6. An object moves along the x-axis. The graph shows its position x as a function of time t. Find

A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi

Lubov Fominskaja [6]

Answer:

The moment of inertia is $I=\frac{M}{12} a^{2}$

Explanation:

The moment of inertia is equal:

$I=\int\limits^a_b {r^{2} } \, dm$

If r is $-\frac{a}{2}$

and $dm=\frac{M}{a} dr$

$I=\int\limits^a_b {r^{2}\frac{M}{a} } \, dr\\a=\frac{a}{2} \\b=-\frac{a}{2}$

$I=\frac{M}{a} \int\limits^a_b {r^{2} } \, dr\\\\I=\frac{M}{a} (\frac{M}{3} )_{b}^{a}\\ I=\frac{M}{3a} (\frac{a^{3} }{8} +\frac{a^{3} }{8} )\\I=\frac{M}{12} a^{2}$

7 0

3 years ago

A permanent magnet has a magnetic dipole moment of 0.160 A · m^2. The magnet is in the presence of an external uniform magnetic

Elena L [17]

Answer:

the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm

the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J

Explanation:

The torque is given by :

$\bar {N} = \bar {m} * \bar {B}$

where ;

m = 0.160 A.m²

B = 0.0800 T

θ = 35°

So the magnitude of the torque N = mBsinθ

N = (0.160)(0.0800)(sin 35°)

N = 0.007341

N = 7.34×10⁻³ Nm

Hence, the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm

b) The potential energy $\bar{U} = \bar{-m} * \bar{B}$

U = -mBcosθ

U = (- 0.160)(0.0800)(cos 45)

U = -0.010485

U = -1.0485 ×10⁻² J

Thus, the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J

6 0

2 years ago

A confined aquifer with a porosity of 0.15 is 30 m thick. The potentiometric surface elevation at two observation wells 1000 m a

AlekseyPX

Answer:

Part (a) The flow rate per unit width of the aquifer is 1.0875 m³/day

Part (b) The specific discharge of the flow is 0.0363 m/day

Part (c) The average linear velocity of the flow is 0.242 m/day

Part (d) The time taken for a tracer to travel the distance between the observation wells is 4132.23 days = 99173.52 hours

Explanation:

Part (a) the flow rate per unit width of the aquifer

From Darcy's law;

$q = -Kb\frac{dh}{dl}$

where;

q is the flow rate

K is the permeability or conductivity of the aquifer = 25 m/day

b is the aquifer thickness

dh is the change in th vertical hight = 50.9m - 52.35m = -1.45 m

dl is the change in the horizontal hight = 1000 m

q = -(25*30)*(-1.45/1000)

q = 1.0875 m³/day

Part (b) the specific discharge of the flow

$V = \frac{Q}{A} = \frac{q}{b} = -K\frac{dh}{dl}\\\\V = -(25 m/d).(\frac{-1.45 m}{1000 m}) = 0.0363 m/day$

V = 0.0363 m/day

Part (c) the average linear velocity of the flow assuming steady unidirectional flow

Va = V/Φ

Φ is the porosity = 0.15

Va = 0.0363 / 0.15

Va = 0.242 m/day

Part (d) the time taken for a tracer to travel the distance between the observation wells

The distance between the two wells = 1000 m

average linear velocity = 0.242 m/day

Time = distance / speed

Time = (1000 m) / (0.242 m/day)

Time = 4132.23 days

$= 4132.23 days *\frac{24 .hrs}{1.day} = 99173.52, hours$

4 0

2 years ago

Rocks from volcanoes are used by scientists to study the interior of the earth<br><br>TRUE OR FALSE

cupoosta [38]

I think it’s false bc A volcano is not in the interior of the earth

6 0

3 years ago

Read 2 more answers

Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed o

kobusy [5.1K]

Answer:

$\Delta \lambda=14.3\ nm$

Explanation:

It is given that,

The number of lines per unit length, N = 900 slits per cm

Distance between the formed pattern and the grating, l = 2.3 m

n the first-order spectrum, maxima for two different wavelengths are separated on the screen by 2.98 mm, $\Delta Y=2.98\ mm = 0.00298\ m$

Let d is the slit width of the grating,

$d=\dfrac{1}{N}$

$d=\dfrac{1}{900\ cm}$

$d=1.11\times 10^{-5}\ m$

For the first wavelength, the position of maxima is given by :

$y_1=\dfrac{L\lambda_1}{d}$

For the other wavelength, the position of maxima is given by :

$y_2=\dfrac{L\lambda_2}{d}$

So,

$\Delta \lambda=\dfrac{\Delta y d}{l}$

$\Delta \lambda=\dfrac{0.00298\times 1.11\times 10^{-5}}{2.3}$

$\Delta \lambda=1.43\times 10^{-8}\ m$

or

$\Delta \lambda=14.3\ nm$

So, the difference between these wavelengths is 14.3 nm. Hence, this is the required solution.

3 0

2 years ago