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user100 [1]
3 years ago
5

el extremo de un brazo robotico se mueve hacia la derecha a 8m/s . cuatro segundos despues. se mueve hacia la izquierda a 2m/s.¿

cual es su cambio de velocidad y cual su cambio de aceleracion?
Physics
1 answer:
rewona [7]3 years ago
3 0

Answer: Change in velocity: -10 m/s, acceleration=-2.5m/s

Explanation:

The question in english is written below:

The end of a robotic arm moves to the right at 8m s. Four seconds later it moves to the left at 2m/s. What is its change in velocity and what is its change in acceleration?

We have the following data:

Initial Velocity of the robotic arm: V_{o}=8 m/s (to the right)

Final Velocity of the robotic arm: V_{f}=-2 m/s (to the left)

Variation in time: \Delta t=4 s

Firstly we have to calculate the change in velocity \Delta V, which is given by the following equation:

\Delta V=V_{f}-V_{o}=-2 m/s-8 m/s

\Delta V=-10 m/s This means the resultant velocity is to the left

Now, we can find the acceleration a, given by:

a=\frac{\Delta V}{\Delta t}

a=\frac{-10 m/s}{4 s}

a=-2.5 m/s^{2} This means the acceleration is also to the left, however its magnitude is always positive: 2.5 m/s^{2}

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What do understand by the efficiency of a machine? By using a block and tackel a man can raise a load of 720 N by an effort of 1
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Answer:

Efficiency of a machine is how well the machine works and what the machine is capable of doing.

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720/180=4

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Assume the following values: d1 = 0.880 m , d2 = 1.11 m , d3 = 0.560 m , d4 = 2.08 m , F1 = 510 N , F2 = 306 N , F3 = 501 N , F4
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Answer:

= 2630.6 N.m

Explanation:

(FR)x = ΣFx = -F4 = -407 N

(FR)y = ΣFy =-F1-F2 -F3 = -510 - 306 - 501 = -1317 N

(MR)B =ΣM + Σ(±Fd)

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= 1504 + 510(0.880+1.11) +306(1.11) - 407(0.560)

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3 years ago
The component of the external magnetic field along the central axis of a 78-turn circular coil of radius 34.0 cm decreases from
grigory [225]

Answer:

Induced current, I = 18.88 A

Explanation:

It is given that,

Number of turns, N = 78

Radius of the circular coil, r = 34 cm = 0.34 m

Magnetic field changes from 2.4 T to 0.4 T in 2 s.

Resistance of the coil, R = 1.5 ohms

We need to find the magnitude of the induced current in the coil. The induced emf is given by :

\epsilon=-N\dfrac{d\phi}{dt}

Where

\dfrac{d\phi}{dt} is the rate of change of magnetic flux,

And \phi=BA

\epsilon=-NA\dfrac{dB}{dt}

\epsilon=-78\times \pi (0.34)^2\dfrac{(0.4-2.4)}{2}

\epsilon=28.32\ V

Using Ohm's law, \epsilon=I\times R

Induced current, I=\dfrac{\epsilon}{R}

I=\dfrac{28.32}{1.5}

I = 18.88 A

So, the magnitude of the induced current in the coil is 18.88 A. Hence, this is the required solution.

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Convert 123,453 to a scientific notation
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Answer:

1.23453*10^5 is scientific way

3 0
3 years ago
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