Answer:
A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes
Explanation:
Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:
2as = Vf² - Vi²
s = (Vf² - Vi²)/2a
where,
Vf = Final Velocity of Car = 0 mi/h
Vi = Initial Velocity of Car = 50 mi/h
a = deceleration of car
s = distance covered
Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.
So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:
s = vt
FOR SOBER DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 0.33 s
s = s₁
Therefore,
s₁ = (73.33 ft/s)(0.33 s)
s₁ = 24.2 ft
FOR DRUNK DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 1 s
s = s₂
Therefore,
s₂ = (73.33 ft/s)(1 s)
s₂ = 73.33 ft
Now, the distance traveled by drunk driver's car further than sober driver's car is given by:
ΔS = s₂ - s₁
ΔS = 73.33 ft - 24.2 ft
<u>ΔS = 49.13 ft</u>
Answer:
9) a = 25 [m/s^2], t = 4 [s]
10) a = 0.0875 [m/s^2], t = 34.3 [s]
11) t = 32 [s]
Explanation:
To solve this problem we must use kinematics equations. In this way we have:
9)
a)
where:
Vf = final velocity = 0
Vi = initial velocity = 100 [m/s]
a = acceleration [m/s^2]
x = distance = 200 [m]
Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.
0 = (100)^2 - (2*a*200)
a = 25 [m/s^2]
b)
Now using the following equation:
0 = 100 - (25*t)
t = 4 [s]
10)
a)
To solve this problem we must use kinematics equations. In this way we have:
Note: The positive sign of the equation means that the car increases his speed.
5^2 = 2^2 + 2*a*(125 - 5)
25 - 4 = 2*a* (120)
a = 0.0875 [m/s^2]
b)
Now using the following equation:
5 = 2 + 0.0875*t
3 = 0.0875*t
t = 34.3 [s]
11)
To solve this problem we must use kinematics equations. In this way we have:
10^2 = 2^2 + 2*a*(200 - 10)
100 - 4 = 2*a* (190)
a = 0.25 [m/s^2]
Now using the following equation:
10 = 2 + 0.25*t
8 = 0.25*t
t = 32 [s]
Answer:
Ф = 239.73 rad
Explanation:
α = 12 + 15×t
W = ∫α×dt
= ∫(12 + 5×t)×dt
= 12×t + 2.5×t^2
then:
Ф = ∫W×dt
= ∫(12×t + 2.5×t^2)dt
= 6×t^2 + 5/6×t^3
therefore the angle at t = 4.88s is:
Ф = 6×(4.88)^2 + 5/6×(4.88)^3
= 239.73 rad
Answer:
Check the first and the third choices:
<u><em /></u>
- <u><em>a. The temperature of a gas is directly proportional to its volume</em></u>
- <u><em>b. The temperature-to-volume ratio of a gas is constant.</em></u>
Explanation:
Rewrite the table for better understanding:
Temperature of gas (K) Volume of gas (L)
298 4.55
315 4.81
325 4.96
335 ?
Calculate the ratios temperature to volume with 3 significant figures:
Then, those numbers show a <u><em>constant temperature-to-volume ratio</em></u>, which may be expressed in a formula as:
- Temperature / Volume = constant, which is a directly proportional variation (the volume increases in a constant proportion to the increase of the temperature).
Hence, the correct choices are:
- The temperature of a gas is directly proportional to its volume (first statement), and
- The emperature-to-volume ratio of a gas is constant (third statement).
Answer:
The magnetic force on a free moving charge depends on the velocity of the charge and the magnetic field, direction of the force is given by the right hand rule. While gravitational depends on the mass and distance of the moving particle and electric forces depends on the magnitude of the charge and distance of separation.
Explanation:
The magnetic force on a free moving charge depends on the velocity of the charge and the magnetic field and direction of the force is given by the right hand rule. While gravitational depends on the mass and distance of the moving particle and electric forces depends on the magnitude of the charge and distance of separation.
The magnetic force is given by the charge times the vector product of velocity and magnetic field. While gravitational force is given by the square of the particle mass divided by the square its distance of separation. Also electric forces is given by the square of the charge magnitude divided by the square its distance separation.
