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allsm [11]
4 years ago
8

На рычаге размещены два противовеса таким образом, что рычаг находится в состоянии равновесия. Вес расположенного слева противов

еса равен P1=13H.
Какой вес P2 расположенного справа противовеса, если все обозначенные на перекладине рычага участки имеют одинаковую длину?

P1=13H
P2=?
Physics
1 answer:
prisoha [69]4 years ago
8 0

Answer:

Вес противовеса, или сила нагрузки, составляет затем 100 000 фунтов-футов, разделенных на 20 футов, или 5000 фунтов.

Explanation:

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Explanation:

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The fluid friction that opposes the motion of objects through air is known as what?
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A student, who weighs 720N, is standing on a bathroom scale and riding an elevator that is moving downwards with a speed that is
jasenka [17]

Answer:

1) The mass of the student is approximately 73.39 kg

2) The net force on the student is approximately 947.523 N

3) The value the scale will read is approximately 96.59 kg

Explanation:

The given parameters are;

The weight of the student = 720 N

The speed at which the elevator is decreasing = 3.1 m/s²

1) The weight of the student = The mass of the student × The acceleration due to gravity

The acceleration due to gravity is a constant = 9.81 m/s²

Substituting the known values gives;

720 N = The mass of the student × 9.81 m/s²

∴ The mass of the student = 720 N/(9.81 m/s²) ≈ 73.39 kg

2) The forces acting on the student are;

i) The force of gravity which is the weight of the student acting downwards

ii) The inertia force of the slowing elevator acting downwards in the same direction as the weight of the student

The net force, F_{net} = The weight of the student + The inertia force of the slowing elevator

∴ The net force, F_{net} = 720 N + 73.39 kg × 3.1 m/s² ≈ 947.523 N

3) The scale will read the mass of the student as follows;

Mass reading of student on the scale = Force on scale/9.81

∴ Mass reading of student on the scale = 947.523/9.81 ≈ 96.59 kg

The value the scale will read = 96.59 kg.

3 0
3 years ago
A uniform, solid sphere of radius 2.50 cm and mass 4.75 kg starts with a purely translational speed of 3.00 m/s at the top of an
allsm [11]

Answer:

The final translational seed at the bottom of the ramp is approximately 4.84 m/s

Explanation:

The given parameters are;

The radius of the sphere, R = 2.50 cm

The mass of the sphere, m = 4.75 kg

The translational speed at the top of the inclined plane, v = 3.00 m/s

The length of the inclined plane, l = 2.75 m

The angle at which the plane is tilted, θ = 22.0°

We have;

K_i + U_i = K_f + U_f

K = (1/2)×m×v²×(1 + I/(m·r²))

I = (2/5)·m·r²

K =  (1/2)×m×v²×(1 + 2/5) = 7/10 × m×v²

U = m·g·h

h = l×sin(θ)

h = 2.75×sin(22.0°)

∴ 7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) = 7/10 × 4.75×v_f² + 0

7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) ≈ 77.93

∴ 77.93 ≈ 7/10 × 4.75×v_f²

v_f² = 77.93/(7/10 × 4.75)

v_f ≈ √(77.93/(7/10 × 4.75)) ≈ 4.84

The final translational seed at the bottom of the ramp, v_f ≈ 4.84 m/s.

3 0
3 years ago
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