Answer:
a. 10.5 s b. 6.6 s
Explanation:
a. The driver's perception/reaction time before drinking.
To find the driver's perception time before drinking, we first find his deceleration from
v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m
So, a = v² - u²/2s
substituting the values of the variables into the equation, we have
a = v² - u²/2s
a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)
a = - 499.52 m²/s²/234.7 m
a = -2.13 m/s²
Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time
So, t = (v - u)/a
Substituting the values of the variables into the equation, we have
t = (0 m/s - 22.35 m/s)/-2.13 m/s²
t = - 22.35 m/s/-2.13 m/s²
t = 10.5 s
b. The driver's perception/reaction time after drinking.
To find the driver's perception time after drinking, we first find his deceleration from
v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m
So, a = v² - u²/2s
substituting the values of the variables into the equation, we have
a = v² - u²/2s
a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)
a = 179.83 m²/s² - 499.52 m²/s²/234.7 m
a = -319.69 m²/s² ÷ 234.7 m
a = -1.36 m/s²
Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time
So, t = (v - u)/a
Substituting the values of the variables into the equation, we have
t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²
t = - 8.94 m/s/-1.36 m/s²
t = 6.6 s
which of the following is used to answer scientific questions?
A. Experiments
B. Intuition
C. Opinion polls
D. Imagination
what is momentum of a train that is 60,000 kg that is moving at velocity of 17m/s?
explain your answer
A 2028 kg Oldsmobile traveling south on Abbott Road at 14.5 m/s is unable to stop on the ice covered intersection for a red light at Saginaw Street. The car collides with a 4146 kg truck hauling animal feed east on Saginaw at 9.7 m/s. The two vehicles remain locked together after the impact. Calculate the velocity of the wreckage immediately after the impact. Give the speed for your first answer and the compass heading for your second answer. (remember, the CAPA abbreviation for degrees is deg) -1.75
Answer:
v = 8.1 m/s
θ = -36.4º (36.4º South of East).
Explanation:
Assuming no external forces acting during the collision (due to the infinitesimal collision time) total momentum must be conserved.Since momentum is a vector, if we project it along two axes perpendicular each other, like the N-S axis (y-axis, positive aiming to the north) and W-E axis (x-axis, positive aiming to the east), momentum must be conserved for these components also.Since the collision is inelastic, we can write these two equations for the momentum conservation, for the x- and the y-axes:We can go with the x-axis first:[tex]p_{ox} = p_{fx} (1)[/tex]
⇒ [tex]m_{tr} * v_{tr}= (m_{olds} + m_{tr}) * v_{fx} (2)[/tex]
Replacing by the givens, we can find vfx as follows:[tex]v_{fx} = \frac{m_{tr}*v_{tr} }{(m_{tr} + m_{olds)} } = \frac{4146kg*9.7m/s}{2028kg+4146 kg} = 6.5 m/s (3)[/tex]
We can repeat the process for the y-axis:[tex]p_{oy} = p_{fy} (4)[/tex]
⇒[tex]m_{olds} * v_{olds}= (m_{olds} + m_{tr}) * v_{fy} (5)[/tex]
Replacing by the givens, we can find vfy as follows:[tex]v_{fy} = \frac{m_{olds}*v_{olds} }{(m_{tr} + m_{olds)} } = \frac{2028kg*(-14.5)m/s}{2028kg+4146 kg} = -4.8 m/s (6)[/tex]
The magnitude of the velocity vector of the wreckage immediately after the impact, can be found applying the Pythagorean Theorem to vfx and vfy, as follows:[tex]v_{f} = \sqrt{v_{fx} ^{2} +v_{fy} ^{2} }} = \sqrt{(6.5m/s)^{2} +(-4.8m/s)^{2}} = 8.1 m/s (7)[/tex]
In order to get the compass heading, we can apply the definition of tangent, as follows:[tex]\frac{v_{fy} }{v_{fx} } = tg \theta (8)[/tex]
⇒ tg θ = vfy/vfx = (-4.8m/s) / (6.5m/s) = -0.738 (9)
⇒ θ = tg⁻¹ (-0.738) = -36.4º
Since it's negative, it's counted clockwise from the positive x-axis, so this means that it's 36.4º South of East.The radius of the Sun is 6.96 x 108 m and the distance between the Sun and the Earth is roughtly 1.50 x 1011 m. You may assume that the Sun is a perfect sphere and that the irradiance arriving on the Earth is the value for AMO, 1,350 W/m2. Calculate the temperature at the surface of the Sun.
Answer:
5766.7 K
Explanation:
We are given that
Radius of Sun , R=[tex]6.96\times 10^{8} m[/tex]
Distance between the Sun and the Earth, D=[tex]1.50\times 10^{11}m[/tex]
Irradiance arriving on the Earth is the value for AMO=[tex]1350W/m^2[/tex]
We have to find the temperature at the surface of the Sun.
We know that
Temperature ,T=[tex](\frac{K_{sc}D^2}{\sigma R^2})^{\frac{1}{4}}[/tex]
Where [tex]K_{sc}=1350 W/m^2[/tex]
[tex]\sigma=5.67\times 10^{-8}watt/m^2k^4[/tex]
Using the formula
[tex]T=(\frac{1350\times (1.5\times 10^{11})^2}{5.67\times 10^{-8}\times (6.96\times 10^{8})^2})^{\frac{1}{4}}[/tex]
T=5766.7 K
Hence, the temperature at the surface of the sun=5766.7 K
A 35 kg box initially sliding at 10 m/s on a rough surface is brought to rest by 25 N
of friction. What distance does the box slide?
