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3 years ago

13

Determine the number of revolutions through which a typical automobile tire turns in 1 yr. Suppose the automobile travels 13500

miles each year on tires with radius 0.220 m.

1 answer:

RSB [31]3 years ago

5 0

Answer:

Number of revolution made by tire is 1.57 x 10⁷

Explanation:

Radius of tire, r = 0.220 m

Circumference of tire, C = 2πr

Substitute the value of r in the above equation.

C = 2 x π x 0.220 m = 1.38 m

Total distance covered by tire in a year, D = 13500 miles

But 1 mile = 1609.34 m

So, D = 13500 x 1609.34 m

Number of revolutions take by tire, N = $\frac{D}{C}$

$N=\frac{13500\times1609.34}{1.38}$

N = 15743543

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Suppose someone pours 0.250 kg of 20.0ºC water (about a cup) into a 0.500-kg aluminum pan with a temperature of 150ºC. Assume th

Troyanec [42]

Answer : The temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of aluminium = $0.90J/g^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_1$ = mass of aluminum = 0.500 kg = 500 g

$m_2$ = mass of water = 0.250 kg = 250 g

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of aluminum = $150^oC$

$T_2$ = initial temperature of water = $20^oC$

Now put all the given values in the above formula, we get:

$500g\times 0.90J/g^oC\times (T_f-150)^oC=-250g\times 4.184J/g^oC\times (T_f-20)^oC$

$T_f=59.10^oC$

Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

8 0

3 years ago

During a science lesson on changes, Ms. Sloan's students made mixtures of sulfur (the yellow substance) and iron filings (the bl

kramer

Answer:

Both substance will melt and become a solution.

Explanation:

When a mixture of iron filings (black substance) and sulphur powder (yellow substance) are heated in a test tube under a bunsen burner, usually they will undergo a chemical reaction where they will melt to form a new solution of ferrous sulphide.

Thus, we can say what happens is that both substances melt to form a solution.

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3 years ago

What happens in the gray zone between solid and liquid?

Reil [10]

Answer:

Hence from liquid to solid or solid to liquid the transition has to cross the grey zone. This grey zone transition is is very crucial which includes the intermolecular forces acting on the molecules and each atoms which makes the change in state from hot to cold and cold to hot.

Explanation:

3 0

3 years ago

La ecuación de la posición de una esferita está dada por: r(t)=(2.Cos(πt) i-3.Sen(πt) j) (m) ¿Cuál es la velocidad de la esferit

katovenus [111]

Answer:

v = (-4.44 i^ + 6.66 j^ ) m/s, a_average =( 0 i^ -2π j^) m/s²

Explanation:

The expression left corresponds to an oscillatory movement (MAS), the speed is defined by

v = dr / dt

the function of position

r = 2 cos πt i^ + 3 sin πt j^

let us note that it is a movement in two dimensions

let's perform the derivative

v = -2π sin πt i^ + 3π cos πt j^

we evaluate this expression for t = 0.25 s, remember that the angle is in radians

v = -2π sin (π 0.25) i^ + 3π cos (π 0.25) j^

v = (-4.44 i^ + 6.66 j^ ) m/s

To calculate the mean acceleration we use the expression

a = ( $v_{f} - v_{o}$ ) / Δt

indicates that the time is the first 3 s

we look for the initial velocity t = 0 s

v₀ = 0 i ^ + 3π j ^

we look for the fine velocity, t = 3 s

v_f = - 2π sin (π 3) + 3π cos (π 3) j ^

v_f = 0 i ^ - 3π j ^

we calculate the average acceleration

Δt = (3 -0) = 3 s

a_average = (0-0) / 3 i ^ + (-3π - 3π) / 3

a_average = (0 i ^ -2π j ^ ) m/s²

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3 years ago

A book sitting on a table is moved horizontally. Describe the

Akimi4 [234]

Frictional force and Applied force has same “magnitude” and “opposite” direction.

Option: B

<u>Explanation</u>:

When a book is moved horizontally by applying “force” on the book, the frictional force is opposed to the book by the table. Here, this “frictional force” is opposing the book has the same force what we applied on the book but this frictional force and the applied force are opposite in direction. Always the “frictional force” is opposite to the “applied force” which stops the object to move. For example, if a force applied leftward to the object the frictional force is acted on the right side of the object.

When two objects are in contact they experience a "frictional force". This "frictional force" acts opposite to the force applied on to move the object.

Formula for "frictional force" is $\mu\times N$

Where, $\mu$ is coefficient of friction and N is normal force.

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3 years ago