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8090 [49]
2 years ago
14

A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimum coefficient of static friction

between the road and the car's tires that will allow the car to travel at this speed without sliding?
A. 1.23B. 0.662C. 0.816D. 0.952E. 0.736
Physics
1 answer:
DanielleElmas [232]2 years ago
3 0

Answer:

The minimum coefficient of static friction between the road and the car's tires is 0.816

Option "C"

Explanation:

Given;

velocity of the car, v = 40.0 m/s

radius of horizontal curve, r =  200 m

For a minimum coefficient of static friction between the road and the car's tires that will allow the car to travel at the given speed without sliding, centripetal force must equal frictional force.

F_{frictional} = F_{centripetal}\\\\\mu mg = \frac{mv^2}{r} \\\\\mu = \frac{v^2}{rg}

where;

μ is the minimum coefficient of static friction

\mu = \frac{40^2}{200*9.8} \\\\\mu = 0.816

Thus, the minimum coefficient of static friction between the road and the car's tires is 0.816

Option "C"

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A ray of light is incident on a plane surface separating two sheets of glass with refractive indexes 1.70 and 1.58. The angel of
dlinn [17]

Answer:

r = 71.8⁰

Explanation:

given,

refractive index of the glass 1 = 1.70

refractive index of glass 2 = 1.58

angle of incidence = 62°

angle of refraction =?

using Snell's law

\dfrac{sin\ i}{sin\ r} = \dfrac{n_2}{n_1}

\dfrac{sin\ 62^0}{sin\ r} = \dfrac{1.58}{1.70}

1.7 ×sin 62 ^0 = 1.58× sin r

sinr = \dfrac{1.7\times sin 62^0}{1.58}

sin r = 0.95

r = sin⁻¹(0.95)

r = 71.8⁰

angle of refraction =r = 71.8⁰

4 0
2 years ago
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A conducting rod (length = 80 cm) rotates at a constant angular rate of 15 revolutions per second about a pivot at one end. A un
Studentka2010 [4]

Answer:

3.62 V

Explanation:

L = 80 cm = 0.8 m

f = 15 rps

B = 60 m T = 0.060 T

ω = 2 x π x f = 2 x 3.14 x 15 = 94.2 rad/s

v = r ω

here, r be the radius of circular path. Here r = length of rod = L

v = 0.80 x 94.2 = 75.36 m/s

The motional emf is given by

e = B  v  L = 0.060 x 75.36 x 0.8 = 3.62 V

4 0
2 years ago
Emmy kicks a soccer ball up at an angle of 45° over a level field. She watches the ball's trajectory and notices that it lands,
Elenna [48]

Let u be the initial velocity of the soccer ball at an angle of inclination of \theta_0 with the positive x-axis.

Given that:

\theta_0=45^{\circ}

The horizontal distance covered by the projectile=20 m

Time of flight, t_f=2 seconds

Acceleration due to gravity, g= 10 m/s^2 downward.

As "north" and "up" as the positive x ‑ and y ‑directions, respectively.

So, g= -10 m/s^2

As the acceleration due to gravity is in the vertical direction, so the horizontal component of the initial velocity remains unchanged.

The x-component of the initial velocity, u_x=u\cos\theta_0.

The horizontal distance covered by the projectile = u_x\times t_f

\Rightarrow u_x\times t_f=20

\Rightarrow u_x\times 2=20

\Rightarrow u_x=10 m/s

So, the horizontal component of the velocity is 10 m/s which is constant and the graph has been shown in the figure (i).

Now,  u\cos(45^{\circ})=10 [as u_x=u\cos\theta_0]

\Rightarrow u=10\sqrt{2} m/s.

The vertical component of the initial velocity,

u_y= u\sin\theta_0

\Rightarrow u_y=10\sqrt{2}\sin(45^{\circ})

\Rightarrow u_y=10 m/s

Let v be the vertical component of the velocity at any time instant t.

From the equation of motion,

v=u+at

where u: initial velocity, v: final velocity, a: constant acceleration, and t: time taken to change the velocity from u to v.

In this case, we have u=u_y, a= -10 m/s^2.

So at any time instant, t.

v=u_y+(-10)t

\Rightarrow v=10-10t

The vertical component of the velocity, v, is the function of time and related as v=10-10t.

This is a linear equation.

At 2 second, the vertical component of the velocity

v=10-10x2=-10 m/s.

The graph has been shown in figure (ii).

7 0
2 years ago
Joel uses a claw hammer to remove a nail from a wall. He applies a force of 40 newtons on the hammer. The hammer applies a force
jarptica [38.1K]

Hi!


Mechanical advantage is defined as the<em> ratio of force produced by an object to the force that is applied to it.</em>

In our case, this would be the ratio of the force applied by the claw hammer on the nail to the force Joel applies to the claw hammer, which is

160:40 or 4:1

So the mechanical advantage of the hammer is four.


Hope this helps!


3 0
2 years ago
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A particular material has an index of refraction of 1.25. What percent of the speed of light in a vacuum is the speed of light i
beks73 [17]

Answer:

80% (Eighty percent)

Explanation:

The material has a refractive index (n) of 1.25

Speed of light in a vacuum (c) is 2.99792458 x 10⁸  m/s

We can find the speed of light in the material (v) using the relationship

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v = c/n

therefore v = 2.99792458 x 10⁸  m/s ÷ (1.25) = 239 833 966 m/s

v = 239 833 966 m/s

Therefore the percentage of the speed of light in a vacuum that is the speed of light in the material can be calculated as

(v/c) × 100 = (1/n) × 100 = (1/1.25) × 100 = 0.8 × 100 = 80%

Therefore speed of light in the material (v) is eighty percent of the speed of light in the vacuum (c)

3 0
2 years ago
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