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storchak [24]
2 years ago
12

Wyatt is moving a box with a mass of 37 kg a distance of 37 meters. Wyatt did 360 J of work in 2 minutes when moving the box. Wh

at was his power output? A) 9.72 W B) 3 W C) 180 W D) 323 W
Physics
2 answers:
pentagon [3]2 years ago
7 0
His power output was 3 Watt (360 Joule/120 seconds). The power output can be calculated by dividing the quantity of work by the amount of second needed for the activity and also by multiplying the force amount with the velocity of the activity. The power output usually used for measuring the ability of machine for doing its job.
enot [183]2 years ago
4 0

Answer:

Power output, P = 3 watts

Explanation:

It is given that,

Work done by Wyatt, W = 360 J

Time taken, t = 2 minutes = 120 s

We need to find the power output. It is given by:

P=\dfrac{W}{t}

Where

W = work done

t = time taken

So, P=\dfrac{360\ J}{120\ s}

P = 3 Watts

So, the power output is 3 to watts. Hence, this is the required solution.

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frez [133]

Answer:

a = Δv/t = (vf - vi)/t = (0 - 5)/4 = -1.25 m/s²

Explanation:

You may or may not need the negative sign, depending on how the question designer was thinking about the problem.

4 0
3 years ago
A particle moving along the x-axis has a position given by m, where t is measured in s. What is the magnitude of the acceleratio
8090 [49]

Question:

A particle moving along the x-axis has a position given by x=(24t - 2.0t³)m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero

Answer:

24 m/s

Explanation:

Given:

x=(24t - 2.0t³)m

First find velocity function v(t):

v(t) = ẋ(t) = 24 - 2*3t²

v(t) = ẋ(t) = 24 - 6t²

Find the acceleration function a(t):

a(t) = Ẍ(t) = V(t) = -6*2t

a(t) = Ẍ(t) = V(t) = -12t

At acceleration = 0, take time as T in velocity function.

0 =v(T) = 24 - 6T²

Solve for T

T = \sqrt{\frac{-24}{6}} = \sqrt{-4} = -2

Substitute -2 for t in acceleration function:

a(t) = a(T) = a(-2) = -12(-2) = 24 m/s

Acceleration = 24m/s

4 0
3 years ago
7. These temperatures were recorded in Pasadena for a week in April. 87 85 80 78 83 86 90 Find each of these. (a) Mean (e) Range
slega [8]

Answer:

a) Mean = 84.14

b) Median = 85

c) Mode = no mode (since there is no variable that appears more than once in this dataset)

d) Midrange = 84

e) Range = 12

f) Variance = 14.69

g) Standard deviation = 3.83

Explanation:

The raw data to be processed is

87 85 80 78 83 86 90

a) Mean = (Σx)/N

The mean is the sum of variables divided by the number of variables

x = each variable

N = number of variables = 7

Mean = (87+85+80+78+83+86+90)/7

Mean = 84.14

b) Median is the number in the middle of the dataset when the variables are arranged in ascending or descending order.

Arranging the data in ascending order

78, 80, 83, 85, 86, 87, 90

The number in the middle is the 4th number = 85

Median = 85

c) Mode is the variable that occurs the most in a distribution.

For this question, all of the variables occur only once, with no variable occurring more than once. Hence, there is no mode for this dataset.

d) Midrange is the arithmetic mean of the highest and lowest number in the dataset.

Mathematically,

Midrange = (Highest + Lowest)/2

Midrange = (90 + 78)/2

Midrange = 84

e) Range is the difference the highest and the lowest numbers in a dataset.

Range = 90 - 78 = 12

f) Variance is an average of the squared deviations from the mean.

Mathematically,

Variance = [Σ(x - xbar)²/N]

xbar = mean

Σ(x - xbar)² = (78 - 84.14)² + (80 - 84.14)² + (83 - 84.14)² + (85 - 84.14)² + (86 - 84.14)² + (87 - 84.14)² + (90 - 84.14)² = 102.8572

Variance = (102.8572)/7

Variance = 14.69

g) Standard deviation = √(variance)

Standard deviation = √(14.69)

Standard deviation = 3.83

Hope this Helps!!!

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