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Anestetic [448]

3 years ago

8

A grain silo is shown below:grain silo formed by cylinder with radius 5 feet and height 175 feet and a half sphere on the topwha

t is the volume of grain that could completely fill this silo, rounded to the nearest whole number? use 22 over 7 for pi.

1 answer:

nekit [7.7K]3 years ago

6 0

Find the volume of the bottom and top separately and then add them.
Cylinder volume is the area of the bottom times the height
(22/7)(5^2)•175=13750 ft^3

The volume of a sphere is
V=(4/3)(22/7)r^3
where r is the radius. Here that's also 5 since it fits on the cylinder.
Also we only want half the sphere so use
V=(2/3)(22/7)•5^3=261.9 ft^3
Which we round upto 262.
Now add the parts together
13750+262=14,012 ft^3

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A 2.0-kg object is attached to a spring (k = 55.6 N/m) that hangs vertically from the ceiling. The object is displaced 0.045 m v

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Maximum acceleration will be $1.251m/sec^2$

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a hiker walks 200 m west and then walks 100 m north in what direction is her resulting displacement draw to show your answer

Sliva [168]

Answer:

The direction of the displacement is in North-West.

Explanation:

Resultant displacement D is

$=\sqrt{(200)^{2} + (100)^{2} } \\=223.60m$

Here the direction is

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3 years ago

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

kati45 [8]

Answer:

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

Explanation:

The average emf induced can be calculated from the formula

$Emf = -N\frac{\Delta \phi}{\Delta t}$

Where $N$ is the number of turns

$\Delta \phi$ is the change in magnetic flux

$\Delta t$ is the time interval

The change in magnetic flux is given by

$\Delta \phi = \phi _{f} - \phi _{i}$

Where $\phi _{f}$ is the final magnetic flux

and $\phi _{i}$ is the initial magnetic flux

Magnetic flux is given by the formula

$\phi = BAcos(\theta)$

Where $B$ is the magnetic field

$A$ is the area

and $\theta$ is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, $\theta = 0^{o}$

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

$A = \pi r^{2}$

∴ $A = \pi (0.15)^{2}$

$A = 0.0225\pi$

∴ $\phi_{i} = (1.5)(0.0225\pi)cos(0^{o} )$

$\phi_{i} = (1.5)(0.0225\pi)$

( NOTE: $cos (0^{o}) = 1$ )

$\phi_{i} = 0.03375\pi$ Wb

For $\phi_{f}$

The field pointed upwards, that is $\theta = 90^{o}$ . Since $cos (90^{o}) = 0$

Then

$\phi_{f} = 0$

Hence,

$\Delta \phi = 0- 0.03375\pi$

$\Delta \phi = - 0.03375\pi$

From the question

$\Delta t = 2.8 ms = 2.8 \times 10^{-3} s$

Here, $N$ = 1

Hence,

$Emf = -N\frac{\Delta \phi}{\Delta t}$ becomes

$Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }$

$Emf = 37.9 V$

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.

5 0

3 years ago