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lawyer [7]
2 years ago
6

Two identical ladders are 3.0 m long and weigh 600 N each. They are connected by a hinge at the top and are held together by a h

orizontal rope, 1.0 m above the smooth floor forming a symmetric "A" arrangement. The angle between the ladders is 60° and both ladders have their center of gravity at their midpoint. What is the tension in the rope?
Physics
1 answer:
ruslelena [56]2 years ago
7 0

Answer:

The tension in the rope is 281.60 N.

Explanation:

Given that,

Length = 3.0 m

Weight = 600 N

Distance = 1.0 m

Angle = 60°

Consider half of the ladder,

let tension be T, normal reaction force at ground be F, vertical reaction at top hinge be Y and horizontal reaction force be X.

Y+F=600....(I)

X=T.....(II)

On taking moment about base

X\times l\cos\theta+Y\times l\sin\theta-F\dfrac{l}{2}\sin\theta-T\times d=0

Put the value into the formula

X\times3\cos30+Y\times3\sin30-600\times1.5\sin30-T\times1=0

3\cos30 T-T=600\times1.5\sin30-Y \times3\sin30

1.598T=450-1.5(600-F)....(III)

We need to calculate the force for ladder

2F=600\trimes  2

F=600\ N

We need to calculate the tension in the rope

From equation (3)

1.598T=450-1.5(600-600)

1.598T=450

T=\dfrac{450}{1.598}

T=281.60\ N

Hence, The tension in the rope is 281.60 N.

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Two fun-loving otters are sliding toward each other on a muddy (and hence frictionless) horizontal surface. One of them, of mass
zvonat [6]

Answer:

(a). The magnitude and direction of the velocity of the otters after collision is 1.35 m/s toward left.

(b). The mechanical energy dissipates during this play is 226.98 J.

Explanation:

Given that,

Mass of one otter = 8.50 kg

Speed = 6.00 m/s

Mass of other = 5.75 kg

Speed = 5.50 m/s

(a). We need to calculate the magnitude and direction of the velocity of these free-spirited otters right after they collide

Using conservation of momentum

m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v

Put the value into the formula

8.50\times(-6.00)+5.75\times5.50=(8.50+5.75)\times v

v=\dfrac{-19.375}{14.25}

v=-1.35\ m/s

Negative sign shows the direction of motion of the object after collision is toward left.

(b). We need to calculate the initial kinetic energy

Using formula of kinetic energy

K.E_{i}=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2

Put the value into the formula

K.E_{i}=\dfrac{1}{2}\times8.50\times(6.00)^2+\dfrac{1}{2}\times5.75\times(5.50)^2

K.E_{i}=239.96\ J

We need to calculate the final kinetic energy

Using formula of kinetic energy

K.E_{f}=\dfrac{1}{2}(m_{1}+m_{2})v^2

Put the value into the formula

K.E_{f}=\dfrac{1}{2}\times(8.50+5.75)\times(-1.35)^2

K.E_{f}=12.98\ J

We need to calculate the mechanical energy dissipates during this play

Using formula of loss of mechanical energy

\Delta K.E=K.E_{f}-K.E_{i}

Put the value into the formula

\Delta K.E=12.98-239.96

\Delta K.E=-226.98\ J

Negative sign shows the loss of mechanical energy

Hence, (a). The magnitude and direction of the velocity of the otters after collision is 1.35 m/s toward left.

(b). The mechanical energy dissipates during this play is 226.98 J.

8 0
2 years ago
Read 2 more answers
In the picture below, explain why the bear fell. Use FRICTION to explain your answer.
algol13
The bear fell because it slides to the surface of ice due to lack of friction.

One of these theories is that friction<span> causes the liquid layer of water to form on </span>ice<span>. </span>Friction<span> is the force that generates heat whenever two objects slide against each other. If you rub your hands together, you can feel them heat up. That's </span>friction<span> at work. When a </span>skate<span> moves over the surface of </span>ice, the friction<span> between the </span>skate<span> and the </span>ice<span> generates heat that melts the </span>outermost<span> layer of </span>ice<span>.</span>
5 0
1 year ago
WallyGPX accelerates from 0 m/s to 8 m/s in 3 seconds. What is his acceleration? Is this acceleration higher than that of a car
olga nikolaevna [1]

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4 0
2 years ago
A copper wire has radius 0.800 mm and carries current I at 20.0°C. A silver wire with radius 0.500 mm carries the same current a
IgorC [24]

Answer:

Ecu/Eag = 0.46

Explanation:

E = PI/A

Ecu = Pcu × I/A

Pcu = 1.72×10^-8 ohm-meter

r = 0.8 mm = 0.8/1000 = 8×10^-4 m

A = πr^2 = π×(8×10^-4)^2 = 6.4×10^-7π

Ecu = 1.72×10^-8I/6.4×10^-7π = 0.026875I/1

Eag = Pag × I/A

Pag = 1.47×10^-8 ohm-meter

r = 0.5 mm = 0.5/1000 = 5×10^-4 m

A = πr^2 = π × (5×10^-4)^2 = 2.5×10^-7π

Eag = 1.47×10^-8I/2.5×10^-7π = 0.0588I/π

Ecu/Eag = 0.026875I/π × π/0.0588I = 0.46

7 0
2 years ago
Suppose that we use a heater to boil liquid nitrogen (N2 molecules). 4480 J of heat turns 20 g of liquid nitrogen into gas. Note
love history [14]

Answer:

The energy is 9.4\times10^{-21}\ J

(a) is correct option

Explanation:

Given that,

Energy = 4480 j

Weight of nitrogen = 20 g

Boil temperature = 77 K

Pressure = 1 atm

We need to calculate the internal energy

Using first law of thermodynamics

Q=\Delta U+W

Q=\Delta U+nRT

Put the value into the formula

4480=\Delta U+\dfrac{20}{28}\times8.314\times77

\Delta U=4480-\dfrac{20}{28}\times8.314\times77

\Delta U=4022.73\ J

We need to calculate the number of molecules in 20 g N₂

Using formula of number of molecules

N=n\times \text{Avogadro number}

Put the value into the formula

N=\dfrac{20}{28}\times6.02\times10^{23}

N=4.3\times10^{23}

We need to calculate the energy

Using formula of energy

E=\dfrac{\Delta U}{N}

Put the value into the formula

E=\dfrac{4022.73}{4.3\times10^{23}}

E=9.4\times10^{-21}\ J

Hence, The energy is 9.4\times10^{-21}\ J

4 0
1 year ago
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