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erma4kov [3.2K]
3 years ago
11

A driver must always stop within 50 ft but not less than ____________ ft from the nearest rail when the signal is flashing and t

he crossing gates are lowered.
Physics
2 answers:
k0ka [10]3 years ago
5 0

Answer:

20 ft

Explanation:

docker41 [41]3 years ago
4 0

Answer:

15ft

Explanation:

As stated by the motor laws a driver must stop within 50ft and not less than 15ft from the railroad.

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A car company wants to ensure its newest model can stop in less than 450 ft when traveling at 60 mph. If we assume constant dece
seraphim [82]

Answer:

The value of acceleration that accomplishes this is 8.61 ft/s² .

Explanation:

Given;

maximum distance to be traveled by the car when the brake is applied, d = 450 ft

initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s

final velocity of the car when it stops, v = 0

Apply the following kinematic equation to solve for the deceleration of the car.

v² = u² + 2as

0 = 88.02² + (2 x 450)a

-900a = 7747.5204

a = -7747.5204 / 900

a = -8.61 ft/s²

|a| = 8.61 ft/s²

Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .

4 0
2 years ago
If a ball is thrown vertically upward with a velocity of 80 ft/s, then its height after t seconds is s = 80t − 16t 2 . (a) what
MAXImum [283]
Max height occurs when v = 0.
v(t) = ds(t)/dt
v(t) = 80 - 32t
0 = 80 - 32t
t = 5/2

s(5/2) = 80(5/2) - 16(5/2)^2
s(5/2) = 100
Answer: 100 ft

96 = 80t - 16t²
t = 3, 2
(80 ± √256) / 32 using the quadratic equation.

v(2) = 16
v(3) = -16
4 0
3 years ago
Two things that electrical force depends upon?
lidiya [134]
Electric force depends on the charge and the strength of the electric field. The equation that relates the three:

F = Eq where q is the charge and E is the electric field strength.
4 0
3 years ago
Consider a box with two gases separated by an impermeable membrane. The membrane can move back and forth, but the gases cannot p
padilas [110]

Answer:

The answer is "0.3336\ m^3"

Explanation:

Using the Promideal gas law:

P_A=P_B\\\\P_A(V_A-\eta_A b)= \eta_A RT......(1)\\\\P_B V_B=\eta_B \bar{R}T........(2)\\\\From (1) \zeta (2)\\\\  

\frac{\eta_A}{V_A-\eta_A b}=\frac{\eta B}{V B}\\\\  \frac{V A- \eta_A b}{V B}=\frac{\eta A}{\eta B }\\\\  \frac{V A-b}{V B}=\frac{1}{2}\\\\V A+V B=1\\\\V B =1- V A\\\\\frac{V A-b}{1-V A}=\frac{1}{2}\\\\2V A-2b=1-V A\\\\3 V A=1+2b\\\\V A=\frac{1+2b}{3}\\\\

      =\frac{1+2(4\times 10^{-4})}{3}\\\\=0.3336\ m^3

8 0
3 years ago
A bus starts from rest. If it's velocity becomes 60 km/hr in 5 minutes, find
dimulka [17.4K]

Answer:

5 miles

Explanation:

the bus is going 60km/hour meaning its going a mile a minute and it went on for 5 minutes meaning it went for 5 minutes/

5 0
3 years ago
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