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3 years ago

5. A massless string passes over a frictionless pulley and carries

Physics

1 answer:

devlian [24]3 years ago

8 0

Answer:

2m₁m₃g / (m₁ + m₂ + m₃)

Explanation:

I assume the figure is the one included in my answer.

Draw a free body diagram for each mass.

m₁ has a force T₁ up and m₁g down.

m₂ has a force T₁ up, T₂ down, and m₂g down.

m₃ has a force T₂ up and m₃g down.

Assume that m₁ accelerates up and m₂ and m₃ accelerate down.

Sum of the forces on m₁:

∑F = ma

T₁ − m₁g = m₁a

T₁ = m₁g + m₁a

Sum of the forces on m₂:

∑F = ma

T₁ − T₂ − m₂g = m₂(-a)

T₁ − T₂ − m₂g = -m₂a

(m₁g + m₁a) − T₂ − m₂g = -m₂a

m₁g + m₁a + m₂a − m₂g = T₂

(m₁ − m₂)g + (m₁ + m₂)a = T₂

Sum of the forces on m₃:

∑F = ma

T₂ − m₃g = m₃(-a)

T₂ − m₃g = -m₃a

a = g − (T₂ / m₃)

Substitute:

(m₁ − m₂)g + (m₁ + m₂) (g − (T₂ / m₃)) = T₂

(m₁ − m₂)g + (m₁ + m₂)g − ((m₁ + m₂) / m₃) T₂ = T₂

(m₁ − m₂)g + (m₁ + m₂)g = ((m₁ + m₂ + m₃) / m₃) T₂

m₁g − m₂g + m₁g + m₂g = ((m₁ + m₂ + m₃) / m₃) T₂

2m₁g = ((m₁ + m₂ + m₃) / m₃) T₂

T₂ = 2m₁m₃g / (m₁ + m₂ + m₃)

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3 years ago

What is 1.0 x 10^9 in standard form?

jasenka [17]

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3 years ago

Give two mathematical examples of Newton's third law and how you get the solution

bagirrra123 [75]

Answer:

1) Any particle moving in a horizontal plane slowed by friction, deceleration = 32 μ

2) The particle moving by acceleration = P/m - 32μ OR The external force = ma + 32μm

Explanation:

* Lets revise Newton’s Third Law:

- For every action there is a reaction, equal in magnitude and opposite

in direction.

- Examples:

# 1) A particle moving freely against friction in a horizontal plane

- When no external forces acts on the particle, then its equation of

motion is;

∵ ∑ forces in direction of motion = mass × acceleration

∵ No external force

∵ The friction force (F) = μR, where μ is coefficient of the frictional force

and R is the normal reaction of the weight of the particle on the

surface

∵ The frictional force is in opposite direction of the motion

∴ ∑ forces in the direction of motion = 0 - F

∴ 0 - F = mass × acceleration

- Substitute F by μR

∴ - μR = mass × acceleration

∵ R = mg where m is the mass of the particle and g is the acceleration

of gravity

∴ - μ(mg) = ma ⇒ a is the acceleration of motion

- By divide both sides by m

∴ - μ(g) = a

∵ The acceleration of gravity ≅ 32 feet/sec²

∴ a = - 32 μ

* Any particle moving in a horizontal plane slowed by friction,

deceleration = 32 μ

# 2) A particle moving under the action of an external force P in a

horizontal plane.

- When an external force P acts on the particle, then its equation

of motion is;

∵ ∑ forces in direction of motion = mass × acceleration

∵ The external force = P

∵ The friction force (F) = μR, where μ is coefficient of the frictional force

and R is the normal reaction of the weight of the particle on the

surface

∵ The frictional force is in opposite direction of the motion

∴ ∑ forces in the direction of motion = P - F

∴ P - F = mass × acceleration

- Substitute F by μR

∴ P - μR = mass × acceleration

∵ R = mg where m is the mass of the particle and g is the acceleration

of gravity

∴ P - μ(mg) = ma ⇒ a is the acceleration of motion

∵ The acceleration of gravity ≅ 32 feet/sec²

∴ P - 32μm = ma ⇒ (1)

- divide both side by m

∴ a = (P - 32μm)/m ⇒ divide the 2 terms in the bracket by m

∴ a = P/m - 32μ

* The particle moving by acceleration = P/m - 32μ

- If you want to fin the external force P use equation (1)

∵ P - 32μm = ma ⇒ add 32μm to both sides

∴ P = ma + 32μm

* The external force = ma + 32μm

7 0

3 years ago

A car covers 400 km in an hour towards west .calculate the velocity

crimeas [40]

Answer:

-400km/hr

Explanation:

Velocity=displacement/time

=400/1

=400Km/hr

=-400km/hr (because west direction)

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2 years ago

We now have an algebraic expression with only one variable, which can be solved. Once we have that, we can plug it back into one

Len [333]

Answer:

A=1

B=-2

Explanation:

Part A and B of the question wasn't given, however, I attached the relevant parts to solve this question as follows.

From part B as attached, it shows that the right option is C which is

2A+3B=-4

Substituting B with 3A-5 then we form the second equation as shown

2A+3(3A-5)=-4

By simplifying the above equation, we obtain

2A+9A-15=-4

Re-arranging, then

11A=-4+15

Finally

11A=11

A=1

To obtain B, we already know that 3A-5 so substituting the value of A into the above then we obtain

B=3(1)-5=-2

Therefore, required values are 1 and -2

3 0

3 years ago