D}{2}\ -------------- eqn(3)

using this value of C from equation (3) in equation (1):

3(\frac{2-5D}{2}) + 4D = 5\\\\\frac{6-15D}{2} + 4D = 5\\\\\frac{6-15D+8D}{2} = 5\\\\6-7D = (5)(2)\\7D = 6-10\\\\D = -\frac{4}{7}

<u>D = - 0.57</u>

Put this value in equation (3), we get:

C = \frac{2-(5)(\frac{-4}{7} )}{2}\\\\C = \frac{\frac{14+20}{7}}{2}\\\\C = \frac{34}{(7)(2)}\\\\C =  \frac{17}{7}\\

<u>C = 2.43</u>

5 0
3 years ago
A scientist suspends a spring from a force sensor. She then pulls on the spring causing the spring to stretch 1.0 cm. She record
Nadusha1986 [10]
This situation describes the Hooke's Law which states that "When an elastic object - such as a spring - is stretched, the increased length is called its extension. The extension of an elastic object is directly proportional to the force applied to it". The formula is <span>F = k × e , F for the force, k for spring constant expressed in N/m, e for extension in m. This equation works for as long the spring is not stretch too much because once it exceeded its limit, the spring will not return to its original length the moment the load is removed.</span>
8 0
3 years ago
Suppose a clay model of a koala bear has a mass of 0.205 kg and slides on ice at a speed of 0.720 m/s. It runs into another clay
Westkost [7]

Answer:

Final velocity, v = 0.28 m/s

Explanation:

Given that,

Mass of the model, m_1=0.205\ kg

Speed of the model, u_1=0.72\ m/s

Mass of another model, m_2=0.32\ kg

Initial speed of another model, u_2=0

To find,

Final velocity

Solution,

Let V is the final velocity. As both being soft clay, they naturally stick together. It is a case of inelastic collision. Using the conservation of linear momentum to find it as :

m_1u_1+m_2u_2=(m_1+m_2)V

V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}

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So, their final velocity is 0.28 m/s. Hence, this is the required solution.

6 0
3 years ago
A track is traveling a. a speed of 25.0 m/s along a level road. A crate is resting on the bed of the truck, and the coefficient
Mice21 [21]

To solve this problem it is necessary to apply the concepts related to

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Where,

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Where,

m = mass

v = velocity

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d = \frac{1}{2} \frac{v^2}{\mu g}

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d = \frac{1}{2} \frac{25^2}{0.65*9.8}

d = 49.05m

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6 0
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