A cruise ship made a trip to Guam and back. The trip there took 12 hours and the trip

Alika [10]

The dot would be the earthquake epicenter. That is where the earthquake originated.

6 0

4 years ago

During electrical storms, a bolt of lightning can transfer 10 C of charge in 2.0 µs (the amount of time can vary considerably).

nydimaria [60]

Answer:

Explanation:

charge, q = 10 C

time, t = 2 micro second

Current, i = q / t

i = 10 / (2 x 10^-6) = 5 x 10^6 A

(a)

distance, d = 1 m

the formula for the magnetic field is given by

$B = \frac{\mu_{0}}{4\pi}\frac{2i}{d}$

$B = 10^{-7}\frac{2\times 5\times 10^{6}}{1}$

B = 1 Tesla

Now the distance is d' = 1 km = 1000 m

$B' = \frac{\mu_{0}}{4\pi}\frac{2i}{d}$

$B' = 10^{-7}\frac{2\times 5\times 10^{6}}{1000}$

B' = 0.001 Tesla

(b) The magnetic field of earth is Bo = 3 x 10^-5 tesla

B / Bo = 3.3 x 10^4

B'/Bo = 33.3

4 0

3 years ago

You drop a steel ball bearing, with a radius of 2.40 mm, into a beaker of honey. Note that honey has a viscosity of 6.00 Pa/s an

Stells [14]

Answer:

The “terminal speed” of the ball bearing is 5.609 m/s

Explanation:

Radius of the steel ball R = 2.40 mm

Viscosity of honey η = 6.0 Pa/s

$\text { Viscosity has Density } \sigma=1360 \mathrm{kg} / \mathrm{m}^{3}$

$\text { Steel has a density } \rho=7800 \mathrm{kg} / \mathrm{m}^{3}$

$\left.\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)$

While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

$V_{t}=\frac{2 \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{9 \eta}$

Substitute the given values to find "terminal speed"

$\mathrm{V}_{\mathrm{t}}=\frac{2 \times 0.0024^{2}(7800-1360) 9.8}{9 \times 6}$

$\mathrm{V}_{\mathrm{t}}=\frac{0.0048 \times 6440 \times 9.8}{54}$

$\mathrm{V}_{\mathrm{t}}=\frac{302.9376}{54}$

$\mathrm{V}_{\mathrm{t}}=5.609 \mathrm{m} / \mathrm{s}$

The “terminal speed” of the ball bearing is 5.609 m/s

7 0

3 years ago

Calculate the critical angle for light going from Glycerine to air.

Georgia [21]

The refractive index for glycerine is $n_g=1.473$ , while for air it is $n_a = 1.00$ .

When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:
$\theta_c = \arcsin ( \frac{n_2}{n_1} )$
where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
$\theta_c = \arcsin ( \frac{1.00}{1.473} )=42.8^{\circ}$

6 0

3 years ago

Which of the following distinguishes electromagnetic waves from mechanical waves

Ivan

Answer:

Explanation:

1. Mechanical waves require material medium for their propagation while electromagnetic waves do not require material medium for their propagation.

2. Mechanical waves can either be transverse or longitudinal while electromagnetic waves are transverse.(Transverse waves are waves in which the vibration of the particules of the medium is perpendicular to the direction of the motion of wave. E.g water waves, waves of a plucked string and all electromagnetic waves RIVUXG . Longitudinal waves are waves whose vibration are parallel to the direction of the motion of the medium e.g waves in strings, sound waves.e.t.c)

6 0

3 years ago