calculate the amount of work done when a grocery store stocker uses 120N of force to lift a sack of flour 1.5m onto a shelf

Predict whether the changes in enthalpy, entropy, and free energy will be positive or negative for the melting of ice, and expla

BlackZzzverrR [31]

<span>Melting of ice is an endothermic process, meaning that energy is absorbed. When ice spontaneously melts, ΔH (change in enthalpy) is "positive". ΔS (entropy change) is also positive, because, becoming a liquid, water molecules lose their fixed position in the ice crystal, and become more disorganized. ΔG (free energy of reaction) is negative when a reaction proceeds spontaneously, as it happens in this case. Ice spontaneously melts at temperatures higher than 0°C. However, liquid water also spontaneously freezes at temperatures below 0°C. Therefore the temperature is instrumental in determining which "melting" of ice, or "freezing" of water becomes spontaneous. The whole process is summarized in the Gibbs free energy equation:
ΔG = ΔH – TΔS</span>

4 0

3 years ago

A force of 5.0 N acts on a 15 kg body initially at rest. Compute the work done by the force in (a) the first, (b) the second, an

Leokris [45]

Answer:

(a) 0.833 j

(b) 2.497 j

(c) 4.1625 j

(d) 4.995 watt

Explanation:

We have given force F = 5 N

Mass of the body m = 15 kg

So acceleration $a=\frac{F}{m}=\frac{5}{15}=0.333m/sec^2$

As the body starts from rest so initial velocity u = 0 m/sec

(a) From second equation of motion $s=ut+\frac{1}{2}at^2$

For t = 1 sec

$s=0\times 1+\frac{1}{2}\times 0.333\times 1^2=0.1666m$

We know that work done W =force × distance = 5×0.1666 =0.833 j

(b) For t = 2 sec

$s=0\times 2+\frac{1}{2}\times 0.333\times 2^2=0.666m$

We know that work done W =force × distance = 5×0.666 =3.33 j

So work done in second second = 3.33-0.833 = 2.497 j

(c) For t = 3 sec

$s=0\times 3+\frac{1}{2}\times 0.333\times 3^2=1.4985m$

We know that work done W =force × distance = 5×1.4985 =7.4925 j

So work done in third second = 7.4925 - 2.497 -0.833 = 4.1625 j

(d) Velocity at the end of third second v = u+at

So v = 0+0.333×3 = 0.999 m /sec

We know that power P = force × velocity

So power = 5× 0.999 = 4.995 watt

4 0

4 years ago

what is the wavelength of radio waves transmitted by an FM station at 90MHz where 1M=10^6, and speed of radiowave is 3*10^8m/s

Igoryamba

Answer:

λ = 3.33 m

Explanation:

<u><em>Given:</em></u>

Frequency = f = 9 × 10⁷ Hz

Speed = v = 3 × 10⁸ m/s

<u><em>Required:</em></u>

Wavelength = λ = ?

<u><em>Formula:</em></u>

v = fλ

<u><em>Solution:</em></u>

<em>Putting the givens in the formula</em>

v = fλ

λ = $\frac{v}{f}$

λ = $\frac{3*10^8}{9*10^7}$

λ = 0.33 × 10¹

λ = 3.33 m

5 0

3 years ago

How far will a freely falling object fall from rest in five seconds? six seconds?

Vanyuwa [196]

Distance = 1/2*gravity*velocity^2
<span>So plugging: 1/2 * 9.8 * 25 = 122.5units
and is six second
</span><span>Distance = 1/2*gravity*velocity^2
</span><span>So plugging: 1/2 * 9.8 * 36= 176.4 units</span>

4 0

3 years ago

A mass attached to a spring oscillates in simple harmonic motion with an amplitude of 10 cm. When the mass is 5.0 cm from its eq

timama [110]

When the mass is 5.0 cm from its equilibrium point, the percentage of its energy that is kinetic is 75%.

<h3>Total energy of the mass</h3>

The total energy possessed by the mass under the simple harmonic motion is calculated as follows;

U = ¹/₂kA²

where;

k is the spring constant
A is the amplitude of the oscillation

<h3>Potential energy of the mass at 5 cm from equilibrium point</h3>

P.E = ¹/₂k(Δx)²

<h3>Kinetic energy of mass</h3>

K.E = U - P.E

K.E = ¹/₂kA² - ¹/₂k(Δx)²

<h3>Percentage of its energy that is kinetic</h3>

$K.E (\%) = \frac{U - P.E}{U} \times 100\%\\\\K.E (\%) =\frac{\frac{1}{2}kA^2 - \frac{1}{2}k(\Delta x)^2 }{\frac{1}{2}kA^2} \times 100\%\\\\K.E (\%) = \frac{A^2 - (\Delta x)^2}{A^2} \times 100\%\\\\K.E (\%) = \frac{10^2 - (10-5)^2}{10^2} \times 100\%\\\\K.E (\%) = \frac{10^2 - 5^2}{10^2} \times 100\%\\\\K.E (\%) = 75\%$

Thus, when the mass is 5.0 cm from its equilibrium point, the percentage of its energy that is kinetic is 75%.

Learn more about kinetic energy here: brainly.com/question/25959744

3 0

2 years ago