The measure of the force with which air molecules push on a surface is called

Human reaction times are worsened by alcohol. How much further (in feet) would a drunk driver's car travel before he hits the br

sweet-ann [11.9K]

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

ΔS = 49.13 ft

6 0

3 years ago

A 45.00 kg person in a 43.00 kg cart is coasting with a speed of 19 m/s before it goes up a hill. there is no friction, what is

HACTEHA [7]

Answer:

the maximum vertical height the person in the cart can reach is 18.42 m

Explanation:

Given;

mass of the person in cart, m₁ = 45 kg

mass of the cart, m₂ = 43 kg

acceleration due to gravity, g = 9.8 m/s²

final speed of the cart before it goes up the hill, v = 19 m/s

Apply the principle of conservation of energy;

$mgh_{max} = \frac{1}{2}mv^2_{max}\\\\ gh_{max} = \frac{1}{2}v^2_{max}\\\\h_{max} = \frac{v^2_{max}}{2g} \\\\h_{max} =\frac{(19)^2}{2\times 9.8} \\\\h_{max} = 18.42 \ m$

Therefore, the maximum vertical height the person in the cart can reach is 18.42 m

5 0

3 years ago

Elements in Group VIIIA (also known as Group 18, or the noble gases) have

Sonbull [250]

The answer is C (the same number of valence electrons)

3 0

3 years ago

A plane is flying due east with the velocity of 90m/s. The wind is blowing out of the north at 4m/s. What is the magnitude of th

Butoxors [25]

as we know that the velocity vectors are at right angles
magnitude = ?
hypotenuse of a right triangle.
v^2 = 90^2 + 4^2
v^2 = 8116
Taking the square root of both sides here we get,
v = 90.1 m/s
hope it helps

4 0

3 years ago

A radio station broadcasts its music with waves at a frequency of 7.34 x 10²Hz. These radio waves travel at a speed of 3.00x 10%

Lorico [155]

Answer:

See below

Explanation:

I will use 3 x 10^8 m/s for speed or wave

speed = wavelength * frequency

3 x 10^8 = w * 7.34 x 10^2 <====== are you sure this isn't KILO Hz ?

w = 408719. 3 meters

4 0

2 years ago