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olchik [2.2K]
2 years ago
12

An elevator held by a single cable is ascending but slowing down. Is the work done by tension positive, negative, or zero? What

about the work done by gravity? Explain
Physics
1 answer:
inessss [21]2 years ago
8 0

Explanation:

1.) Work = Force*Distance

2.) The direction of motion lies in the same direction as the tension in the cable. So the work done by tension would be positive.

3.) The direction of the weight (due to gravity) would act opposite to the direction of motion of the elevator, so work done by gravity becomes negative.

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A merry-go-round with a a radius of R = 1.63 m and moment of inertia I = 196 kg-m2 is spinning with an initial angular speed of
kondor19780726 [428]

Answer:

1) L = 299.88 kg-m²/s

2) L = 613.2 kg-m²/s

3) L = 499.758 kg-m²/s

4) ω₁ = 0.769 rad/s

5) Fc = 70.3686 N

6) v = 1.2535 m/s

7) ω₀ = 1.53 rad/s

Explanation:

Given

R = 1.63 m

I₀ = 196 kg-m²

ω₀ = 1.53 rad/s

m = 73 kg

v = 4.2 m/s

1) What is the magnitude of the initial angular momentum of the merry-go-round?

We use the equation

L = I₀*ω₀ = 196 kg-m²*1.53 rad/s = 299.88 kg-m²/s

2) What is the magnitude of the angular momentum of the person 2 meters before she jumps on the merry-go-round?

We use the equation

L = m*v*Rp = 73 kg*4.2 m/s*2.00 m = 613.2 kg-m²/s

3) What is the magnitude of the angular momentum of the person just before she jumps on to the merry-go-round?

We use the equation

L = m*v*R = 73 kg*4.2 m/s*1.63 m = 499.758 kg-m²/s

4) What is the angular speed of the merry-go-round after the person jumps on?

We can apply The Principle of Conservation of Angular Momentum

L in = L fin

⇒ I₀*ω₀ = I₁*ω₁

where

I₁ = I₀ + m*R²

⇒  I₀*ω₀ = (I₀ + m*R²)*ω₁

Now, we can get ω₁

⇒  ω₁ = I₀*ω₀ / (I₀ + m*R²)

⇒  ω₁ = 196 kg-m²*1.53 rad/s / (196 kg-m² + 73 kg*(1.63 m)²)

⇒  ω₁ = 0.769 rad/s

5) Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on?

We have to get the centripetal force as follows

Fc = m*ω²*R  

⇒  Fc = 73 kg*(0.769 rad/s)²*1.63 m = 70.3686 N

6) Once the person gets half way around, they decide to simply let go of the merry-go-round to exit the ride.

What is the linear velocity of the person right as they leave the merry-go-round?

we can use the equation

v = ω₁*R = 0.769 rad/s*1.63 m = 1.2535 m/s

7) What is the angular speed of the merry-go-round after the person lets go?

ω₀ = 1.53 rad/s

It comes back to its initial angular speed

8 0
2 years ago
A closed and elevated vertical cylindrical tank with diameter 2.00 m contains water to a depth of 0.800 m. A worker accidentally
vazorg [7]

Answer:

a. v1 = 5.06 m/s,  v2 = 3.96 m/s ,  R = 1.27

b. t = 1 hr, 11 min, 26 sec  

Explanation:

Using the Bernoulli's laws to use the conserved energy

a. Solve the speed and the radio of this speed of the tank is open to the air

p₀ + ρgh₀ + ½ρv₀² = p₁ + ρgh₁ + ½ρv₁²

5000Pa + 1000kg/m³ * 9.8m/s² * 0.800m + 0 = 0 + 0 + ½ * 1000kg/m³ * v²

v² = 25.68  m²/s²

v1 = 5.06 m/s

Because it is open the tank so P=0 pa so:

0 Pa + 1000kg/m³ * 9.8m/s² * 0.800m + 0 = 0 + 0 + ½ * 1000kg/m³ * v²

v² = 15.68  m²/s²

v2 = 3.96 m/s

The ratio on the air is solve using both velocities so:

R = v1/v2 = 5.06 m/s / 3.96 m/s

R = 1.27

b. Now to find the time it takes for the tank to drain if the tank is open to the air

dh/dt = -u

dh/dt = -v * A/A'

dh/dt = v*(.02m)²/(2.0m)² = -v / 10000

and we can further substitute for v:

dh/dt = -(1/1e4)*√[(p+9800h)/500]

Solve replacing

-(1000/49)*√(49000h) = t + C

-(1000/49)*√(49000*0.8) = 0 + C

C = - 4040.6

Then when h = 0,

t = 4286 s

t = 1 hr, 11 min, 26 sec  

6 0
2 years ago
A motorcycle of mass 100 kilograms travels around a flat, circular track of radius 10 meters with a constant speed of 20 meters
JulijaS [17]

Answer:

100/10 = 10 , 10 × 10 = 100÷20 = 5

I'm pretty sure its wrong

8 0
2 years ago
Read 2 more answers
The boom hoisting sheave must have pitch diameters of no less than _______times the nominal diameter of the rope used.
alexira [117]

Answer:

18 times

Explanation:

According to the security purposes which is set under the rules and regulation OSHA, which describes all the rights to the worker.

In the boom hoist receiving system all the sheaves which are used should have a pitch diameter of rope not less than 18 times the diameter of the nominal rope which is used.

7 0
2 years ago
A hot air balloon of total mass M (including passengers and luggage) is moving with a downward acceleration of magnitude a. As i
LUCKY_DIMON [66]

Answer:

The fraction of mass that was thrown out is calculated by the following Formula:

M - m = (3a/2)/(g²- (a²/2) - (ag/2))

Explanation:

We know that Force on a moving object is equal to the product of its mass and acceleration given as:

F = ma

And there is gravitational force always acting on an object in the downward direction which is equal to g = 9.8 ms⁻²

Here as a convention we will use positive sign with acceleration to represent downward acceleration and negative sign with acceleration represent upward acceleration.

Case 1:

Hot balloon of mass = M

acceleration = a

Upward force due to hot air = F = constant

Gravitational force downwards = Mg

Net force on balloon is given as:

Ma = Gravitational force - Upward Force                              

Ma = Mg - F                      (balloon is moving downwards so Mg > F)

F = Mg - Ma

F = M (g-a)

M = F/(g-a)

Case 2:

After the ballast has thrown out,the new mass is m. The new acceleration is -a/2 in the upward direction:

Net Force is given as:

-m(a/2) = mg - F        (Balloon is moving upwards so F > mg)

F = mg + m(a/2)

F = m(g + (a/2))

m = F/(g + (a/2))

Calculating the fraction of the initial mass dropped:

M-m = \frac{F}{g-a} - \frac{F}{g+\frac{a}{2} }\\M-m = F*[\frac{1}{g-a} - \frac{1}{g+\frac{a}{2} }]\\M-m = F*[\frac{(g+(a/2)) - (g-a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{g+(a/2) - g + a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{(3a/2)}{g^{2}-\frac{a^{2}}{2}-\frac{ag}{2}} ]

5 0
2 years ago
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