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mafiozo [28]
1 year ago
13

If a participant were holding two different weights in their hands and the jnd for a 10-gram weight was 1 gram, what should the

jnd be for a 100-gram weight, according to weber's law?
Physics
1 answer:
Nataliya [291]1 year ago
7 0

The jnd for a 100-gram weight, according to Weber's law will be 10 gram.

<h3>What is Weber's law?</h3>

It should be noted that Weber's law asserts that the nature of any given stimulus will always affect how change is perceived. In other words, the size, weight, importance, etc. of the prior situation and the significance of the change both influence whether a change will be observed.

In this case, it was given that the jnd for a 10-gram weight was 1 gram, therefore, the jnd for 100 gram will be;

= 100 / 10

= 10 gram

Therefore, jnd for a 100-gram weight, according to Weber's law will be 10 grams.

Learn more about weight on:

brainly.com/question/19753744

#SPJ1

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(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as meas
aleksley [76]

Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = \frac{k*q}{r}

We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.

If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:

V = \frac{k*q1}{x} +\frac{k*q2}{(1.63m-x)} = 0

⇒k*q1* (1.63m - x) = -k*q2*x:

Replacing by the values of q1, q2, and k, and solving for x, we get:

⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.

3 0
3 years ago
Read 2 more answers
? Which statement is true of an object in equilibrium?
Degger [83]
The answer is C,<span> The sum of all forces acting on the object is zero. hope that helps!!</span>
7 0
3 years ago
8. An effort force of 15 Newtons is applied to an ideal pulley system to lift up a 16 Newton object. If the effort force is exer
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Answer:

the distance that the object is raised above its initial position is 5.625 m.​

Explanation:

Given;

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load lifted by the ideal pulley system, L = 16 N

distance moved by the effort, d₁ = 6 m

let the distance moved by the object = d₂

For an ideal machine, the mechanical advantage is equal to the velocity ratio of the machine.

M.A = V.R

M.A = \frac{Load}{Effort} = \frac{L}{E} \\\\V.R = \frac{disatnce \ moved \  by \ the \ effort}{disatnce \ moved \  by \ the \ load} = \frac{d_1}{d_2} \\\\For \ ideal \ machine; \ M.A = V.R\\\\\frac{L}{E} = \frac{d_1}{d_2} \\\\d_2 = \frac{E \times d_1}{L} \\\\d_2 = \frac{15 \times 6}{16} \\\\d_2 = 5.625 \ m

Therefore, the distance that the object is raised above its initial position is 5.625 m.​

3 0
2 years ago
1. While John is traveling along an interstate highway, he notices a 160-mile marker as he passes through town. Later John passe
pogonyaev

Number of miles that marker shows when passes through town= 160 miles.

Number of miles that marker shows currently to John = 115 miles.

We need to find the distance between town and John's current location.

For the problem, we can clearly see that Town is at 160 miles away but when John passes the marker shows 115 miles.

So, it's just the difference between 160 miles and 115 miles.

In order to find that difference, we need to subtract those two numbers.

160miles - 115miles = 45 miles.

So, we could say the distance between town and John's current location is 45 miles.

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Stick a fork into each end of the root vegetable as shown. The forks should be on the same side of the vegetable, and
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Answer:

Yes

Explanation:

There is a position that works better than this and that is switching the sides of the forks.

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