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bezimeni [28]
3 years ago
8

Uzupełnij zdania właściwymi sformułowaniami. Wyobraź sobie, że między linę a siodełko karuzeli łańcuchowej wmontowany jest siłom

ierz. Jeśli na postoju nie dotykasz nogami do ziemi, to siłomierz wskazuje A / B. Kiedy karuzela się kręci, na siłomierzu odczytasz C / D. A. Twój ciężar wraz z siodełkiem C. Wartość wytrzymałości liny B. Twój ciężar D. Wartość siły dośrodkowej
Physics
1 answer:
iragen [17]3 years ago
3 0

Explanation:

Here's a clearer rendering of the question requirements;

Complete the sentences with the correct wording. Imagine that a force gauge is mounted between the rope and the chain carousel saddle. If you do not touch your feet to the ground when the vehicle is stationary, the dynamometer indicates A / B. When the carousel turns, you will read C / D on the dynamometer.

A. Your weight with the saddle

C. Rope strength value

B. Your weight

D. Centripetal force value

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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
3 years ago
Movement of a molecule against its concentration gradient can occur through Select one or more: a. facilitated diffusion b. pass
uranmaximum [27]

Answer:

The answer to your question is:

Explanation:

There are two kinds of cell transport passive transportation and active transportation.

Passive transportation does not need energy because molecules move from higher concentration to lower concentration.

Active transportation needs energy because molecules moves against concentration.  

a. facilitated diffusion It's an example of passive transportation so this answer is wrong.

b. passive transport Molecules move in favor of concentration so this answer is wrong.

c. osmosis is another example of passive transport so this answer is wrong.

d. simple diffusion it's another example of passive transport, so it's wrong this answer.

e. active transport this is the right answer.

5 0
3 years ago
Which of the following materials is necessary to stop an alpha particle? a. three feet of concrete c. single sheet of aluminum f
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A single sheet of paper can stop an alpha particle
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3 years ago
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What’s the equalibrium rule?
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The vector sum of forces acting on a non-accelerating object equals zero.
equation form: ΣF = 0
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A pilot can withstand an acceleration of up to 9g, which is about 88 m/s2, before blacking out. (a) What is the acceleration exp
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Yes the plot dose make it with out baking out
7 0
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