All categories

2 years ago

Explain how power and work are related

2 answers:

8 0

Work is an activity involving a force on an object and the distance traversed by the object due to the force. It can be measured by the change in energy. Energy is capacity of doing work.For example if a something has energy of x joule it can perform x joule work. Power is the work done in unit time. Hope this helps and please mark me as brainlest and like:)

pochemuha2 years ago

3 0

Without power you can't work .

You might be interested in

With what force will a car hit a tree if the car has a mass of 3,000 kg and it is accelerating at a rate of 2m/s2

Arisa [49]

<span>F=ma = 3000x2m/sec^2 =6000 newtons. </span>

6 0

3 years ago

Rosalind is an impulsive child who seldom waits for her turn when playing with other children. Rosalind's father is also an impu

wariber [46]

Answer:

Alburt Bandura

Explanation:

Rosalind has learned to be impulsive and seldom waits for her turn when playing with other children, this trait she has learned from her father, who is also an impulsive man.

The theorist who would most likely suggest that Rosalind has learned to be impulsive from watching her father is Alburt Bandura. Alburt in his famous Bebo Doll experiment confirmed that children learn from the adults behavior in their life.

3 0

2 years ago

An insulated rigid tank contains 3 kg of H2O in the form of a saturated mixture of liquid and vapor at a pressure of 150 kPa and

Andre45 [30]

Answer:

change in entropy is 1.44 kJ/ K

Explanation:

from steam tables

At 150 kPa

specific volume

Vf = 0.001053 m^3/kg

vg = 1.1594 m^3/kg

specific entropy values are

Sf = 1.4337 kJ/kg K

Sfg = 5.789 kJ/kg

initial specific volume is calculated as

$v_1 = vf + x(vg - vf)$

$= 0.001053 + 0.25(1.1594 - 0.001053)$

$v_1 = 0.20964 m^3/kg$

$s_1 = Sf + x(Sfg)$

$= 1.4337 + 0.25 \times 5.7894 = 2.88 kJ/kg K$

FROM STEAM Table

at 200 kPa

specific volume

Vf = 0.001061 m^3/kg

vg = 0.88578 m^3/kg

specific entropy values are

Sf = 1.5302 kJ/kg K

Sfg = 5.5698 kJ/kg

constant volume so $v_1 - v_2 = 0.29064 m^3/kg$

$v_2 = v_1 = vf + x(vg - vf)$

$=0.29064 = x_2(0.88578 - 0.001061)$

$x_2 = 0.327$

$s_2 = 1.5302 + 0.32 \times 5.5968 = 3.36035 kJ/kg K$

Change in entropy $\Delta s = m(s_2 - s_1)$

=3( 3.36035 - 2.88) = 1.44 kJ/kg

8 0

3 years ago

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical ener

defon

Complete question:

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find

(a) the force constant of the spring and (b) the amplitude of the motion.

Answer:

(a) the force constant of the spring = 47 N/m

(b) the amplitude of the motion = 0.292 m

Explanation:

Given;

mass of the spring, m = 200g = 0.2 kg

period of oscillation, T = 0.410 s

total mechanical energy of the spring, E = 2 J

The angular speed is calculated as follows;

$\omega = \frac{2\pi}{T} \\\\\omega = \frac{2\pi}{0.41} \\\\\omega = 15.33 \ rad/s$

(a) the force constant of the spring

$\omega = \sqrt{\frac{k}{m} } \\\\\omega^2 = \frac{k}{m} \\\\k = m \omega^2\\\\k = (0.2)(15.33)^2\\\\k = 47 \ N/m$

(b) the amplitude of the motion

E = ¹/₂kA²

2E = kA²

A² = 2E/k

$A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2\times 2}{47} }\\\\A = 0.292 \ m$

7 0

3 years ago

Please give best dialogue in photo that is attached And you are marked as Brainliest

coldgirl [10]

Answer:

The correct answer is

Explanation:

Character 1: Why are these comic books so hard to read? Can u read them?

Character 2: No it's easy to read Tigger. It's because you are an animal an I am a human being. U don't know how to read.

Hope this helps....

Have a nice day!!!!

8 0

3 years ago

Read 2 more answers