If I Throw a bowling ball and a tennis ball from a building, describe if one would hit the ground first and why.

If the frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum speed of the oscillat

Maksim231197 [3]

Answer:

If the frequency of the motion of a simple harmonic oscillator is doubled , then maximum speed of the oscillator changes by the factor 2

Explanation:

We know that in a simple harmonic oscillator the maximum speed is given by

$v_{max}$ = $Aw$

Here A is amplitude which is constant , so from above equation we see that maximum speed is directly proportional to $w\\$ of the oscillation .

Since $w = 2 \pi f$

$v_{max}^{|}/v_{max} = 2f/f$ = 2

Where $v_{max}^{|}$ is the maximum speed when frequency is doubled .

6 0

3 years ago

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest

shutvik [7]

This is note the complete question, the complete question is:

One of the lousy things about getting old (prepare yourself!) is that you can be both near-sighted and farsighted at once. Some original defect in the lens of your eye may cause you to only be able to focus on some objects a limited distance away (near-sighted). At the same time, as you age, the lens of your eye becomes more rigid and less able to change its shape. This will stop you from being able to focus on objects that are too close to your eye (far-sighted). Correcting both of these problems at once can be done by using bi-focals, or by placing two lenses in the same set of frames. An old physicist instructor can only focus on objects that lie at distance between 0.47 meters and 5.4 meters.

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest 2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness?

Answer: 3.04 D

Explanation:

when an object is held 21 cm away from the instructor's eyes, the spectacle lens must produce 0.47m ( the near point) away.

An image of 0.47m from the eye will be ( 47 - 2 )

i.e 45 cm from the spectacle lens since the spectacle lens is 2cm away from the eye.

Also, the image distance will become negative

gap between lense and eye = 2cm

Therefore;

image distance d₁ = - 45cm = - 0.45m

object distance d₀ = 21 - 2 = 19cm = 0.19m

P = 1/f = 1/ d = 1/d₀ + 1/d₁ = 1/0.19 + (-1/0.45)

P = 1/f = 5.26315789 - 2.22222222

P = 1/f = 3.04093567 ≈ 3.04 D

5 0

3 years ago

Clois what is the weight of a body in the earth, if its weig is 5Nin moon? <br>

Debora [2.8K]

Explanation:

because the moon has less mass than earth, the force due to gravity at the lunar surface is only about 1/6 that on earthso,the weight of a body on earth is 6×5N =30N

8 0

2 years ago

Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slid

nataly862011 [7]

Answer:

A)

the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

B)

data that were not necessary to the solution are;

a) mass of truck and b) mass of load

Explanation:

Given that;

mass of load $m_{LS}$ = 10000 kg

mass of flat bed $m_{FB}$ = 20000 kg

initial speed of truck $v_{0}$ = 12 m/s

coefficient of friction between the load sits and flat bed μs = 0.5

A) the minimum stopping distance for which the load will not slide forward relative to the truck.

Now, using the expression

Fs,max = μs $F_{N}$ -------------let this be equation 1

where $F_{N}$ = normal force = mg

so

Fs,max = μs mg

ma $_{max}$ = μs mg

divide through by mass

a $_{max}$ = μs g ---------- let this be equation 2

in equation 2, we substitute in our values

a $_{max}$ = 0.5 × 9.8 m/s²

a $_{max}$ = 4.9 m/s²

now, from the third equation of motion

v² = u² + 2as

$v_{f}$ ² = $v_{0}$ ² + 2aΔx

where $v_{f}$ is final velocity ( 0 m/s )

a is acceleration( - 4.9 m/s² )

so we substitute

(0)² = (12 m/s)² + 2(- 4.9 m/s² )Δx

0 = 144 m²/s² - 9.8 m/s²Δx

9.8 m/s²Δx = 144 m²/s²

Δx = 144 m²/s² / 9.8 m/s²

Δx = 14 m

Therefore, the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

B) data that were not necessary to the solution are;

a) mass of truck and b) mass of load

3 0

3 years ago

Your friend, Tim, is playing with his sled. He ties a rope to his sled and attaches it to Lar's snowmobile. Tim has a mass of 71

kykrilka [37]

The maximum speed of Tim is 16.95 m/s.

The given parameters:

Mass of the rope, m = 71 kg
Tension on the rope, T = 220 N
Coefficient of kinetic friction, = 0.1
Time of motion, t = 8 s

<h3>What is Newton's second law of motion?</h3>

Newton's second law of motion states that, the force applied to an object is directly proportional to the product of mass and acceleration of the object.

The net force on Tim is calculated by applying Newton's second law of motion as follows;

$T - \mu _k F_n = ma\\\\T - \mu _k F_n = m\frac{v}{t} \\\\T - \mu_k mg = m\frac{v}{t} \\\\t(\frac{T - \mu_k mg}{m} )= v\\\\8 (\frac{220 \ -\ 0.1 \times 71 \times 9.8}{71} ) =v \\\\ 16.95 \ m/s = v$

Thus, the maximum speed of Tim is 16.95 m/s.

Learn more about net horizontal force here: brainly.com/question/21684583

7 0

2 years ago