Ask question

Ask question

All categories

Nimfa-mama [501]

3 years ago

7

A company is interested in buying a new machine to replace their outdated equipment. However, before committing to the purchase,

the company asked the manufacturer to provide data on a certain machine's efficiency. The data suggests that with a supplied energy at a rate of 1600 w, the machine will do useful work at a rate of 1540 w. What's the efficiency of this particular machine

2 answers:

Helen [10]3 years ago

8 0

The mechanical efficiency = actual work / ideal work

So ζ = 1540 / 1600 * 100% = 96.25%

s2008m [1.1K]3 years ago

4 0

Answer:

96.25%

Explanation:

You might be interested in

Calculate the ratio of the resistance of 12.0 m of aluminum wire 2.5 mm in diameter, to 30.0 m of copper wire 1.6 mm in diameter

alukav5142 [94]

Answer: 0.258

Explanation:

The resistance $R$ of a wire is calculated by the following formula:

$R=\rho\frac{l}{s}$ (1)

Where:

$\rho$ is the resistivity of the material the wire is made of. For aluminium is $\rho_{Al}=2.65(10)^{-8}m\Omega$ and for copper is $\rho_{Cu}=1.68(10)^{-8}m\Omega$

$l$ is the length of the wire, which in the case of aluminium is $l_{Al}=12m$ , and in the case of copper is $l_{Cu}=30m$

$s$ is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:

$s=\pi{(\frac{d}{2})}^{2}$ (2) Where $d$ is the diameter of the circumference.

For aluminium wire the diameter is $d_{Al}=2.5mm=0.0025m$ and for copper is $d_{Cu}=1.6mm=0.0016m$

So, in this problem we have two transversal areas:

<u>For aluminium:</u>

$s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}$

$s_{Al}=0.000004908m^{2}$ (3)

<u>For copper:</u>

$s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}$

$s_{Cu}=0.00000201m^{2}$ (4)

Now we have to calculate the resistance for each wire:

<u>Aluminium wire:</u>

$R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}$ (5)

$R_{Al}=0.0647\Omega$ (6) Resistance of aluminium wire

<u>Copper wire:</u>

$R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}$ (6)

$R_{Cu}=0.250\Omega$ (7) Resistance of copper wire

At this point we are able to calculate the ratio of the resistance of both wires:

$Ratio=\frac{R_{Al}}{R_{Cu}}$ (8)

$\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}$ (9)

Finally:

$\frac{R_{Al}}{R_{Cu}}=0.258$ This is the ratio

3 0

3 years ago

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,00

Aleksandr-060686 [28]

Answer:

Explanation:

We have the following relation between power, P and intensity, I

$I = P/(4*pi*r^2)$

= $10^3/(4*pi*(35000*10^3))$

= $6.5*10^-14 W/M^2$

We also have the following relationship between electric field and electromagnetic radiation thus

$I = (ceE^2)/2$

Hence $E = \sqrt{2I/ce}$

substituting the values of I, c and e, we have

$7*10^-6 V/m$

3 0

2 years ago

Everything we know about stars comes from studying their _____.

denis-greek [22]

Light / Radiation. The light radiation defines its color. Thermal radiation for its temperature and the overall appearance of the wavelengths the star emits gives off the info from where astronomers and scientist are able to build up knowledge on the age and current state of the stars

5 0

3 years ago

Read 2 more answers

A small glass bead has been charged to + 30.0 nC . A small metal ball bearing 2.60 cm above the bead feels a 1.80×10−2 N downwar

BlackZzzverrR [31]

Answer:

The charge on the ball bearing 4.507 × 10^-8 C

Explanation:

From Coulomb's law

F = kq1q2/r²

make q2 the subject

q2 = Fr²/kq1

q2 = (1.8×10^-2 × 0.026²) ÷ (9×10^9 × 30×10^-9)

q2 = 4.507 × 10^-8 C

8 0

2 years ago

What are the seven types of electromagnetic waves?

Sunny_sXe [5.5K]

Radio waves. Giant satellite-dish antennas pick up long-wavelength, high-frequency radio waves. ...
Microwaves. Because cosmic microwaves can't get through the whole of Earth's atmosphere, we have to study them from space. ...
Infrared. ...
Visible light. ...
Ultraviolet light. ...
X rays. ...
Gamma rays.

5 0

2 years ago