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anygoal [31]
3 years ago
10

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

between the road and the car's tires that will allow the car to travel atthis speed without sliding?
a. 1.23
b. 0.662
c. 0.816
d. 0.952
e. 0.736
Physics
1 answer:
zhenek [66]3 years ago
7 0

Answer:

c. 0.816

Explanation:

Let the mass of car be 'm' and coefficient of static friction be 'μ'.

Given:

Speed of the car (v) = 40.0 m/s

Radius of the curve (R) = 200 m

As the car is making a circular turn, the force acting on it is centripetal force which is given as:

Centripetal force is, F_c=\frac{mv^2}{R}

The frictional force is given as:

Friction = Normal force × Coefficient of static friction

f=\mu N

As there is no vertical motion, therefore, N=mg. So,

f=\mu mg

Now, the centripetal force is provided by the frictional force. Therefore,

Frictional force = Centripetal force

f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu=\frac{v^2}{Rg}

Plug in the given values and solve for 'μ'. This gives,

\mu = \frac{(40\ m/s)^2}{200\ m\times 9.8\ m/s^2}\\\\\mu=\frac{1600\ m^2/s^2}{1960\ m^2/s^2}\\\\\mu=0.816

Therefore, option (c) is correct.

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A 0.0382-kg bullet is fired horizontally into a 3.78-kg wooden block attached to one end of a massless, horizontal spring (k = 8
wel

Answer:

280.87 ms⁻¹

Explanation:

Consider the motion of the bullet-block combination after collision

m = mass of the bullet = 0.0382 kg

M = mass of wooden block = 3.78 kg

V = velocity of the bullet-block combination after collision

k = spring constant of the spring = 833 N m⁻¹

A = Amplitude of oscillation = 0.190 m

Using conservation of energy

Kinetic energy of  bullet-block combination after collision = Spring potential energy gained due to compression of spring

(0.5)(m + M)V^{2} = (0.5)kA^{2}

(0.0382 + 3.78)V^{2} = (833)(0.190)^{2}

V = 2.81 ms⁻¹

v_{o} = initial velocity of the bullet before striking the block

Using conservation of momentum for the collision between bullet and block

m v_{o} = (m + M) V

(0.0382) v_{o} = (0.0382 + 3.78) (2.81)

v_{o} = 280.87 ms⁻¹

7 0
3 years ago
For a short period of time, the frictional driving force acting on the wheels of the 2.5-Mg van is N, where t is in seconds. If
KATRIN_1 [288]

Answer:

15.6m/s

Completed Question;

For a short period of time, the frictional driving force acting on the wheels of the 2.5-Mg van is N= 600t^2 , where t is in seconds. If the van has a speed of 20 km/h when t = 0, determine its speed when t = 5

Explanation:

Mass m = 2500kg

Speed v1 = 20km/h = 20/3.6 m/s = 5.556 m/s

To determine speed v2;

Using the principle of momentum and impulse;

mv1 + ∫₀⁵ F dt = mv2

8 0
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The HL theorem is also known as the Hypothenus Leg theorem, it states that "the hypotenuse and leg of one right triangle are congruent to the hypotenuse and leg of another right triangle, then the two triangles are congruent."

Learn more about the postulates of the HL theorem here:

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2 years ago
if an object is moving at a constant speed in a changing direction, What is the acceleration of the object (zero or not zero)​
Andreas93 [3]
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  • Hence, acceleration is both dependant upon the speed as well as the direction.
  • So, if an object is moving at a constant speed in a changing direction, the acceleration will also change. It will not be zero.
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if an object is moving at a constant speed in a changing direction, the acceleration of the object will not be zero.

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Buhrs atomic model differed from ruthofords because it explained that
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