Answer:
the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15
Explanation:
Given that;
speed of car V = 120 km/h = 33.3333 m/s
Reaction time of an alert driver = 0.8 sec
Reaction time of an alert driver = 3 sec
extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec
now, extra distance that car will travel in case of sleepy driver will be'
S_d = V × 2.2 sec
S_d = 33.3333 m/s × 2.2 sec
S_d = 73.3333 m
hence, number of car of additional car length n will be;
n = S_n / car length
n = 73.3333 m / 5m
n = 14.666 ≈ 15
Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15
Answer:
0.94 m³/s
Explanation:
From the question given above, the following data were obtained:
Air flow (in ft³/min) = 2×10³ ft³/min
Air flow (in m³/s) =.?
Next, we shall convert 2×10³ ft³/min to m³/min. This can be obtained as follow:
35.315 ft³/min = 1 m³/min
Therefore,
2×10³ ft³/min = 2×10³ ft³/min × 1 m³/min / 35.315 ft³/min
2×10³ ft³/min = 56.63 m³/min
Finally, we shall convert 56.63 m³/min to m³/s. This can be obtained as follow:
1 m³/min = 1/60 m³/s
Therefore,
56.63 m³/min = 56.63 m³/min × 1/60 m³/s ÷ 1 m³/min
56.63 m³/min = 0.94 m³/s
Thus, 2×10³ ft³/minis equivalent to 0.94 m³/s.
Answer:
<u><em>0.03 m/s</em></u>
Explanation:
<em>Applying law of conservation of momentum, </em>
- <em>m₁v₁ + m₂v₂ = (m₁ + m₂)v</em>
- <em>0.105(24) + 75(0) = (0.105 + 75)v</em>
- <em>75.105v = 2.52</em>
- <em>v = 2.52/75.105</em>
- <em>v = </em><u><em>0.03 m/s</em></u>
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Answer:
0.217 m/s
Explanation:
The protons in the beam passes undeflected when the electric force is equal to the magnetic force:
qE = qvB
where
q is the proton's charge
E is the magnitude of the electric field
v is the speed of the protons
B is the magnitude of the magnetic field
Re-arranging the equation,

And by substituting
E = 0.5 N/C
B = 2.3 T
We find
