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Mariana [72]
3 years ago
6

5 description of a motion​

Physics
1 answer:
yuradex [85]3 years ago
8 0

Answer:

<h2>Displacement</h2><h2>Distance</h2><h2>Velocity</h2><h2> Acceleration</h2><h2>Speed</h2><h2> Time</h2>

Explanation:

HOPE IT HELPS

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The machine in the figure is ideal and an effort force of
DENIUS [597]

Answer:

1.5 m

Explanation:

Let the distance from the box to the pivot be c.

Let the distance from the pivot to the effort be y.

From the question given above, the following data were obtained:

Effort force (Fₑ) = 7 N

Force of resistance (Fᵣ) = 14 N

Distance from the box to the pivot (c) = 0.75 m

Distance from the pivot to the effort (y) =?

Clockwise moment = Fₑ × y

Anticlock wise moment = Fᵣ × c

Clockwise moment = Anticlock wise moment

Fₑ × y = Fᵣ × c

7 × y = 14 × 0.75

7 × y = 10.5

Divide both side by 7

y = 10.5 / 7

y = 1.5 m

Therefore, the distance from the pivot to the effort is 1.5 m

5 0
2 years ago
A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top o
masha68 [24]

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let <em>T</em> represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, F_{f2} = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = F_{f2} × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27

Frictional \ force, F_{f2} = \left (981 -  \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8}  \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times  0.25 \right) \approx 219.92

The frictional force on the block W₂, F_{f2} ≈ 219.92 N

Therefore;

The force acting the block w₁, due to w₂ F_{w2} = 219.92/0.25 ≈ 879.68

The total normal force acting on the ground, N = W₁ + \mathbf{F_{w2}}

The frictional force from the ground, \mathbf{F_{f1}} = N×μ + \mathbf{F_{f2}} = P

Where;

P = The horizontal force applied to the block

P = (W₁ + \mathbf{F_{w2}}) × μ + \mathbf{F_{f2}}

Therefore;

P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84

The horizontal force applied to the block, P ≈ 1,420.84 N

Learn more about friction force here;

brainly.com/question/18038995

3 0
3 years ago
Joffrey talks and moves slowly. When asked a question, he answers slowly in monotone monosyllables, if he answers at all. Joffre
lana [24]

Answer:

R.E.T.A.R.D.A.T.I.O.N

Explanation:

It won't let me spell it normal

7 0
3 years ago
What geological feature can be created by a divergent boundary
VARVARA [1.3K]
<span>Divergent plate boundaries are locations where plates are moving away from one another. This occurs above rising convection currents. The rising current pushes up on the bottom of the lithosphere, lifting it and flowing laterally beneath it. This lateral flow causes the plate material above to be dragged along in the direction of flow. At the crest of the uplift, the overlying plate is stretched thin, breaks and pulls apart. hope this helps

</span>
5 0
3 years ago
Read 2 more answers
A supersaturated solution is one which A. has less solute dissolved than the solution should hold at that temperature. B. has mo
Anna35 [415]

Answer:

Option C is correct

Explanation:

A supersaturated solution is one that has more solute dissolved than the solution should hold at that temperature.

Examples include carbonated water, sugar syrup, honey.

A solution of a chemical compound can be dissolved in heated water to prepare a supersaturated solution. A solution becomes supersaturated as its temperature is changed.

3 0
3 years ago
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