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Andre45 [30]
2 years ago
7

Two blocks, with masses m and 3m , are attached to the ends of a string with negligible mass that passes over a pulley, as shown

above. The pulley has negligible mass and friction and is attached to the ceiling by a bracket. The blocks are simultaneously released from rest.
(b) Determine the work done by the string on the two-block system as each block moves a distance d.
Physics
1 answer:
liq [111]2 years ago
6 0

Answer:

(a)v = √gd (b) Td - Td = 0 (c) -g/4 m/s²

Explanation:

Solution

Given that:

By the work energy theorem,

ΔKE = W = work done

1/2 *3m * v² + 1/2 * m * v² = 3mgd - mgd

so,

2 v² = 2gd

v = √gd

(b)W tension system = Td - Td = 0

(c) a = 3mg - mg/ 4m = g/2

Thus,

Acm = 3 m (g/2) + m (g/2)/4m

= -g/4 m/s²

Note: The complete question to this example is shown below

Two blocks, with masses m and 3m, are attached to the ends of a string with negligible mass that passes over a pulley, as shown above. The pulley has negligible mass and friction and is attached to the ceiling by a bracket. The blocks are simultaneously released from rest.

(a) Derive an equation for the speed v of the block of mass 3m after it falls a distance d in terms of m, d, and physical constants, as appropriate.

(b) Determine the work done by the string on the two-block system as each block moves a distance d.

(c) The acceleration of the center of mass of the blocks-string-pulley system has magnitude aCOM . Briefly  explain, in terms of any external forces acting on the system, why aCOM

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BabaBlast [244]

Answer: Thermal comductivity (K) is 3.964x 10 ^-3 W/m.k

Explanation:

Thermal comductivity K = QL/A∆T

Q= Amount of heat transferred through the material in watts = 75W

L= Distance between two isothermal planes = 0.740mm

A= Area of the surface in square metres = 2m^2

∆T= Temperature change = (37-30) °C.

Solving this : K =( 75 x 0.740 x 10^-3)/ 2 x (37-30)

K = 3.964x 10 ^-3 W/m.k

7 0
2 years ago
A 25N force is applied to a bar that can pivot around its end. The force is r=0.75 m away from the end of an angle at 0= 30. wha
Alecsey [184]

Answer:

<h2>9.375Nm</h2>

Explanation:

The formula for calculating torque τ = Frsin∅ where;

F = applied force (in newton)

r = radius (in metres)

∅ = angle that the force made with the bar.

Given  F= 25N, r = 0.75m and ∅ = 30°

torque on the bar τ  = 25*0.75*sin30°

τ = 25*0.75*0.5

τ = 9.375Nm

The torque on the bar is 9.375Nm

6 0
2 years ago
: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop
Sholpan [36]

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

  1. On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
  2. On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒Fa*sin(50)-Fb*sin(20)=0

⇒Fa*sin(50)=Fb*sin(20)

⇒Fa=2.24Fb

Given that resultant force on car is 950N in positive x-direction

⇒∑Fx=950  

⇒Fa*cos(20)+Fb*cos(50)=950

⇒2.24*Fb*cos(20)+Fb(50)=950

⇒Fb*(2.24*cos(20)+cos(50))=950

⇒Fb=\frac{950}{2.24*cos(20)+cos(50)}

⇒Fb=\frac{950}{2.24*0.94+0.64}

⇒ Fb=\frac{950}{2.75}=345.5

⇒Fa=2.24*Fb

      =2.24*345.5

      =773.93

Therefore approximately, Fa=774 N and Fb=346 N

5 0
2 years ago
The weight of Earth's atmosphere exerts an average pressure of 1.01 ✕ 105 Pa on the ground at sea level. Use the definition of p
zloy xaker [14]

Answer:

The weight of Earth's atmosphere exert is 516.6\times10^{17}\ N

Explanation:

Given that,

Average pressure P=1.01\times10^{5}\ Pa

Radius of earth R_{E}=6.38\times10^{6}\ m

Pressure :

Pressure is equal to the force upon area.

We need to calculate the weight of earth's atmosphere

Using formula of pressure

P=\dfrac{F}{A}  

F=PA

F=P\times 4\pi\times R_{E}^2

Where, P = pressure

A = area

Put the value into the formula

F=1.01\times10^{5}\times4\times\pi\times(6.38\times10^{6})^2

F=516.6\times10^{17}\ N

Hence, The weight of Earth's atmosphere exert is 516.6\times10^{17}\ N

8 0
2 years ago
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The one that is loaded worst. The overall weight is not important; tongue weight is a matter of loading. Our 12,000 lb snow cat trailer, which has stops to position the cat properly, has under 100 lbs tongue weight. Excessive tongue weight is a Bad Thing because it reduces weight on the towing vehicle's front wheels, leading to instability.

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