Answer:
the distance moved by the box is 70.03 m.
Explanation:
Given;
mass of the box, m = 35 kg
initial velocity of the box, u = 10 m/s
frictional force, F = 25 N
Apply Newton's second law of motion to determine the deceleration of the box;
-F = ma
a = -F / m
a = (-25 ) / 35
a = -0.714 m/s²
The distance moved by the box is calculated as follows;
v² = u² + 2ad
where;
v is the final velocity of the box when it comes to rest = 0
0 = 10² + (2 x - 0.714)d
0 = 100 - 1.428d
1.428d = 100
d = 100 / 1.428
d = 70.03 m
Therefore, the distance moved by the box is 70.03 m.
What energy store is in the human
BEFORE he/she lifts the hammer?
I believe the answer would be protentional because they have the potential energy in them to lift the hammer.
Energy from the Sun is transferred from the Earth’s surface to the atmosphere, resulting in
atmospheric convection currents that produce winds. How do physical properties of the air
contribute to convection currents?
a -The warmer air sinks because it is more dense than cooler air.
b -The warmer air rises because it is more dense than cooler air.
c- The warmer air sinks because it is less dense than cooler air.
d -The warmer air rises because it is less dense than cooler air.
Anyone can help me out with this question ? Just number 2,
Answer:
- 21⁰C .
Explanation:
Speed of jet = 2.05 x 10³ km /h
= 2050 x 1000 / (60 x 60 ) m /s
= 569.44 m / s
Mach no represents times of speed of sound , the speed of jet
1.79 x speed of sound = 569.44
speed of sound = 318.12 m /s
speed of sound at 20⁰C = 343 m /s
Difference = 343 - 318.12 = 24.88⁰C
We know that 1 ⁰C change in temperature changes speed of sound
by .61 m /s
So a change in speed of 24.88 will be produced by a change in temperature of
24.88 / .61
= 41⁰C
temperature = 20 - 41 = - 21⁰C .
He throws a second ball (B2) upward with the same initial velocity at the instant that the first ball is at the ceiling. c. How long after the second ball is thrown do the two balls pass each other? d. When the balls pass each other how far are they above the juggler’s hands? e. When they pass each other what are their velocities?
Answer:
hello your question has some missing parts
A juggler performs in a room whose ceiling is 3 m above the level of his hands. He throws a ball vertically upward so that it just reaches the ceiling.
answer : c) 0.39 sec
d) 2.25 m
e) 1.92 m/sec
Explanation:
The initial velocity of the first ball = 7.67 m/sec ( calculated )
Time required for first ball to reach ceiling = 0.78 secs ( calculated )
Determine how long after the second ball is thrown do the two balls pass each other
Distance travelled by first ball downwards when it meets second ball can be expressed as : d = 1/2 gt^2 = 9.8t^2 / 2
hence d = 4.9t^2 ----- ( 1 )
Initial speed of second ball = first ball initial speed = 7.67 m/sec
3 - d = 7.67t - 4.9t ---- ( 2 )
equating equation 1 and 2
3 = 7.67t therefore t = 0.39 sec
Determine how far the balls are above the Juggler's hands ( when the balls pass each other )
form equation 1 ;
d = 4.9 t^2 = 4.9 *(0.39)^2 = 0.75 m
therefore the height the balls are above the Juggler's hands is
3 - d = 3 - 0.75 = 2.25 m
determine their velocities when the pass each other
velocity = displacement / time
velocity = d / t = 0.75 / 0.39 sec = 1.92 m/sec
Which of the following is a mixture?
a air
biron
Chydrogen
d nickel
Answer:
it will option option A hope it helps
According to Newton's law of universal gravitation, which statements are true?
As we move to higher altitudes, the force of gravity on us decreases.
O As we move to higher altitudes, the force of gravity on us increases,
O As we gain mass, the force of gravity on us decreases.
O Aswe gain mass, the force of gravity on us increases.
DAs we move faster, the force of gravity on us increases.
g Incandescent bulbs generate visible light by heating up a thin metal filament to a very high temperature so that the thermal radiation from the filament becomes visible. One bulb filament has a surface area of 30 mm2 and emits 60 W when operating. If the bulb filament has an emissivity of 0.8, what is the operating temperature of the filament
Answer:
2577 K
Explanation:
Power radiated , P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.6704 × 10⁻⁸ W/m²K⁴, ε = emissivity of bulb filament = 0.8, A = surface area of bulb = 30 mm² = 30 × 10⁻⁶ m² and T = operating temperature of filament.
So, T = ⁴√(P/σεA)
Since P = 60 W, we substitute the vales of the variables into T. So,
T = ⁴√(P/σεA)
= ⁴√(60 W/(5.6704 × 10⁻⁸ W/m²K⁴ × 0.8 × 30 × 10⁻⁶ m²)
= ⁴√(60 W/(136.0896 × 10⁻¹⁴ W/K⁴)
= ⁴√(60 W/(13608.96 × 10⁻¹⁶ W/K⁴)
= ⁴√(0.00441 × 10¹⁶K⁴)
= 0.2577 × 10⁴ K
= 2577 K