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2 years ago

13

When light waves hit ice, most of them bounce off and radiate back into space. Because of this, ice is best described as which k

ind of surface?
A.) reflective

B.) absorbent

C.) dense

D.) rough

2 answers:

viktelen [127]2 years ago

8 0

A. Reflective
The light reflects off the ice

Hope this helped

liberstina [14]2 years ago

7 0

A: Reflective. You can tell this because the light waves are bouncing off of the ice, rather than being absorbed. Hope this helps!

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What are gears and what is a lever ?

blagie [28]

Answer:

agear lever or agear stick is the lever that you use to change gear in acar or other vehicle

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2 years ago

You drop your cell phone. Prior to hitting the ground, the phone's kinetic energy will ________ and its potential energy will __

SVETLANKA909090 [29]

Answer:

The kinetic energy of the phone would increase. The gravitational potential energy of the phone would decrease.

Explanation:

The kinetic energy ${\rm KE}$ of an object is proportional to the square of the speed of that object. If air resistance is negligible, the phone would accelerate under gravitational pull and speed up. Hence, the kinetic energy of the phone would increase.

The gravitational field near the surface of the earth is approximately constant. Hence, the gravitational potential energy ${\rm GPE}$ of the phone would be proportional to its height. As the phone approaches the ground, the height of the phone becomes lower and the gravitational potential energy of the phone would decrease.

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1 year ago

The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the t

AVprozaik [17]

Answer:

v=115 m/s

or

v=414 km/h

Explanation:

Given data

$A_{area}=0.140m^{2}\\ p_{air}=1.21 kg/m^{3}\\ m_{mass}=80kg$

To find

Terminal velocity (in meters per second and kilometers per hour)

Solution

At terminal speed the weight equal the drag force

$mg=1/2*C*p_{air}*v^{2}*A_{area}\\ v=\sqrt{\frac{2*m*g}{C**p_{air}*A_{area}} }\\ Where C=0.7\\v=\sqrt{\frac{2*9.8*80}{1.21*0.14*0.7} }\\ v=115m/s$

For speed in km/h(kilometers per hour)

To convert m/s to km/h you need to multiply the speed value by 3.6

$v=(115*3.6)km/h\\v=414km/h$

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2 years ago

A square loop of side 7 cm is placed with the nearest side 2 cm from a long wire carrying a current that varies with time at a c

ioda

Answer:

Explanation:

side of the square loop, a = 7 cm

distance of the nearest side from long wire, r = 2 cm = 0.02 m

di/dt = 9 A/s

Integrate on both the sides

$\int _{0}^{i}di =9\int _{0}^{t}dt$

i = 9t

(a) The magnetic field due to the current carrying wire at a distance r is given by

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

$B = \frac{\mu_{0}\times 9t}{2\pi r}$

(b)

Magnetic flux,

$\phi=\int B\times a dr$

$\phi=\int \frac{\mu_{0}\times 9t}{2\pi r}\times a dr$

$\phi=\frac{\mu_{0}\times 9t\times a}{2\pi}\times ln\left ( \frac{2 + 7}{2} \right )$

$\phi=\frac{\mu_{0}\times 9t\times 0.07}{2\pi}\times ln(4.5)$

$\phi = 1.89 \times 10^{-7}t$

(c)

R = 3 ohm

$e = -\frac{d\phi}{dt}$

magnitude of voltage is

e = 1.89 x 10^-7 V

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8 0

2 years ago

A small logo is embedded in a thick block of crown glass (n = 1.52), 4.70 cm beneath the top surface of the glass. The block is

harkovskaia [24]

The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

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Where,

$d_g =$ Depth of glass

$n_w =$ Refraction index of water

$n_g =$ Refraction index of glass

$n_{air} =$ Refraction index of air

$d_w =$ Depth of water

I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to

$d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})$

$d'w = (1.7cm) (\frac{1}{1.33})+(4.2cm)(\frac{1}{1.52})$

$d'w = 4.041cm$

Therefore the distance below the upper surface of the water that appears to be the logo is 4.041cm

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2 years ago