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3 years ago

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To complete a project, 200,000 Joules of work is needed. The time taken to complete the project is 20 seconds. How much power is

needed?

2 answers:

Leno4ka [110]3 years ago

5 0

Answer:

10,000

Explanation:

200,000/20 = power needed

200,000/20 = 10,000

Hope this Helps!

MArishka [77]3 years ago

3 0

Answer:

10,000 Watts

Explanation:

Power is the ratio of work to time, or work over time. It can be found using the following formula.

p=w/t

where p is the power, w is the work, and t is the time.

The project needed 200,000 Joules, and it took 20 seconds to complete the project. Therefore, w=200,000 and t=20. Substitute these values into the formula.

p=200,000/20

Divide

p=10,000

Add appropriate units. The units for power is Watts.

p=10,000 Watts

10,000 Watts of power is needed.

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Answer: D

The final velocity of the object will be directed north of east

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Read 2 more answers

If it takes you 20 joules of work to move a couch 10 meters in 10 seconds, what is the power?

Debora [2.8K]

Answer:

Power = 2 j/s

Explanation:

Power = Work / Time

= 20/10

= 2 j/s

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2 years ago

PLEASE HELP ME 45 POINTS

sergij07 [2.7K]

Answer:

a) We kindly invite you to see the explanation and the image attached below.

b) The acceleration of the masses is 4.203 meters per square second.

c) The tension force in the cord is 28.02 newtons.

d) The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is 3.551 meters per second.

Explanation:

a) At first we assume that pulley and cord are both ideal, that is, masses are negligible and include the free body diagrams of each mass and the pulley in the image attached below.

b) Both masses are connected to each other by the same cord, the direction of acceleration will be dominated by the mass of greater mass (mass A) and both masses have the same magnitude of acceleration. By the 2nd Newton's Law, we create the following equation of equilibrium:

Mass A

$\Sigma F = T - m_{A}\cdot g = -m_{A}\cdot a$ (1)

Mass B

$\Sigma F = T - m_{B}\cdot g = m_{B}\cdot a$ (2)

Where:

$T$ - Tension force in the cord, measured in newtons.

$m_{A}$ , $m_{B}$ - Masses of blocks A and B, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$a$ - Net acceleration of the each block, measured in meters per square second.

By subtracting (2) by (1), we get an expression for the acceleration of each mass:

$m_{B}\cdot a +m_{A}\cdot a = T-m_{B}\cdot g -T + m_{A}\cdot g$

$(m_{B}+m_{A})\cdot a = (m_{A}-m_{B})\cdot g$

$a = \frac{m_{A}-m_{B}}{m_{B}+m_{A}} \cdot g$

If we know that $m_{A} = 5\,kg$ , $m_{B} = 2\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the acceleration of the masses is:

$a = \left(\frac{5\,kg-2\,kg}{5\,kg+2\,kg}\right) \cdot\left(9.807\,\frac{m}{s^{2}} \right)$

$a = 4.203\,\frac{m}{s^{2}}$

The acceleration of the masses is 4.203 meters per square second.

c) From (2) we get the following expression for the tension force in the cord:

$T = m_{B}\cdot (a+g)$

If we know that $m_{B} = 2\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $a = 4.203\,\frac{m}{s^{2}}$ , then the tension force in the cord:

$T = (2\,kg)\cdot \left(4.203\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}} \right)$

$T = 28.02\,N$

The tension force in the cord is 28.02 newtons.

d) Given that system starts from rest and net acceleration is constant, we determine the time taken by the block to cover a distance of 1.5 meters through the following kinematic formula:

$\Delta y = \frac{1}{2}\cdot a\cdot t^{2}$ (3)

Where:

$a$ - Net acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

$\Delta y$ - Covered distance, measured in meters.

If we know that $a = 4.203\,\frac{m}{s^{2}}$ and $\Delta y = 1.5\,m$ , then the time taken by the system is:

$t = \sqrt{\frac{2\cdot \Delta y}{a} }$

$t = \sqrt{\frac{2\cdot (1.5\,m)}{4.203\,\frac{m}{s^{2}} } }$

$t \approx 0.845\,s$

The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is calculated by the following formula:

$v = a\cdot t$ (4)

Where $v$ is the final speed of the system, measured in meters per second.

If we know that $a = 4.203\,\frac{m}{s^{2}}$ and $t \approx 0.845\,s$ , then the final speed of the system is:

$v = \left(4.203\,\frac{m}{s^{2}} \right)\cdot (0.845\,s)$

$v = 3.551\,\frac{m}{s}$

The final speed of the system is 3.551 meters per second.

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How much is required to accelerate a 5kg mass at 20 m/s2

zhannawk [14.2K]

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3 years ago

Rearrange the equation to solve for the permeability of free space (μ0). Remember that the slope is the ratio of magnetic field

IRINA_888 [86]

Complete Question

The complete question is shown on the first uploaded image

Compare the percent error between your value and the accepted value of 1.26 times 10-6 T · m/A. (Use the accepted value given to three significant figures in your calculation.). 100% % error = %

Answer:

The permeability of free space is $\mu_0 = 1.32*10^{-6} \ Tm/A$

The percentage error is % error = 5.25%

Explanation:

From the question we are told that

The slope is $s= 3.9\ G/A$ , and

Generally $1 \ gauss = 10^{-4 } tesla$

So $s = 3.9 *10^ {-4} T /A$

From the relation given in the question

$\mu_0 = \frac{2R}{N} [\frac{B}{I} ]$

Where R is the radius of the coil $=\frac{Diameter \ of \ coil }{2} = 0.017m$

N is the number of loops of the coil = 10

Now from the question we are told that

$s = \frac{B}{I}$

substituting into the equation above

$\mu_0 = \frac{2R }{N} s$

Substituting values

$\mu_0= \frac{2 * 0.017}{10 } * 3.9*10^{-4}$

$= 1.326 *10^ {-6} \ Tm/A$

Generally the % error is mathematically represented as

% $error = \frac{Measured \ value - accepted \ value}{accepted \ value } *100$

Given that the accepted value is $\mu_o_ {acc} = 1.26 *10 ^{-6} \ T \cdot m /A$

Hence substituting values

% $error = \frac{(1.326-1.26)*10^{-6}}{1.26 *10 ^ {-6}} *100$

$= 5.24$

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3 years ago