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jok3333 [9.3K]
3 years ago
9

Devonte pushes a wheelbarrow with 830 W of power. How much work is required to get the wheelbarrow across the yard in 11 s? Roun

d your answer to two significant figures.
Physics
2 answers:
zaharov [31]3 years ago
8 0
9,100 is the answer if you do the work so i failed for yall you welcome
Maslowich3 years ago
6 0

Answer: To get the wheelbarrow across the yard in 11 s Is required 9,100 J of work.

Explanation:

Hi, to solve this problem we have to apply the formula:

Power (w) = work (J) / time (sec)

So, replacing the variables with the values given:

830 w = work / 11sec

Isolating work we have:

830 w / 11 sec = work

Work = 9,130 J  = 9,100 j (rounded to two significant figures)

To get the wheelbarrow across the yard in 11 s Is required 9,100 J of work.

Feel free to ask for more if it´s necessary or if you did not understand something.

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is this true or false? an object with a mass of 9.0kg has an acceleration of 3m/s2. the force acting on it is 3N
ludmilkaskok [199]

The answer is "False". The force acting on the object is 27 N.

According to Newton's second law, when a force <em>F</em> acts on am object of mass <em>m</em>, it produces an acceleration <em>a</em>. The force is given by the expression,

F=ma

Thus, if the body has a mass of 9.0 kg and if it has an acceleration of 3 m/s², then, on substituting the values in the equation for force,

F=ma\\ =(9.0kg)(3m/s^2)\\ =27N

Thus, it can be seen that the force acting on the body is 27 N and not 3 N as is mentioned in the statement. Hence the statement is false.

5 0
3 years ago
A company is interested in buying a new machine to replace their outdated equipment. However, before committing to the purchase,
Helen [10]
The mechanical efficiency = actual work / ideal work

So ζ = 1540 / 1600 * 100% = 96.25%
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22. A ball is thrown horizontally from the roof of a building 12 m tall with a speed of 3.1 m/s.
zysi [14]

Answer:

a) t = 1.6 s

b) d = 4.9 m

c) v = 16 m/s

d) θ = 79°

Explanation:

time of fall

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d = vt = 3.1(1.56) = 4.8512...

vertical velocity vy = at = 9.8(1.56) = 15.336... m/s

v = √(15.336² + 3.1²) = 15.6464... m/s

θ = arctan(15.336/3.1) = 78.5724...°

5 0
3 years ago
A rectangular dam is 101 ft long and 54 ft high. If the water is 35 ft deep, find the force of the water on the dam (the density
blsea [12.9K]

To solve this problem we will begin by finding the pressure through density and average depth. Later we will find the Force, by means of the relation of the pressure and the area.

P = \rho h

Here,

h = Depth average

\rho = Density

Moreover,

\text{Density of water}= \rho = 62.4lb/ft^3

Replacing,

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