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AveGali [126]
3 years ago
14

What is the efficiency of a machine?​

Physics
1 answer:
Julli [10]3 years ago
7 0

Answer:

Efficiency is the percent of work put into a machine by the user (input work) that becomes work done by the machine (output work).

Explanation:

It is a measure of how well a machine reduces friction.

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A car traveled 1025 km from El Paso to Dallas iin 13.5 hr. What was its average velocity
JulsSmile [24]
1) 1025km per 13.5hr = 1025km/13.5hr = 75.9 km/hr

2) 400km per 30min = 400km/30min = 13.3 km/min
13.3km/min X (60min per hr) = 13.3km/min(60min/hr) = 800km/hr

3) 45km/hr in 40min = 45km/hr X 40min X (1hr/60min) = 30km.

<span> 4) (30m/s - 10m/s) per 10s = 20m/s/10s = 2ms/s = 2m/s/s = 2m/s^2 </span>
5 0
3 years ago
A 0.130 m radius, 485-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a
Lady_Fox [76]

Answer:

The magnetic field strength needed is 1.619 T

Explanation:

Given;

Number of turns, N = 485-turn

Radius of coil, r = 0.130 m

time of revolution, t =  4.17 ms = 0.00417 s

average induced emf, V = 10,000 V.

Average induced emf is given as;

V = -ΔФ/Δt

where;

ΔФ is change in flux

Δt is change in time

ΔФ = -NBA(Cos \theta_f - Cos \theta_i)

where;

N is the number of turns

B is the magnetic field strength

A is the area of the coil = πr²

θ is the angle of inclination of the coil and the magnetic field,

\theta_f = 90^o\\\theta_i = 0^o

V = NBACos0/t

V = NBA/t

B = (Vt)/NA

B = (10,000 x 0.00417) / (485 x π x 0.13²)

B =1.619 T

Thus, the magnetic field strength needed is 1.619 T

5 0
3 years ago
Read 2 more answers
What acceleration does the force of earth's gravity peoduce
maksim [4K]
Near Earth's surface, gravitational acceleration is approximately 9.81 m/s2, which means that, ignoring the effects of air resistance, the speed of an object falling freely will increase by about 9.81 metres per second every second.
8 0
3 years ago
A substance contains only one type of atom. The substance is a/an ____________. A. solution B. element C. mixture D. compound
Alex17521 [72]
The answer to this question is:B. Element









3 0
3 years ago
Read 2 more answers
A short current element dl⃗ = (0.500 mm)j^ carries a current of 5.40 A in the same direction as dl⃗ . Point P is located at r⃗ =
SashulF [63]

Answer:

The magnetic field along x axis is

B_{x}=1.670\times10^{-10}\ T

The magnetic field along y axis is zero.

The magnetic field along z axis is

B_{z}=3.484\times10^{-10}\ T

Explanation:

Given that,

Length of the current element dl=(0.5\times10^{-3})j

Current in y direction = 5.40 A

Point P located at \vec{r}=(-0.730)i+(0.390)k

The distance is

|\vec{r}|=\sqrt{(0.730)^2+(0.390)^2}

|\vec{r}|=0.827\ m

We need to calculate the magnetic field

Using Biot-savart law

B=\dfrac{\mu_{0}}{4\pi}\timesI\times\dfrac{\vec{dl}\times\vec{r}}{|\vec{r}|^3}

Put the value into the formula

B=10^{-7}\times5.40\times\dfrac{(0.5\times10^{-3})\times(-0.730)i+(0.390)k}{(0.827)^3}

We need to calculate the value of \vec{dl}\times\vec{r}

\vec{dl}\times\vec{r}=(0.5\times10^{-3})\times(-0.730)i+(0.390)k

\vec{dl}\times\vec{r}=i(0.350\times0.5\times10^{-3}-0)+k(0+0.730\times0.5\times10^{-3})

\vec{dl}\times\vec{r}=0.000175i+0.000365k

Put the value into the formula of magnetic field

B=10^{-7}\times5.40\times\dfrac{(0.000175i+0.000365k)}{(0.827)^3}

B=1.670\times10^{-10}i+3.484\times10^{-10}k

Hence, The magnetic field along x axis is

B_{x}=1.670\times10^{-10}\ T

The magnetic field along y axis is zero.

The magnetic field along z axis is

B_{z}=3.484\times10^{-10}\ T

7 0
3 years ago
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