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Savatey [412]
3 years ago
12

Chapter 05, Problem 15 Multiple-Concept Example 7 and Concept Simulation 5.2 review the concepts that play a role in this proble

m. Car A uses tires for which the coefficient of static friction is 0.335 on a particular unbanked curve. The maximum speed at which the car can negotiate this curve is 26.8 m/s. Car B uses tires for which the coefficient of static friction is 0.683 on the same curve. What is the maximum speed at which car B can negotiate the curve?
Physics
1 answer:
Llana [10]3 years ago
3 0

Answer:

Explanation:

The question relates to motion on a circular path .

Let the radius of the circular path be R .

The centripetal force for circular motion is provided by frictional force

frictional force is equal to μmg , where μ is coefficient of friction and mg is weight

Equating cenrtipetal force and frictionl force in the case of car A

mv² / R = μmg

R = v² /μg

= 26.8 x 26.8 / .335 x 9.8

= 218.77 m

In case of moton of car B

mv² / R = μmg

v²  = μRg

= .683  x 218.77x 9.8

= 1464.35

v = 38.26 m /s .

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two flat mirrors are connected to each other such that they make an angle of ψ. a laser enters the system and first reflects off
Firdavs [7]

The angle is 53.5 degrees.

What is a flat mirror?

A plane mirror is a mirror with a flat (planar) reflective surface. For light rays striking a plane mirror, the angle of reflection equals the angle of incidence.

This ques have been solved in two steps:

Step 1:

given that;

Θ=107 degrees

from figure,

∠BAC=180-ψ-(90-φ)

∠BAC= 90-ψ-φ

∠AOC=180-2(ψ-φ)-2φ

∠AOC=180-2φ

Step 2: part(a)

writing an expression for the psi at point o:

Θ+180-2ψ=180

2ψ=Θ

ψ=Θ/2

above is the expression for ψ

Step 3: part(b)

The angle of ψ is,

ψ=Θ/2

ψ=107/2

ψ=53.5 degrees

Hence, the angle is 53.5 degrees.

To more know about flat mirrors and phi and psi angles the link is given below:

brainly.com/question/15970406?

#SPJ4

7 0
2 years ago
Round 1468.12277 to the nearest whole number
docker41 [41]
1468 :) lol
you’re welcome
3 0
3 years ago
You are driving a passenger car on a two-lane highway. unless otherwise posted, the speed limit is
riadik2000 [5.3K]
The speed limit is: 55 mph
5 0
3 years ago
Read 2 more answers
Suppose that the speed of an electron traveling 2.0 km/s is known to an accuracy of 1 part in 105 (i.e., within 0.0010%). What i
OverLord2011 [107]

Answer: 2.89(10)^{-3} m

Explanation:

The <u>Heisenberg uncertainty principle</u> postulates that the fact each particle has a wave associated with it, imposes restrictions on the ability to determine its position and speed at the same time.  

In other words:  

It is impossible to measure simultaneously (according to quantum physics), and with absolute precision, the value of the position and the momentum (linear momentum) of a particle. Thus, in general, the greater the precision in the measurement of one of these magnitudes, the greater the uncertainty in the measure of the other complementary variable.

Mathematically this principle is written as:

\Delta x \geq \frac{h}{4 \pi m \Delta V} (1)

Where:

\Delta x is the uncertainty in the position of the electron

h=6.626(10)^{-34}J.s is the Planck constant

m=9.11(10)^{-31}kg is the mass of the electron

\Delta V is the uncertainty in the velocity of the electron.

If we know the accuracy of the velocity is 0.001\% of the velocity of the electron V=2 km/s=2000 m/s, then \Delta V is:

\Delta V=2000 m/s(0.001\%)

\Delta V=2000 m/s(\frac{0.001}{100})

\Delta V=2(10)^{-2} m/s (2)

Now, the least possible uncertainty in position \Delta x_{min} is:

\Delta x_{min}=\frac{h}{4 \pi m \Delta V} (3)

\Delta x_{min}=\frac{6.626(10)^{-34}J.s}{4 \pi (9.11(10)^{-31}kg) (2(10)^{-2} m/s)} (4)

Finally:

\Delta x_{min}=2.89(10)^{-3} m

5 0
3 years ago
A street light is at the top of a 13.0 ft. tall pole. A man 6.3 ft tall walks away from the pole with a speed of 3.5 feet/sec al
amm1812

Answer:

\dfrac{dL}{dt}=5.82 \ ft/s

Explanation:

given,

street light height = 13 ft

man height = 6.3 ft

speed of the man = 3.5 ft/sec

\dfrac{H}{L} = \dfrac{h}{l}

\dfrac{H}{L} = \dfrac{h}{L-x}

\dfrac{L}{H} = \dfrac{L-x}{h}

hL = H(L-x)

hL = HL-Hx

L = \dfrac{Hx}{H-h}

L = \dfrac{13x}{13-6.3}

L = 1.94 x

\dfrac{dL}{dt}=\dfrac{dL}{dx}\dfrac{dx}{dt}

\dfrac{dL}{dt}=1.94\times 3

\dfrac{dL}{dt}=5.82 \ ft/s

7 0
3 years ago
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