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jek_recluse [69]
3 years ago
15

Airbrakes on some trucks have how many second lag time with for the engage properly?

Physics
1 answer:
olga2289 [7]3 years ago
6 0
The answer that best completes the given statement above is ONE SECOND (1). There should only be one second of lag time to engage properly for some of the airbrakes of the trucks. One half to one second is the normal lag time and it should not exceed beyond this time. Hope this helps.
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How much force is required to accelerate a 2 kg mass at 3 m/s2
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The parachute on a drag racing car deploys at the end of a run. If the car has a mass of 820 kg and the car is moving 36 m/s, wh
Lelechka [254]

In order to determine the required force to stop the car, proceed as follow:

Calculate the deceleration of the car, by using the following formula:

v^2=v^2_o-2ax

where,

v: final speed = 0m/s (the car stops)

vo: initial speed = 36m/s

x: distance traveled = 980m

a: deceleration of the car= ?

Solve the equation above for a, replace the values of the other parameters and simplify:

\begin{gathered} a=\frac{v^2_o-v^2}{2x} \\ a=\frac{(36\frac{m}{s})^2-(0\frac{m}{s})^2}{2(980m)}=0.66\frac{m}{s^2} \end{gathered}

Next, consider that the formula for the force is:

F=ma

where,

m: mass of the car = 820 kg

a: deceleration of the car = 0.66m/s^2

Replace the previous values and simplify:

F=(820kg)(0.66\frac{m}{s^2})=542.20N

Hence, the required force to stop the car is 542.20N

4 0
1 year ago
The AC voltage source is connected to an inductor and a resistor in series. If the frequency of the source is increased the curr
DaniilM [7]

Answer:

If the frequency of the source is increased the current in the circuit will decrease.

Explanation:

The current through the circuit is given as;

I = \frac{V}{Z}

Where;

V is the voltage in the AC circuit

Z is the impedance

Z = \sqrt{R^2 + X_L^2}

Where;

R is the resistance

X_L is the inductive reactance

X_L = ωL = 2πfL

where;

L is the inductance

f is the frequency of the source

Finally, the current in the circuit is given as;

I = \frac{V}{\sqrt{R^2 + (2\pi fL)^2} }

From the equation above, an increase in frequency (f) will cause a decrease in current (I).

Therefore, If the frequency of the source is increased the current in the circuit will decrease.

5 0
3 years ago
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