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2 years ago

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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?

1 answer:

Llana [10]2 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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A marble slides without friction in a vertical loop around the inside of a smooth, 28.6 cm diameter horizontal pipe. The marble'

Leviafan [203]

Answer:3.49 m/s

Explanation:

Given

Speed of marble at Bottom $v=4.22 m/s$

Diameter of loop $d=28.6 cm$

As Energy is conserved therefore Energy at top is equal to energy at bottom

$E_T=E_B$

$\frac{mv^2}{2}+mgh=\frac{mv_0^2}{2}$ ,where $v_0$ is the velocity at bottom

$\frac{v^2}{2}+gh=\frac{v_0^2}{2}$

$v_0^2=v^2+2gh$

$v^2=v_0^2-2gh$

$v=\sqrt{v_0^2-2gh}$

$v=\sqrt{4.22^2-2\times 9.8\times 0.286}$

$v=\sqrt{17.8084-5.6056}$

$v=3.49 m/s$

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2 years ago

How can doing work on an object increase its kinetic energy?

Lyrx [107]

Answer:If the object's speed increases.

Explanation:

If the object's speed increases, then its kinetic energy will increase. If the kinetic energy increases, the change in kinetic energy will be positive.

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3 years ago

Find the magnitude of the sum of two vectors; A is 5 km , and B is 7 , when the angle btween them is 120

Dimas [21]

Answer: 6.24 km

Explanation:

Given

The magnitude of the first vector(say) $\left | a\right |=5\ km$

the magnitude of the second vector(say) $\left | b\right |=7\ km$

the angle between them is $120^{\circ}$

The resultant vector magnitude is given by

$\left | \vec{R}\right |=\sqrt{a^2+b^2+2ab\cos \theta}$

$\left | \vec{R}\right |=\sqrt{5^2+7^2+2\times 5\times 7\cdot \cos 120^{\circ}}\\\left | \vec{R}\right |=\sqrt{74-35}=\sqrt{39}\\\left | \vec{R}\right |=6.24\ km$

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2 years ago

A bee wants to fly to a flower located due North of the hive on a windy day. The wind blows from East to West at speed 6.68 m/s.

Aleonysh [2.5K]

Answer: $53.31\°$ East of North

Explanation:

We have the following data:

Speed of the wind from East to West: $6.68 m/s$

Speed of the bee relative to the air: $8.33 m/s$

If we graph these speeds (which in fact are velocities because are vectors) in a vector diagram, we will have a right triangle in which the airspeed of the bee (its speed relative to te air) is the hypotense and the two sides of the triangle will be the <u>Speed of the wind from East to West</u> (in the horintal part) and the <u>speed due North relative to the ground</u> (in the vertical part).

Now, we need to find the direction the bee should fly directly to the flower (due North):

$sin \theta=\frac{Windspeed-from-East-to-West}{Speed-bee-relative-to-air}$

$sin \theta=\frac{6.68 m/s}{8.33 m/s}$

Clearing $\theta$ :

$\theta=sin^{-1} (\frac{6.68 m/s}{8.33 m/s})$

$\theta=53.31\°$

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3 years ago

Question 10 of 34

labwork [276]

Julia walks from the park, which is six blocks east of her house, to the store, which is three blocks east of her house. Julia walks for 5 minutes. This walk's average speed will be 1.2 blocks per minute. Option B is correct.

<h3>What is displacement?</h3>

Displacement is defined as the shortest distance between the two points. Distance is the horizontal length covered by the body. While displacement is the shortest distance between the two points.

Displacement is a vector quantity .its unit is m.

The average velocity on this walk will be;

$\rm v_{avg}= \frac{d}{t} \\\\ \rm v_{avg}= \frac{6 \ block+ 3 \ block }{5 \ minute } \\\\ v_{avg}=1.4 \ block /min$

Hence option B is correct.

To learn more about displacement refer to the link; brainly.com/question/10919017

#SPJ1

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1 year ago