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Readme [11.4K]
4 years ago
10

The maximum mass the can be hung vertically from a string without breaking the string is 10kg. A length of this string that is 2

m long is used to rotate a 0.5kg object in a circle on a frictionless table with the string horizontal. The maximum speed that the mass can attain under these conditions without the string breaking is most nearly:
Answer is 20 m/s
Please show me how I can get the answer. Thank you.
Physics
1 answer:
vichka [17]4 years ago
0 0

Answer:

Explanation:

Think about circular motion when answering this question. First determine the maximum force that can be applied on the string. F = mg so F = (10)(10) = 100 N. Then determine the centripetal acceleration of the .5 kg mass, a = F/m so a = 100/.5 = 200 m/s².  On the equation sheet, use equation a(centripetal acceleration) = v²/r so 200 = v²/2 therefore v = 20 m/s. Hope this helps!

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Analyze how friction can be both a positive and negative aspect on our lives
denpristay [2]
Positive - Friction allows us to create heat in a desperate situation, like being lost in the woods. If we didn't have friction, we would probably freeze to death.
Negative - Friction can also cause unwanted fires, such as forest fires. If friction didn't exist, we wouldn't have these.
8 0
3 years ago
An object of mass 6 kg. is resting on a horizontal surface. A horizontal force
son4ous [18]

Answer:

a) The work done by the applied force is 1500 joules.

b) The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) 300 joules of energy are lost during motion.

Explanation:

a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force (W), measured in joules, is defined by the following expression:

W = F\cdot \Delta x (1)

Where:

F - Force, measured in newtons.

\Delta x - Distance, measured in meters.

If we know that F = 15\,N and \Delta x = 100\,m, then the work done by the force exerted on the object is:

W = (15\,N)\cdot (100\,m)

W = 1500\,J

The work done by the applied force is 1500 joules.

b) At first we need to calculate the net acceleration of the object (a), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:

\Delta x = v_{o}\cdot t +\frac{1}{2}\cdot a\cdot t^{2} (2)

Where v_{o} is the initial velocity of the object, measured in meters per second.

We clear the acceleration within the equation above:

\frac{1}{2}\cdot a \cdot t^{2} =  \Delta x-v_{o}\cdot t

a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}

If we know that \Delta x = 100\,m, v_{o} = 0\,\frac{m}{s} and t = 10\,s, then the net acceleration experimented by the object is:

a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}

a = 2\,\frac{m}{s^{2}}

By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:

\Sigma F = F - f = m\cdot a (3)

Where:

F - External force exerted on the object, measured in newtons.

f - Kinetic friction force, measured in newtons.

If we know that F = 15\,N, m = 6\,kg and a = 2\,\frac{m}{s^{2}}, the kinetic friction force is:

f = F-m\cdot a

f = 15\,N-(6\,kg)\cdot \left(2\,\frac{m}{s^{2}} \right)

f = 3\,N

The work done by friction (W'), measured in joules, is:

W' = f\cdot \Delta x (4)

W' = (3\,N) \cdot (100\,m)

W' = 300\,J

And the net work experimented by the object is:

\Delta W = 1500\,J - 300\,J

\Delta W = 1200\,J

By the Work-Energy Theorem we understand that change in translational kinetic energy (\Delta K), measured in joules, is equal to the change in net work. That is:

\Delta K = \Delta W (5)

If we know that \Delta W = 1200\,J, then the change in translational kinetic energy is:

\Delta K = 1200\,J

The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.

7 0
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Alenkinab [10]
As one moves farther and farther from the Sun, the distance between adjacent planets is greater.
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3 years ago
What is the effect of pressure on the solubility of gases in liquids?
34kurt
The solubility of gases in liquids increases with the increase in pressure.
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3 years ago
Si pudieras viajar a la Luna:
Nostrana [21]

Answer:

i) Distancia, ii) La cinta métrica es impracticable.

Explanation:

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La distancia puede involucrar trayectorias curvilíneas entre los puntos mencionados.

ii) Por último, el uso de una cinta métrica es impracticable debido a la cantidad de material a utilizar y los efectos gravitacionales, electromagnéticos y mecánicos que inducen a una deflexión o una ruptura de esa cinta debido a la magnitud de la distancia entre las superficies del planeta y el satélite, respectivamente.

En este caso, es mejor utilizar la medición con tecnología láser, basadas en el fenómeno del electromagnetismo.

4 0
3 years ago
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