The maximum efficiency of a heat engine operating between 2 degrees c and 200 degrees c will be about:

A solenoid of length 0.700m having a circular cross-section of radius 5.00cm stores 6.00 μJ of energy when a 0.400-A current run

AURORKA [14]

Answer:

The winding density of the solenoid, n = 104 turns/m

Explanation:

Given that,

Length of the solenoid, l = 0.7 m

Radius of the circular cross section, r = 5 cm = 0.05 m

Energy stored in the solenoid, $E=6\ \mu J=6\times 10^{-6}\ J$

Current, I = 0.4 A

To find,

The winding density of the solenoid.

Solution,

The expression for the energy stored in the solenoid is given by :

$U=\dfrac{1}{2}LI^2$

Where

L is the self inductance of the solenoid

$L=\mu_on^2lA$

n is the winding density of the solenoid

$n=\sqrt{\dfrac{2U}{\mu_oI^2l\pi r^2}}$

$n=\sqrt{\dfrac{2\times 6\times 10^{-6}}{4\pi \times 10^{-7}\times 0.7\times (0.4)^2\pi (0.05)^2}}$

n = 104 turns/m

So, the winding density of the solenoid is 104 turns/m

8 0

1 year ago

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In the absence of air resistance, at what other angle will a thrown ball go the same distance as one thrown at an angle of 75 de

snow_tiger [21]

As we know that range of the projectile motion is given by

$R = \frac{v^2 sin(2\theta)}{g}$

here we know that range will be same for two different angles

so here we can say the two angle must be complementary angles

so the two angles must be

$\theta, 90 - \theta$

so it is given that one of the projection angle is 75 degree

so other angle for same range must be 90 - 75 = 15 degree

so other projection angle must be 15 degree

5 0

1 year ago

10. A girl pulls a wagon along a level path for a distance of 44 m. The handle of

Rina8888 [55]

The work done on the wagon is 3549 J

Explanation:

The work done by a force when moving an object is given by

$W=Fd cos \theta$

where :

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and of the displacement

In this problem we have the following data:

F = 87 N is the magnitude of the force

d = 44 m is the displacement of the wagon

$\theta=22^{\circ}$ is the angle between the direction of the force and the displacement

Substituting, we find the work done

$W=(87)(44)(cos 22^{\circ})=3549 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0

2 years ago

The density of nuclear matter is about 1018 kg/m3. Given that 1 mL is equal in volume to 1 cm3, what is the density of nuclear m

Sonbull [250]

Answer:

density is $10^{6}$ Mg/µL

Explanation:

given data

density of nuclear = $10^{18}$ kg/m³

1 ml = 1 cm³

to find out

density of nuclear matter in Mg/µL

solution

we know here

1 Mg = 1000 kg

so

1 m³ is equal to $10^{6}$ cm³

and here 1 cm³ is equal to 1 mL

so we can say 1 mL is equal to 10³ µL

so by these we can convert density

density = $10^{18}$ kg/m³

density = $10^{18}$ kg/m³ × $\frac{10^{-3} }{10^{6} }$ Mg/µL

density = $10^{6}$ Mg/µL

8 0

2 years ago

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A segment of wire of total length 2.0 m is formed into a circular loop having 5.0 turns. If the wire carries a 1.2-A current, de

docker41 [41]

Answer:

Magnetic field at the center of the loop $B=5.89\times 10^{-5}\ T.$

Explanation:

It is given that total length of wire is 2 m and number of circular loop is 5 turns.

Therefore ,

$5\times ( 2\pi r)=2 \ m .\\\\r=\dfrac{1}{5 \pi}=0.064\ m.$

We know , magnetic field at the center of loop is given by :

$B=N\dfrac{\mu_o i}{2r}$

Putting all values in above equation we get :

$B=5\times \dfrac{4\pi\times 10^{-7}\times 1.2}{2\times 0.064}\\\\B=5.89\times 10^{-5}\ T.$

Hence , this is the required solution.

8 0

1 year ago