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3 years ago

Which two factors determine the amount of work done?

2 answers:

5 0

Distance and force.
You need to know how much force was needed and how far it went to calculate how much work was done. The answer is C.

dimaraw [331]3 years ago

5 0

<h3><u>Answer;</u></h3>

C.distance and force

<h3><u>Explanation;</u></h3>

Work done is defined as the product of force and distance. Work is measured in Joules, like energy.
<em><u>Work is done when a given force that is applied to an object moves that object from its initial position to another position. The work is calculated by multiplying the force by the distance of movement by an object </u></em>

Work done = Force × Distance

Therefore, <em><u>the work done on a given object depends on the force applied and the distance moved by the object.</u></em>

You might be interested in

13. Austin rode his bike 10 m/s for two minutes. How far did he travel? A. 200 meters B. 1200 meters C. 1000 meters D. 20 meters

Semmy [17]

Answer:

B. 1200

Explanation:

60 sec in one min in 2 min there will be 120 sec. 10x120=1200

5 0

3 years ago

determine the loudness (in decibels) of the sound at a rock concert if the intensity of the sound is 1 x 10–1 w/m2. remember, th

EleoNora [17]

The loudness of the sound at the rock concert, where the intensity of the sound is1 x 10⁻¹ Wm⁻² is 110 dB.

Here we are dealing with loudness which is the perception of the Intensity of the sound.

The formula to refer to in order to find the value of the loudness of a sound is ,

db= 10log(I/I₀)

As we are provided with the current intensity which is 1 x 10⁻¹ Wm⁻². and the initial intensity which is 1 x 10⁻¹² Wm⁻².

So, by substituting the required values in the formula we get

db= 10 * log( 1 x 10⁻¹ /1 x 10⁻¹²)

= 10 * 11 log(10)

= 110

So, the result is 110 dB.

To know more about the intensity of sound refer to the link brainly.com/question/9323731?referrer=searchResults.

To know more about questions related to loudness refer to the link brainly.com/question/21094511?referrer=searchResults.

#SPJ4

4 0

2 years ago

One of the world's largest Ferris wheels, the Cosmo Clock 21 with a radius of 50.0 m is located in Yokohama City, Japan. Each of

STatiana [176]

Answer:

a = 0.55 m / s²

Explanation:

The centripetal acceleration is given by the relation

a = v² / r

angular and linear velocities are related

v = w r

we substitute

a = w² r

In the exercise they indicate the angular velocity w = 1 rev/min, let's reduce to the SI system

w = 1 rev / min (2pi rad / 1rev) (1min / 60s) = 0.105 rad/ s

let's calculate

a = 0.105² 50.0

a = 0.55 m / s²

4 0

3 years ago

A disk of radius 10 cm speeds up from rest. it turns 60 radians reaching an angular velocity of 15 rad/s. what was the angular a

Marianna [84]

Answer:

α = 1.875 rad/s²

t = 8 s

Explanation:

α = ω²/2θ = 15²/(2(60)) = 1.875 rad/s²

t = ω/a = 15 / 1.875 = 8 s

6 0

3 years ago

So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th

shutvik [7]

Answer:

Answer is explained in the explanation section.

Explanation:

Solution:

For this to find, we need to calculate the centripetal acceleration on the equator.

The centripetal acceleration of the equator:

a = 4 $\pi ^{2}$ RcosФ/ $T^{2}$ ,

where,

R is the radius of the earth

R = 6378 KM = 6.3 x $10^{6}$ and

T is the time period

T = 24 h = 86164.1 s

At Equator, Ф = 0°

So, CosФ = 1

Hence,

a = 4 $\pi ^{2}$ R/ $T^{2}$

By plugging in the values, we get:

a = 4 x ( $3.14^{2}$ ) x (6.3 x $10^{6}$ ) / $86164.1^{2}$

a = 0.03 m/ $s^{2}$

Hence, this is the centripetal acceleration on the equator. And we also know that, acceleration due to gravity is 9.8 m/ $s^{2}$ which is very higher than the centripetal acceleration on the equator.

B) Normal force exerted by chair will always be equal and opposite to the mass times gravitational acceleration (F = mg). Otherwise, I would be thrown away from chair in case the normal force is not equal and opposite or I would be drag down to the earth due to greater mass times gravitational acceleration. Hence, both are equal and opposite.

C) Of course, this is not a lie, it is true because the acceleration due to gravity is 9.8 m/ $s^{2}$ and as we calculated the acceleration on the equator is 0.03 m/ $s^{2}$ which way too low to experience.

For percentage difference,

9.8 - 0.03 = 9.77

So, % diff = (9.8 - 9.77)/9.8 x 100

% diff = 0.00306 x 100

% diff = 0.306%

Obviously, this is way too low to experience.

D) With the help of same formula as discussed above, we have:

a = 4 $\pi ^{2}$ RcosФ/ $T^{2}$ ,

Here, Ф = 44.4°

Just putting the values. we get

a = 0.0241 m/ $s^{2}$

Acceleration while sitting in my chair.

6 0

2 years ago