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ELEN [110]
2 years ago
11

A woman pushes a 3.2kg textbook on a table as she walks forward 3.5 meters. How much work does the vertical component of the for

ce from her hand do on the textbook?in j
Physics
1 answer:
Effectus [21]2 years ago
4 0

Answer:

The done by the vertical component of force is zero.

Explanation:

Given data,

The mass of the textbook the women pushes horizontally, m = 3.2 kg

The displacement of the textbook, S = 3.5 m

Let the force of the women acts on the book horizontally,

Therefore, the horizontal component of force is maximum and the vertical component is zero.

If  F is the force applied by the women, then the horizontal and vertical component of the force is,

                                         F_{x}=F Cos\theta

                                         F_{y}=F Sin\theta

Since the force is acting along with the horizontal x component, the vertical component of the force is zero.

Hence, the done by the vertical component of force is zero.

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Based on the information given, it can be inferred that the favor doesn't fall within the AAMA guidelines of her responsibilities.

From the information given, it should be noted that the guidelines of CMA as stipulated under the American Association of Medical Assistant prohibits the CMA from interpreting the medical data of the patient. Therefore, the favor that was asked by Dr. Hsu of Kayla is simply against the guidelines.

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If you were to move to the Canadian North Woods, what adaptations or behavioral changes would you make?
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1. What did you observe about the magnitudes of the forces on the two charges? Were they the same or different? Does your answer
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Answer:

Following are the solution to the given question:

Explanation:

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F=\frac{kq_1q_2}{r^2}

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|F_{12}| = |F_{21}|

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3 years ago
Through both refraction and diffraction, the atmosphere alters the apparent speed and, to a lesser extent, the direction of the
musickatia [10]

Answer:

The Ionospheric Effect

Explanation:

One of the largest errors in GPS positioning is attributable to the atmosphere. The long, relatively unhindered travel of the GPS signal through the virtual vacuum of space changes as it passes through the earth’s atmosphere. Through both refraction and diffraction, the atmosphere alters the apparent speed and, to a lesser extent, the direction of the signal. This causes an apparent delay in the signal's transit from the satellite to the receiver.

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3 years ago
The length of aluminum rods produced by a company are approximated by a Gaussian distribution with a mean of 10 cm and a standar
ExtremeBDS [4]

Given Information:

Mean length of aluminum rods = μ = 10 cm

Standard deviation of length of aluminum rods = σ = 0.02 cm

Required Information:

a) P(9.98 < X < 10.02) = ?

b) P(9.90 < X < 10.1) = ?

Answer:

a) P(9.98 < X < 10.02) = 68.27%

b) P(9.90 < X < 10.1) = 100%

Explanation:

What is Normal Distribution?

Normal Distribution or also known as Gaussian Distribution, is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability

a) We want to find out the probability that the length of aluminum rods is between 9.98 and 10.02 cm.

P(9.98 < X < 10.02) = P( \frac{x - \mu}{\sigma} < Z < \frac{x - \mu}{\sigma} )\\\\P(9.98 < X < 10.02) = P( \frac{9.98- 10}{0.02} < Z < \frac{10.02 - 10}{0.02} )\\\\P(9.98 < X < 10.02) = P( \frac{-0.02}{0.02} < Z < \frac{0.02}{0.02} )\\\\P(9.98 < X < 10.02) = P( -1 < Z < 1 )\\\\

The z-score corresponding to -1 is 0.15866 and 1 is 0.84134

P(9.98 < X < 10.02) = P( Z < 1 ) - P( Z < -1 ) \\\\P(9.98 < X < 10.02) = 0.84134 - 0.15866 \\\\P(9.98 < X < 10.02) = 0.6827\\\\P(9.98 < X < 10.02) = 68.27 \%

Therefore, the probability that the length of aluminum rods is between 9.98 and 10.02 cm is 68.27%

b) We want to find out the probability that the length of aluminum rods is between 9.90 and 10.1 cm.

P(9.90 < X < 10.1) = P( \frac{x - \mu}{\sigma} < Z < \frac{x - \mu}{\sigma} )\\\\P(9.90 < X < 10.1) = P( \frac{9.90- 10}{0.02} < Z < \frac{10.1 - 10}{0.02} )\\\\P(9.90 < X < 10.1) = P( \frac{-0.1}{0.02} < Z < \frac{0.1}{0.02} )\\\\P(9.90 < X < 10.1) = P( -5 < Z < 5 )\\\\

The z-score corresponding to -5 is 0 and 5 is 1

P(9.90 < X < 10.1) = P( Z < 5 ) - P( Z < -5 ) \\\\P(9.90 < X < 10.1) = 1 - 0 \\\\P(9.90 < X < 10.1) = 1\\\\P(9.90 < X < 10.1) = 100 \%

Therefore, the probability that the length of aluminum rods is between 9.90 and 10.1 cm is 100%

How to use z-table?

Step 1:

In the z-table, find the two-digit number on the left side corresponding to your z-score. (e.g 1.0, 2.2, 0.5 etc.)

Step 2:

Then look up at the top of z-table to find the remaining decimal point in the range of 0.00 to 0.09. (e.g. if you are looking for 1.00 then go for 0.00 column)

Step 3:

Finally, find the corresponding probability from the z-table at the intersection of step 1 and step 2.

8 0
3 years ago
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