All categories

2 years ago

A machine produces 4,000 J of work in 5 seconds. how much power does the machine produce.

Physics

1 answer:

ad-work [718]2 years ago

5 0

Explanation:

Power is defined as the work done per unit time or

$P = \dfrac{\Delta W}{\Delta t}$

$\;\;\;\;= \dfrac{4000\:\text{J}}{5\:\text{s}} = 800\:\text{Watts}$

You might be interested in

An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of the compression process,

Sedbober [7]

Answer:

$a.T_3=1723.8kPa\\b.n=0.563\\c.MEP=674.95kPa$

Explanation:

a. Internal energy and the relative specific volume at $s_1$ are determined from A-17: $u_1=214.07kJ/kg, \ \alpha_r_1=621.2$ .

The relative specific volume at $s_2$ is calculated from the compression ratio:

$\alpha_r_2=\frac{\alpha_r_1}{r}\\=\frac{621.2}{16}\\=38.825$

#from this, the temperature and enthalpy at state 2, $s_2$ can be determined using interpolations $T_2=862K$ and $h_2=890.9kJ/kg$ . The specific volume at $s_1$ can then be determined as:

$\alpha_1=\frac{RT_1}{P_1}\\\\=\frac{0.287\times 300}{95} m^3/kg\\0.906316m^3/kg$

Specific volume, $s_2$ :

$\alpha_2=\frac{\alpha_1}{r}\\=\frac{0.906316}{16}m^3/kg\\=0.05664m^3/kg$

The pressures at $s_2 \ and\ s_3$ is:

$P_2=P_3=\frac{RT_2}{\alpha_2}\\\\=\frac{0.287\times862}{0.05664}\\=4367.06kPa$

.The thermal efficiency=> maximum temperature at $s_3$ can be obtained from the expansion work at constant pressure during $s_2-s_3$

$\bigtriangleup \omega_2_-_3=P(\alpha_3-\alpha_2)\\R(T_3-T_2)=P\alpha(r_c-1)\\T_3=T_2+\frac{P\alpha_2}{R}(r_c-1)\\\\=(862+\frac{4367\times 0.05664}{0.287}(2-1))K\\=1723.84K$

b.Relative SV and enthalpy at $s_3$ are obtained for the given temperature with interpolation with data from A-17 : $a_r_3=4.553 \ and\ h_3=1909.62kJ/kg$

Relative SV at $s_4$ is

$a_r_4=\frac{r}{r_c}\alpha _r_3$

= $=\frac{16}{2}\times4.533\\=36.424$

Thermal efficiency occurs when the heat loss is equal to the internal energy decrease and heat gain equal to enthalpy increase;

$n=1-\frac{q_o}{q_i}\\=1-\frac{u_4-u_1}{h_3-h_2}\\=1-\frac{65903-214.07}{1909.62-890.9}\\=0.563$

Hence, the thermal efficiency is 0.563

c. The mean relative pressure is calculated from its standard definition:

$MEP=\frac{\omega}{\alpa_1-\alpa_2}\\=\frac{q_i-q_o}{\alpha_1(1-1/r)}\\=\frac{1909.62-890.9-(65903-214.7)}{0.90632(1-1/16)}\\=674.95kPa$

Hence, the mean effective relative pressure is 674.95kPa

3 0

3 years ago

A rocket rises vertically, from rest, with an acceleration of 3.2 m/s2 until it runs out of fuel at an altitude of 1300 m. After

svetlana [45]

Explanation:

Below is an attachment containing the solution.

6 0

3 years ago

180 J of work is done when lifting a box up to a shelf that is 3 m high. What is the mass of the box?

ale4655 [162]

Explanation:

Work done= MGH

180=m×10×3

180/30=m

m=6kg

4 0

2 years ago

If all this energy is added to 50 kg of water (the amount of water in a 165-lb person) at 37 ∘C, what is the final state of the

Reika [66]

Answer:

Vapors

Explanation:

We take into account that all the energy from the lightning has been transformed into steam.

$\Delta U = Q - W\\Q = mC \Delta T\\Q = mL$

We calculate the amount of energy required by water to convert into steam.

$Q_{water} = 50 \times \times 4180 \times (100-37)\\= 1.132 \times 10^7 \ J$

$Q_{change\ to\ steam} = 50 \times 2.256 \times 106 \\= 1.128 \times 10^8 \ J$

$Q_{total} = 1.132 \times 10^7 + 1.128 \times 10^8\\= 1.126 \times 10^8 \ J$

From the lightning we received $10^{10} \ J$ of energy, out of which $1.126 \times 10^8$ has been used to convert the water into steam.

Energy left = $10^{10} - 1.126 \times 10^8 = 9.88 \times 10^9 \ J$

We use this energy to convert steam into vapors.

$Q = \Delta E$

$Q = \Delta E = mc (T_{f} - T{i})\\T_{f} = \frac {\Delta E}{mc} + T_{i}\\ \\T_{f} = 100 + \frac{9.88 \times 10^{10}}{50 \times 1970}\\T_{f} = 100 + 10^8\\T_{f} = 10 ^{ \ 8} \ {^ \circ } C$

With this temperature, we can easily interpret that the vapors will be dissociated in hydrogen and oxygen particles.

5 0

2 years ago

How many cities could 8 TeV power for a week?

HACTEHA [7]

science hasnt figured it out yet

6 0

2 years ago

Read 2 more answers