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Zinaida [17]
3 years ago
6

A box is being pulled by two ropes. Eduardo pulls to the left with a force of 500 N, and Clara pulls to the right with a force o

f 200 N. The box moves because of the two forces applied to it. Leon records the forces and direction of the forces acting on the box in his lab notebook. A 2 column table with 5 rows. Column 1 is labeled Force with entries Normal, Tension by Eduardo, Tension by Clara, Kinetic friction, Gravity. Column 2 is labeled Direction of Force Vector with entries Upward, Left, Right, Left, Downward. In the table, which force has the wrong direction? Tension by Eduardo Tension by Clara Kinetic friction Gravity
Physics
1 answer:
sergejj [24]3 years ago
8 0

Answer:

kinetic friction or something else

Explanation:

Draw a diagram to illustrate the problem, as shown below.

m = mass of the box, kg

mg =  weight of the box, N, acting downward

Because of Newton's 3rd law, a normal force of N = mg acts upward on the box.

C = 200 N, the force applied by Clara, horizontally to the right

E = 500 N, the force applied by Eduardo, horizontally to the left.

Because E is greater than C, the box will move left.

The frictional force, F, will resist the initial motion. Its magnitude is

F = μN = μmg, where μ = kinetic coefficient of friction.

F acts horizontally to the right.

From the given table, we can conclude that

Normal force acting upward is correct

Tension by Eduardoacting left is correct.

Tension by Clara acting right is correct.

Kinetic friction acting left is Incorrect.

Gravity acting downward is correct.correct

Answer:

The direction of the kinetic friction force is incorrect

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A metal ball attached to a spring moves in simple harmonic motion. The amplitude of the ball's motion is 11.0 cm, and the spring
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Answer:

a)0.674 kg b) 2.2 s c) 0.9 m/s²

Explanation:

The amplitude of the ball (xo) = 11.0cm, half way between its equilibrium point its maximum displacement x = 11 cm / 2 = 5.5 cm = 5.5 / 100 in meters = 0.055 meters, speed at this point = 27.2 cm /s = (27.2 / 100) in m/s = 0.272 m/s,

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v² = (xo² - x²) K/m

m = K ( xo² - x²) / v²

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b) The period of the oscillation can be calculated by the following formula

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T = 2 × 3.142 × √ (0.674/ 5.5) = 2.2s

c) The maximum acceleration of the ball which occurs at the maximum displacement of the ball can be calculated by the following formula

a = K / m × x ( maximum displacement of the body) = 5.5 / 0.674 × 0.11 = 0.9 m/s²

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Answer:

The wavelength of next line in the series will be 397.05 nm

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\frac{1}{\lambda_{max}R_H} = \frac{1}{n_1^2} -\frac{1}{(n_1 +1)^2} \\\\

\frac{1}{\lambda_{max}R_H} = \frac{2n_1+1}{n_1^2(n_1 +1)^2} \\\\\lambda_{max}R_H} = \frac{n_1^2(n_1 +1)^2}{2n_1+1}

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\lambda_{max}R_H} = \frac{n_1^2(n_1 +1)^2}{2n_1+1}\\\\656.46 *10^{-9}*1.097*10^7 = \frac{n_1^2(n_1 +1)^2}{2n_1+1}\\\\7.2 = \frac{n_1^2(n_1 +1)^2}{2n_1+1}

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\frac{n_1^2(n_1 +1)^2}{2n_1+1} =  \frac{(2)^2(2 +1)^2}{2(2)+1} = 7.2\\

∴ n₁(2) = 7.2

For the least wavelength given as 410.29 nm, n = ?

\frac{1}{\lambda} = R_H(\frac{1}{n_1^2} -\frac{1}{n_2^2})\\\\\frac{1}{410.29*10^{-9}} = 1.097*10^7(\frac{1}{2^2} -\frac{1}{n_2^2})\\\\\frac{1}{4.5} =\frac{1}{4} -\frac{1}{n_2^2}\\\\\frac{1}{n_2^2} =\frac{1}{4}  -\frac{1}{4.5} \\\\\frac{1}{n_2^2}  = 0.0277778\\\\n_2^2 = \frac{1}{0.0277778} \\\\n_2^2 = 36\\\\n_2= \sqrt{36}\  = 6

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Therefore, the wavelength of next line in the series will be 397.05 nm

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