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3 years ago

If we decrease the amount of force, and keep all other factors the same, what will happen to the amount of work?

Physics

2 answers:

kozerog [31]3 years ago

8 0

The answer is A. I am sure of it.

Travka [436]3 years ago

7 0

Answer:

<em>A: The amount of work will go down because the distance will decrease.</em>

Explanation:

A force is a push or pulls that is capable of changing the state of motion of an object. Force is measured in Newton

Work done is the amount of energy produced by a body when acted by a force over a distance. A product of force and distance, it is measured in joule

Amount of Work done is a product of force and distance, the formula for work done is given in equation 1;

W = f x d .......................................................1

where W is work done

f is the force

d is distance

from equation 1 the work done by an object is directly proportional to the force applied on the body. This means that an increase in the force will always result in an equivalent increase in work done by the body.

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A projectile of mass m is fired straight upward from the surface of an airless planet of radius R and mass M with an initial spe

motikmotik

Answer:

K = G Mm / 9R

Explanation:

Expression for escape velocity V_e = $\sqrt{\frac{2GM}{R} }$

Kinetic energy at the surface = 1/2 m V_e ²

= 1/2 x m x 2GM/R

GMm/R

Potential energy at the surface

= - GMm/R

Total energy = 0

At height 9R ( 8R from the surface )

potential energy

= - G Mm / 9R

Kinetic energy = K

Total energy will be zero according to law of conservation of mechanical energy

K - G Mm / 9R = 0

K = G Mm / 9R

6 0

3 years ago

Resistance increases with the length of the wire

Marta_Voda [28]

If the length of the wire increases, then the amount of resistance will also increase.

1. Take a long piece of wire and cut it 10 pieces. Those pieces should all be different sizes, one should be 5___ (units in meter, cm, inches, etc.), and the next should be 5 ___ (units in meter, cm, inches, etc.) more than the one before.
2. Take one piece of wire and measure the resistance using ___ and record the results in the data table.
3. Repeat the previous step with all the pieces of wire.
4. Compare and contrast the results you have found.

I hope this helps a bit :)

4 0

3 years ago

How should the student use the data collected and the known quantities from the experiment to determine the initial total mechan

fiasKO [112]

It should be noted that the student should use K = 1/2mv² with the initial speed of the block for one trial.

<h3>Method of using the data collected.</h3>

From the complete information, the student wants to use the data collected and the known quantities from the experiment to determine the initial total mechanical energy of the block-ramp-Earth system for all trials in the experiment.

In this case, it's important to use K = 1/2mv² with the block's initial speed for one trial due to the fact that the initial speed is the same in all the trials.

Learn more about experiments on:

brainly.com/question/17274244

7 0

2 years ago

A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho

fiasKO [112]

Answer:

$v_f=8.17\frac{m}{s}$

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

$W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m$

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

$W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$

The box is initially at rest, so $v_i=0$ . Solving for $v_f$ :

$v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}$

5 0

3 years ago

Two small stereo speakers are driven in step by the same variable-frequency oscillator. Their sound is picked up by a microphone

anygoal [31]

Answer:

The fundamental frequency at which the sound of speakers at the microphone produce constructive interference is 801.076458 Hz

Explanation:

For a given arrangement having constructive interference, we have;

R₁ - R₂ = 2·x = 0 + n·λ

The distance from one speaker to the microphone. R₁ = 4.50 m

The distance between the two speakers = 2.00 m

The angle formed between the direction from the microphone to the speaker closest and the directional path between the speakers = 90°

Therefore, by Pythagoras's theorem, the distance from the speaker furthest from the microphone, to the microphone, 'R₂' is given as follows;

R₂ = √(4.50² + 2.00²) = √(24.25) = 4.9244289009 ≈ 4.924

∴ R₂ ≈ 4.9244289 m

R₂ - R₁ = 4.9244289 m - 4.5 m = 0.4244289 m

For constructive interference, R₂ - R₁ =0.4244289 m = n·λ

For n = 1, we have;

R₂ - R₁ =0.4244289 m = n·λ = 1 × λ = λ

λ = 0.4244589 m

f = v/λ = 340 m/sec/(0.4244289 m) ≈ 801.076458 Hz

Therefore the lowest possible fundamental frequency at which the speakers produce constructive interference, f = 801.076458 Hz

7 0

3 years ago