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2 years ago

What is the main force that must be overcome in order to push an object

Physics

1 answer:

My name is Ann [436]2 years ago

4 0

Answer: Friction

Explanation:

Friction and the normal force would be the two initial forces to overcome.

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As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil

galina1969 [7]

Answer:

Second option 6.3 N at 162° counterclockwise from

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

$-F_3sin(b) + F_1 = 0$

For the address and we have:

$-F_3cos(b) + F_2 = 0$

The forces $F_1$ and $F_2$ are known

$F_1 = 5.7\ N\\\\F_2 = 1.9\ N$

We have 2 unknowns ( $F_3$ and b) and we have 2 equations.

Now we clear $F_3$ from the second equation and introduce it into the first equation.

$F_3 = \frac{F_2}{cos (b)}$

Then

$-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°$

Then we find the value of $F_3$

$F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N$

Finally the answer is 6.3 N at 162° counterclockwise from

F1->

7 0

3 years ago

A flywheel of mass 182 kg has an effective radius of 0.62 m (assume the mass is concentrated along a circumference located at th

musickatia [10]

Answer:

A)5524J,

B) 29.2Nm

Explanation:

This question can be treated using work- energy theorem

Work= change in Kinectic energy

W= Δ KE

Work= difference between the final Kinectic energy and intial Kinectic energy.

We know that

Kinectic energy= 1/2 mv^2 .............eqn(1)

This can be written in term of angular velocity, as

KE= 1/2 I

4 0

3 years ago

4. A car accelerating at 60 meters/second is hit in an accident by a bus. The net force exerted on

Licemer1 [7]

Answer:

500kg

Explanation:

mass = newtons/force divided by the acceleration rate

m = 30,000/60

m = 500

3 0

2 years ago

Ayuda por favor (archivo adjunto) con un ejercicio de expresión sobre periodo de oscilación de esta figura:

vodka [1.7K]

Answer:

nolo se

Explanation:

no lo se

8 0

3 years ago

Water flows through a horizontal pipe of varying cross-section. In the first section, the cross-sectional area is 10 cm2 and flo

Stels [109]

Answer:

(a) the flow speed of the second section is 11 m/s

(b) the pressure of the second section is 6.33 x 10⁴ Pa

Explanation:

Given;

flow rate in the first section, Q₁ = 2750 cm³/sec

area of the first cross section, A₁ = 10 cm²

pressure in the first cross section, P₁ = 1.2 x 10⁵ Pa

area of the second section, A₂ = 2.5 cm²

(a) the flow speed of the second section (V₂)

Apply continuity equation;

Q₁ = Q₂

Q₁ = A₂V₂

V₂ = Q₁ / A₂

V₂ = (2750) / (2.5)

V₂ = 1100 cm/s = 11 m/s

(b) the pressure of the second section (P₂)

Apply Bernoulli's equation;

P₁ + ¹/₂ρV₁² = P₂ + ¹/₂ρV₂²

where;

ρ is density of water = 1000 kg/m³

V₁ is the speed of water in the first section;

Q₁ = A₁V₁

V₁ = Q₁ / A₁

V₁ = (2750) / (10)

V₁ = 275 cm/s = 2.75 m/s

P₂ = P₁ + ¹/₂ρV₁² - ¹/₂ρV₂²

P₂ = P₁ + ¹/₂ρ(V₁² - V₂²)

P₂ = 1.2 x 10⁵ Pa + ¹/₂ x 1000 (2.75² - 11²)

P₂ = 1.2 x 10⁵ Pa + 500(-113.438)

P₂ = 1.2 x 10⁵ Pa - 0.567 x 10⁵ Pa

P₂ = 0.633 x 10⁵ Pa

P₂ = 6.33 x 10⁴ Pa

8 0

2 years ago