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s344n2d4d5 [400]
4 years ago
10

A shovel is the third class lever​

Physics
1 answer:
Nookie1986 [14]4 years ago
8 0

Answer:

Yes

Explanation:

In a third-class lever, the effort force lies between the resistance force and the fulcrum. Some kinds of garden tools are examples of third-class levers. When you use a shovel, for example, you hold one end steady to act as the fulcrum, and you use your other hand to pull up on a load of dirt.

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In can expermient why the can is chrushed​
schepotkina [342]

Answer:

Force is applied to the point where the can can't handle it and starts to get crushed

Explanation:

5 0
3 years ago
A poodle racing across a lawn is an example of _____.
sergey [27]


it is an example of kinetic energy

so the answer is b kinetic energy

6 0
3 years ago
A block of mass 9.5kg rests on a slope an angle of 23.0∘ relative to the horizontal. What is the size of the contact force norma
Rudiy27

The Normal Force = M x G x Cos(theta)

= 9.5 Kg x 9.8 m/s^2 x cos 23

= 9.5 Kg x 9.8 m/s^2 x 0.9205

Converting Kg to Newton,

1 Kg  = 9.81 N

= 9.5 Kg x 9.81 N x 9.8 m/s^2 x 0.9205

= 840.702 N

6 0
3 years ago
A 70 kg student jumps down to form a 1 m high platform. She forgets to bend her knees and her downward motion stops in 0.02 seco
34kurt

Answer:

15,505 N

Explanation:

Using the principle of conservation of energy, the potential energy loss of the student equals the kinetic energy gain of the student

-ΔU = ΔK

-(U₂ - U₁) = K₂ - K₁ where U₁ = initial potential energy = mgh , U₂ = final potential energy = 0, K₁ = initial kinetic energy = 0 and K₂ = final kinetic energy = 1/2mv²

-(0 - mgh) = 1/2mv² - 0

mgh = 1/2mv² where m = mass of student = 70kg, h = height of platform  = 1 m, g = acceleration due to gravity = 9.8 m/s² and v = final velocity of student as he hits the ground.

mgh = 1/2mv²

gh = 1/2v²

v² = 2gh

v = √(2gh)

v = √(2 × 9.8 m/s² × 1 m)

v = √(19.6 m²/s²)

v = 4.43 m/s

Upon impact on the ground and stopping, impulse I = Ft = m(v' - v) where F = force, t = time = 0.02 s, m =mass of student = 70 kg, v = initial velocity on impact = 4.43 m/s and v'= final velocity at stopping = 0 m/s

So Ft = m(v' - v)

F = m(v' - v)/t

substituting the values of the variables, we have

F = 70 kg(0 m/s - 4.43 m/s)/0.02 s

= 70 kg(- 4.43 m/s)/0.02 s

= -310.1 kgm/s ÷ 0.02 s

= -15,505 N

So, the force transmitted to her bones is 15,505 N

3 0
3 years ago
A direct result of the Newlands reclamation Act was the
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I think it is building and managing of irrigation systems 
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4 years ago
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