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babunello [35]
4 years ago
9

Identify two simple machines in this compound machine. A. screw, lever B. screw, incline plane C. double pulley D. lever, pulley

Physics
2 answers:
seropon [69]4 years ago
5 0
It's A, the Screw leaver 
irinina [24]4 years ago
3 0

Answer: correct option is (A) screw and lever.


Explanation :


Machines make our work easier. There are many types of simple machines i.e. lever, pulley, wheel and axle, screw, inclined plane and wedge.  

In the given picture, there are two simple machines as screw and lever.


Nutcracker, stapler, tongs and tweezers are the example of lever.


Examples of use of screw are in a jar lid and ballpoint pens.

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A 100.0-kg bakery sign hangs from two thin cables as shown.
harina [27]

Answer:

1. T₁ = 500 N

2. T₂ = 866 N

Explanation:

Please see attached photo for the diagram.

Thus, we can obtain obtained the value of T₁ and T₂ as follow:

1. Determination of T₁

Angle θ = 30

Hypothenus = 100 kg

Opposite = T₁ =?

Sine θ = Opposite /Hypothenus

Sine 30 = T₁ / 100

Cross multiply

T₁ = 100 × Sine 30

T₁ = 100 × 0.5

T₁ = 50 Kg

Multiply by 10 to express in Newton

T₁ = 50 × 10

T₁ = 500 N

2. Determination of T₂

Angle θ = 60

Hypothenus = 100 kg

Opposite = T₂ = ?

Sine θ = Opposite /Hypothenus

Sine 60 = T₂ / 100

Cross multiply

T₂ = 100 × Sine 60

T₂ = 100 × 0.8660

T₂ = 86.6 Kg

Multiply by 10 to express in Newton

T₂ = 86.6 × 10

T₂ = 866 N

5 0
3 years ago
A typical laboratory centrifuge rotates at 4000 rpm. Testtubes have to be placed into a centrifuge very carefully because ofthe
Bogdan [553]

Answer:

A)  a_c = 1.75 10⁴ m / s², B) a = 4.43 10³ m / s²

Explanation:

Part A) The relation of the test tube is centripetal

               a_c = v² / r

the angular and linear variables are related

              v = w r

we substitute

               a_c = w² r

let's reduce the magnitudes to the SI system

              w = 4000 rpm (2pi rad / 1 rev) (1 min / 60s) = 418.88 rad / s

               r = 1 cm (1 m / 100 cm) = 0.10 m

let's calculate

              a_c = 418.88² 0.1

               a_c = 1.75 10⁴ m / s²

part B) for this part let's use kinematics relations, let's start looking for the velocity just when we hit the floor

as part of rest the initial velocity is zero and on the floor the height is zero

                v² = v₀² - 2g (y- y₀)

                v² = 0 - 2 9.8 (0 + 1)

                v =√19.6

                v = -4.427 m / s

now let's look for the applied steel to stop the test tube

                v_f = v + a t

                0 = v + at

                a = -v / t

                a = 4.427 / 0.001

                a = 4.43 10³ m / s²

4 0
3 years ago
An object of mass m is placed on Spring A, which is compressed by distance x. The spring is released and the velocity of the pus
ss7ja [257]

Answer:

C

Explanation:

3 0
3 years ago
What is the term for the circular movement of material inside Earth's mantle?
Gala2k [10]

Answer:

convection currents is the answer

5 0
3 years ago
What is the net force needed to accelerate a 5 kg object at 3 m/s2? Suppose that in this situation you discovered that there is
Rudik [331]

The Force need to accelerate the object is by 3 m/s² is 15 N.  Suppose a friction force of 5 N acts on the motion of the object, the force needed to be applied to the object is 20 N

<h3>Force:</h3>

This can be defined as the product of the mass and the acceleration of a body. The S.I unit of force is kgm/s or Newton(N)

To calculate the force needed to accelerate a mass of 5 kg object at 3 m/s² we use the formula below.

Formula:

  • F = ma........ equation 1

Where:

  • F = Net force needed to accelerate the object
  • m = mass of the object
  • a = acceleration of the object

From the question,

Given:

  • m = 5 kg
  • a = 3 m/s²

Substitute these values into equation 1

  • F = 5(3)
  • F = 15 N

Suppose a frictional force of 5 N acts on the motion, The force applied is

  • F = F'+ma............ Equation 2

Where:

  • F = Frictional force = 5N

Substitute into equation 2

  • F = 5(3)+5
  • F' = 15+5
  • F = 20 N.

Hence, The Force need to accelerate the object is by 3 m/s² is 15 N. Suppose a friction force of 5 N acts on the motion of the object, the force needed to be applied to the object is 20 N

Learn more about force here: brainly.com/question/12970081

4 0
3 years ago
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