<span>The Climate Friendly Farming Project was designed to provide funding for long term research into the sustainability and adaptability of agricultural systems(wheat, vegetables, and dairy systems) as increasing greenhouse gases propagate and decline the efficiency of current agricultural practice.</span>

4 0

3 years ago

The magnetic field strength at the north pole of a 2.0-cmcm-diameter, 8-cmcm-long Alnico magnet is 0.10 TT. To produce the same

horrorfan [7]

Answer:

2653 turns

Explanation:

We are given that

Diameter,d=2 cm

Length of magnet,l=8 cm= $8\times 10^{-2} m$

1m=100 cm

Magnetic field,B=0.1 T

Current,I=2.4 A

We are given that

Magnetic field of solenoid and magnetic are same and size of both solenoid and magnetic are also same.

Length of solenoid= $8\times 10^{-2} m$

Magnetic field of solenoid

$B=\frac{\mu_0NI}{l}$

Using the formula

$0.1=\frac{4\pi\times 10^{-7}\times 2.4\times N}{8\times 10^{-2}}$

Where $\mu_0=4\pi\times 10^{-7}$

$N=\frac{0.1\times 8\times 10^{-2}}{4\pi\times 10^{-7}\times 2.4}=2653 turns$

6 0

3 years ago

An automobile with a standard differential turns sharply to the left. The left driving wheel turns on a 20-m radius. Distance be

Inessa05 [86]

Explanation:

The given data is as follows.

Inner wheel Radius = 20 m,

Distance between left and right wheel = 1.5m,

Let us assume speed of drive shaft is N rpm.

Formula to calculate angular velocity is as follows.

Angular velocity of automobile = w = $\frac{V}{R}$

where, V = linear velocity of automobile m/min,

R = turning radius from automobile center in meter

In the given case, angular velocity remains same for inner and outer wheel but there is change in linear velocity of inner wheel and outer wheel.

Now, we assume that

u = linear velocity of inner wheel

and, u' = linear velocity of outer wheel.

Formula for angular velocity of inner wheel w = ,

Formula for angular velocity of outer wheel w =

Now, for inner wheels

w =

= $\frac{u}{(R - d)}$

u = $V \times \frac{(R - d)}{R}$

= $V \times (1 - \frac{d}{R})$

If radius of wheel is r it will cover distance in one min.

Since, velocity of wheel is u it will cover distance u in unit time(min)

Thus, u = $2\pi rn$ = $V \times (1 - \frac{d}{R})$

Now, rotation per minute of inner wheel is calculated as follows.

n = $\frac{V}{2 \pi r \times (1 - \frac{d}{R})}$

= $\frac{V}{2 \pi r \times (1 - \frac{0.75}{20})}$ (since 2d = 1.5m given, d = 0.75m),

= $\frac{V}{r} \times 0.1532$

So, rotation per minute of outer wheel; n' =

= $\frac{V}{2 \pi r \times (1 + \frac{0.75}{20})}$

= $\frac{V}{r} \times 0.1651$

5 0

4 years ago

A projectile is launched with an initial speed of 60.0 m/s at an angle of 30.0° above the horizontal. The projectile lands on a

vodka [1.7K]

Answer:

51.96 m/s^-1

Explanation:

a) see the attachment

b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be

v_ox=v_o*cosФ

=60*cos (30)

= 51.96 m/s^-1

3 0

3 years ago

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