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Annette [7]
2 years ago
5

During a lab investigation, students added four 50 g masses to two boxes and arranged the boxes so that they were motionless on

a pulley, as shown in the diagram. The students then followed the procedure described in the box. The students recorded their observations after each procedure and reset the pulley system to the original conditions. During which procedures did students observe an unbalanced upward force on Box 1? A) Procedures 1, 2, and 3 B) Procedures 4 and 5 C) Procedures 3 and 4 D) Procedures 3 and 4.
Physics
1 answer:
IceJOKER [234]2 years ago
3 0

When the resultant force is not equal to zero termed an unbalanced force. By procedures 4 and 5 students observe an unbalanced upward force on Box 1. Hence option 1 is right for the problem.

<h3>What is an unbalanced force?</h3>

The forces operating on a body are known as unbalanced forces when the resulting force exerted on it is not equal to zero.

Unbalanced forces acting on the body, causing it to modify its state of motion. To further grasp the nature of imbalanced forces.

<h3 />

The following reasons by which we can understand the unbalanced force caused by the box.

Due to these two reasons, books will move up.

By adding another mass to box 2. The box becomes lighter. As the box becomes lighter the gravity force acting on the box will be less due to which the box easily can move up.

By removing the two masses from box 1. Due to which other become heavier other becomes heavier pulling it down causing box 1 one to go up.

Hence by procedures 4 and 5 students observe an unbalanced upward force on Box 1. Hence option 1 is right for the problem.

To learn more about the unbalanced force refer to the link;

brainly.com/question/227461

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A 500N person stands 2.5m from a wall against which a horizontal beam is attached. The beam is 6m long and weighs 200N. A cable
Vinil7 [7]

PART A)

Here by force balance along Y direction

Tsin45  + F_y = 500 + 200

Force balance along X direction

Tcos45 = F_x

now by torque balance

Tsin45 (6) = 500 (2.5) + 200 (3)

T(4.24) = 1250 + 600

T = 436.3 N

PART B)

now from above equations

Tsin45 + F_y = 700

F_y = 391.5 N

F_x = Tcos45 = 308.5 N

now net reaction force of wall is given as

F = \sqrt{F_x^2 + F_y^2}

F = 498.4 N

5 0
3 years ago
For the material in the previous question that yields at 200 MPa, what is the maximum mass, in kg, that a cylindrical bar with d
Sedaia [141]

Answer:

The maximum mass the bar can support without yielding = 32408.26 kg

Explanation:

Yield stress of the material (\sigma) = 200 M Pa

Diameter of the bar = 4.5 cm = 45 mm

We know that yield stress of the bar is given by the formula

                Yield Stress = \frac{Maximum load}{Area of the bar}

⇒                                \sigma = \frac{P_{max} }{A}  ---------------- (1)

⇒ Area of the bar (A) = \frac{\pi}{4} ×D^{2}

⇒                            A  = \frac{\pi}{4} × 45^{2}

⇒                            A = 1589.625 mm^{2}

Put all the values in equation (1) we get

⇒ P_{max} = 200 × 1589.625

⇒ P_{max} = 317925 N

In this bar the P_{max} is equal to the weight of the bar.

⇒ P_{max} = M_{max} × g

Where M_{max} is the maximum mass the bar can support.

⇒ M_{max} = \frac{P_{max} }{g}

Put all the values in the above formula we get

⇒ M_{max} = \frac{317925}{9.81}

⇒ M_{max} = 32408.26 Kg

There fore the maximum mass the bar can support without yielding = 32408.26 kg

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3 years ago
If a person believes that dreams have hidden meaning he or she would agree with Freud's ideas about _ content
uysha [10]
A.lucid because it makes more sense yo the answer how about you do it yourself
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A force of 20N changes the position of a body. If mass of the body is 2kg, find the acceleration produced in the body.2. A ball
shepuryov [24]

Explanation:

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<em>where</em><em> </em><em>m</em><em>=</em><em> </em><em>mass</em>

<em>v</em><em>=</em><em> </em><em>velocity</em><em>.</em>

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5 0
3 years ago
Please help! thank you​
BlackZzzverrR [31]

Answer:

poor, too precise

good

good

good

Explanation:

8 0
2 years ago
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