All categories

3 years ago

How is high speed internet transmitted around the world?

Physics

1 answer:

Degger [83]3 years ago

7 0

Answer:

How is high-speed internet typically transmitted around the world? over telephone wires b. through fiber optic cables c. beamed wirelessly to homes d via satellites

Explanation:

You might be interested in

Help I need a answer to this

natta225 [31]

Answer:

1700 Joules

Explanation:

Work=force x distance

Force = 170 kg

Distance= 10 Meters

170 x 10 = 1700 Joules of work

3 0

3 years ago

A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the lens strength (a.k.a, lens p

Kipish [7]

Answer:

20.0 cm

Explanation:

Here is the complete question

The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?

Solution

Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.

Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.

Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m

Now, P' = 1/u + 1/v

1/u = P'- 1/v

1/u = 55.0 D - 1/0.02 m

1/u = 55.0 m⁻¹ - 1/0.02 m

1/u = 55.0 m⁻¹ - 50.0 m⁻¹

1/u = 5.0 m⁻¹

u = 1/5.0 m⁻¹

u = 0.2 m

u = 20 cm

So, at 55.0 dioptres, the closet object she can see is 20 cm from her eye.

8 0

2 years ago

a foul ball is hit into the stands at a baseball game. the ball rises to a height of 38 meters and is caught on its way down by

lisov135 [29]

The velocity of the ball when it was caught is 12.52 m/s.

<em>"Your question is not complete it seems to be missing the following, information"</em>,

find the velocity of the ball when it was caught.

The given parameters;

maximum height above the ground reached by the ball, H = 38 m

height above the ground where the ball was caught, h = 30 m

The height traveled by the ball when it was caught is calculated as follows;

y = H - h

y = 38 - 30 = 8 m

The velocity of the ball when it was caught is calculated as;

$v_f^2 = v_0 + 2gh\\\\v_f^2 = 0 + (2\times 9.8 \times 8)\\\\v_f^2 = 156.8\\\\v_f = \sqrt{156.8} \\\\v_f = 12.52 \ m/s$

Thus, the velocity of the ball when it was caught is 12.52 m/s.

Learn more here: brainly.com/question/14582703

4 0

3 years ago

Because of your success in physics class you are selected for an internship at a prestigious bicycle company in its research and

tiny-mole [99]

To develop the problem it is necessary to apply the equations related to the moment of inertia.

The given values can be defined as,

$M = 1.0 kg$

$r = 0.5 m$

$m = 10 g$

$I = 0.280 kg.m^2$

According to the definition of the moment of inertia applied to the exercise we can arrive at the equation that,

$I = I_{rim} + n * I_{spoke}$

Where n is the number of spokes necessary to construct the wheel.

$I_{rim} = M*r^2 = 1.0 * 0.5^2$

$I_{spoke} = \frac{1}{3} * m * r^2 = \frac{1}{3}* 10 * 10^-3 * 0.5^2$

Replacing the values at the general equation we have,

$0.280 = 1.0 * 0.5^2 + n * (1/3 * 10 * 10^-3 * 0.5^2 )$

Solving for n,

$n = 36$

Therefore the number of spokes necessary to construct the wheel is 36

PART B) The mass of the wheel is given by the sum of all masses and the total spokes, then

$M_w= M + n*m$

$M_w = 1.0 + 36* 10 * 10^{-3} Kg$

$M_w = 1.36 Kg$

Therefore the mass of the wheel must be of 1.36Kg

4 0

3 years ago

This is a question on my physics test :)

Licemer1 [7]

Answer:

119.6 J/Kg°C

Explanation:

Data obtained from the question include:

Mass of substance (ms) = 170 g

Initial temperature of substance (Ts) = 120 °C

Volume of water = 200 mL

Initial temperature of water (Ts) = 10 °C

Temperature of the mixture (T2) = 12.6 °C

Density of water = 1 g/mL

Specific heat capacity of water (Cw) = 4200J/Kg°C

Specific heat capacity of substance (Cs) =..?

Next, we shall determine the mass of water. This can be obtained as follow:

Volume of water = 200 mL

Density of water = 1 g/mL

Mass of water =..?

Density = mass /volume

1 = mass /200

Cross multiply

Mass of water = 1 x 200

Mass of water = 200 g

Convert 200 g of water to Kg

Mass of water = 200/1000 0.2 Kg

Mass of water = 0.2 Kg

Now, we obtained the specific heat capacity of the substance using the following formula:

MwCw(T2 – Tw) + MsCs(T2 – Ts) = 0

Mass of water = 0.2 Kg

Initial temperature of water (Ts) = 10 °C

Specific heat capacity of water (Cw) = 4200J/Kg°C

Temperature of the mixture (T2) = 12.6 °C

Mass of substance (ms) = 170 g = 170/1000 = 0.17 Kg

Initial temperature of substance (Ts) = 120 °C

Specific heat capacity of substance (Cs) =..?

MwCw(T2 – Tw) + MsCs(T2 – Ts) = 0

0.2× 4200(12.6 – 10) + 0.17×Cs×(12.6 – 120) = 0

840(2.6) + 0.17Cs(– 107.4) = 0

2184 – 18.258Cs = 0

Rearrange

2184 = 18.258Cs

Divide both side by the coefficient of Cs i.e 18258

Cs = 2184/18.258

Cs = 119.6 J/Kg°C

Therefore, the specific heat capacity of the substance is 119.6 J/Kg°C

7 0

3 years ago