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Zolol [24]
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.
Physics
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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A 1-kg block is lifted vertically 1 m at constant speed by a boy. The work done by the boy is about:
Igoryamba

Answer:

9.8\; {\rm J}, assuming that the gravitational field strength is g = 9.8\; {\rm N \cdot kg^{-1}}.

Explanation:

Notice that both the speed and the direction of motion of this block are constant. In other words, the velocity of this block is constant.

By Newton's Second Law, the net force on this block would be 0. External forces on this block should be balanced. Thus, the magnitude of the (downward) weight of this block should be equal to the magnitude of the (upward) force that the boy applies on this block.

Let m denote the mass of this block. It is given that m = 1\; {\rm kg}. The weight of this block would be:

\begin{aligned}\text{weight} &= m\, g \\ &= 1\; {\rm kg} \times 9.8\; {\rm N \cdot kg^{-1}} \\ &= 9.8\; {\rm N}\end{aligned}.

Hence, the force that the boy applies on this block would be upward with a magnitude of F = 9.8\; {\rm N}.

The mechanical work that a force did is equal to the product of:

  • the magnitude of the force, and
  • the displacement of the object in the direction of the force.

The displacement of this block (upward by s = 1\; {\rm m}) is in the same direction as the (upward) force that this boy had applied. Thus, the work that this boy had done would be the product of:

  • the magnitude of the force that this boy exerted, F = 9.8\; {\rm N}, and
  • the displacement of this block in the direction, s = 1\; {\rm m}.

\begin{aligned}\text{work} &= F\, s \\ &= 9.8\; {\rm N} \times 1\; {\rm m} \\ &= 9.8\; {\rm J}\end{aligned}.

5 0
2 years ago
Which year was pluto no longer considered a planet?.
katrin2010 [14]
2006, i hope this helps
6 0
1 year ago
Josh tests an unknown liquid substance using litmus paper. When he compared the used litmus paper to a pH scale, the color match
Hunter-Best [27]

Answer:

acid

Explanation:

because ph below 7 is acid

8 0
2 years ago
Oxygen is the most abundant gas in Earth’s atmosphere true or false
iogann1982 [59]

Answer: false

Explanation:

Nitrogen is actually the most abundant gas in the atmosphere.

The answer to the question is false. Hope this helps you!

3 0
3 years ago
Read 2 more answers
An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see
mash [69]

Answer:

(7.90 × 10⁻¹⁵) J

Explanation:

The electric force exerted on the elecrron by rhe electric field is given by

F = qE

where |q| = charge on the particle = (1.602 × 10⁻¹⁹) C

E = magnitude of the electric field = (2.9 × 10⁶) V/m or N/C

F = 1.602 × 10⁻¹⁹ × 2.9 × 10⁶ = (4.646 × 10⁻¹³) N

From Newton's first law of motion relation, we can obtain the acceleration this force confers on the electron

F = ma

m = mass of the electron = (9.11 × 10⁻³¹) kg

a = acceleration of the electron caused by the electric force = ?

(4.646 × 10⁻¹³) = (9.11 × 10⁻³¹) × a

a = (4.646 × 10⁻¹³)/(9.11 × 10⁻³¹)

a = (5.10 × 10¹⁷) m/s²

Now, using the equations of motion, we can obtain the velocity with which the electron reaches the positive plate

u = initial velocity of the electron = 0 m/s (since the electron was initially at rest)

v = final velocity of the electron = ?

a = acceleration of the electron = (5.10 × 10¹⁷) m/s²

y = distance covered by the electron = 1.7 cm = 0.017 m

v² = u² + 2ay

v² = 0² + 2(5.10 × 10¹⁷)(0.017)

v² = (1.734 × 10¹⁶)

v = 131,677,182.5 m/s = (1.32 × 10⁸) m/s

Kinetic energy with which the electron hits the positive plate = (1/2)(m)(v²) = (1/2)(9.11 × 10⁻³¹)(1.32 × 10⁸)² = (7.90 × 10⁻¹⁵) J

Hope this Helps!!!

3 0
3 years ago
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