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sveticcg [70]
3 years ago
13

Thiết kế mạch điện tử sử dụng chân ngõ ra (I/O) của arduino UNO để lái relay 12VDC/500mA sao cho chân Arduino xuất ra tín hiệu 5

V thì relay đóng, ngược lại chân I/O arduino xuất tín hiệu 0V thì relay hở. Mạch lái dùng BJT
Physics
1 answer:
myrzilka [38]3 years ago
6 0

Answer:

Mình cũng không biết làm bạn ơi =)))

Explanation:

You might be interested in
The question and answer options are in the photo!
Bezzdna [24]
Friction pushes her hands!!
6 0
2 years ago
EXPERTS/ACE and people that wanna help 4 sure only!
mario62 [17]
That first one you have selected (3,-3) works in both equations so it's correct.
good job.

you can do this guess and test method with multiple choice answers. If it works in both equations it is the solution. Otherwise use substitution or elimination to combine the two into one equation in only one variable. Then you can solve for the one variable first and use it to solve for the other.

3 0
3 years ago
A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen
PilotLPTM [1.2K]

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})

Where:

\omega_{o}, \omega - Initial and final angular velocities, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

\theta_{o}, \theta - Initial and final angular position, measured in radians.

Then,

\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}

Given that \omega_{o} = 0\,\frac{rad}{s}, \omega = 0.70\,\frac{rad}{s} and \theta-\theta_{o} = 4.9\,rad, the angular acceleration is:

\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}

\alpha = 0.05\,\frac{rad}{s^{2}}

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

\omega = \omega_{o} + \alpha \cdot t

Where t is the time measured in seconds.

The time is cleared and obtain after replacing every value:

t = \frac{\omega-\omega_{o}}{\alpha}

If \omega_{o} = 0\,\frac{rad}{s},  \omega = 0.70\,\frac{rad}{s} and \alpha = 0.05\,\frac{rad}{s^{2}}, the required time is:

t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }

t = 14\,s

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

\bar \alpha = \frac{\omega-\omega_{o}}{t}

If \omega_{o} = 0\,\frac{rad}{s},  \omega = 0.70\,\frac{rad}{s} and t = 14\,s, the average angular acceleration is:

\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}

\bar \alpha = 0.05\,\frac{rad}{s^{2}}

The average angular acceleration is 0.05 radians per square second.

4 0
2 years ago
Determine the volume displaced and then calculate the density of this 54 g sample of brass.
inessss [21]

Answer:

DETAILS IN THE QUESTION INSUFFICIENT TO ANSWER

Explanation:

Assuming the liquid to be water ,

the density d_{w} of water is : 1000kgm^{-3}=1gcm^{-3}

Buoyant force exerted by a liquid on an object with V_{imm} of it's volume immersed is :

F_{B}=V_{imm}*d_{l}*g

where ,

  • F_{B} is the buoyant force
  • d_{l} is the density of the liquid
  • g is the acceleration due to gravity

Thus at equilibrium:

m_{brass}*g=V_{imm}*d_{l}*g\\m_{brass}=V_{imm}*d_{l}\\54=V_{imm}*1\\V_{imm}=54cm^{3}

from these , we get the density of brass to be 1gcm^{-3}

which is not possible

7 0
3 years ago
A 7.0kg object rests on a horizontal frictionless surface. What is the magnitude of the horizontal
sukhopar [10]

Answer:

16.1 N

Explanation:

From the question,

F = ma.............................. Equation 1

Where F = horizontal force, m = mass of the object, a = acceleration .

Given: m = 7.0 kg, a = 2.3 m/s²

Substitute this values into equation 1

F = (7.0×2.3)

F = 16.1 N.

Hence the magnitude of the horizontal force is 16.1 N

6 0
2 years ago
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