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2 years ago

Cell phones are technology that is popular throughout the world. Would this technology be of use in rural areas of the world

2 answers:

6 0

Unless there is service it wifi, no. To access most features on a cellphone you need wifi or service. These features can include communicating with others or watching a video. Cellphones are practically useless without internet connection.

katovenus [111]2 years ago

6 0

Answer:

Explanation:

Cell phone is not only a device to set up a call, but it also provides various features now a days which keeps us connected to the world. It even help one to learn by providing various means of applications.

This technology is mostly used in urban area as they are equipped with wifi connectivity and antennas. If rural areas are provided with such technologies and advancement, then cell phones are well suited for such areas.

The advancement in technology in terms of mobile is vast and connects people through out the world.

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Coming to the bottom of a mountain, a skier moving with speed v collides with a barrier and is brought to a stop in an amount of

IRISSAK [1]

Answer:

Explanation:

Impulse of a force is measured by force x time or F X t

Impulse also equals change in momentum or

F x t = m v₂ - m v₁

The given case is as follows

in the first case

F x t = mv - o = mv

F = mv / t

in the second case

F₁ x 4 t = mv

F₁ = 1/4 x mv /t

F₁ = F / 4

option a) is correct .

iii )

In the last case

F₂ X t = m v/2 -0

F₂ = 1/2 x mv / t

= 1/2 x F

F₂ = F/2

Option e ) is correct.

8 0

2 years ago

The atoms in a solid move about freely

ivolga24 [154]

No, not exactly. They jiggle and tremble and vibrate a lot, but
they always basically stay in very nearly the same place.

It's like if you're allowed to go anywhere you want in your jail cell,
you wouldn't exactly call that "moving about freely".

6 0

3 years ago

What is the approximate terminal velocity of a sky diver before the parachute opens

anzhelika [568]

Answer:

The approximate terminal velocity of a sky diver before the parachute opens is 320 km/h.

Explanation:

The terminal velocity is the maximum magnitude of velocity that is attained by the diver when he or she falls in the air.
The terminal velocity of the person diving in air before opening parachute is 320 km/h that means the velocity when the person is experiencing free fall is 320 km/h.
During terminal velocity, we can represent mathematical equation as;

Buoyancy force + drag force = Gravity

6 0

3 years ago

Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an

ddd [48]

Answer:

Al's mass is 102.92 kg

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

$F_{net} = 0$

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

$F_{net} = \frac{dp_{horizontal}}{dt} = 0$

$p_{horizontal_i }= p_{horizontal_f}$

where the suffix i and f means initial and final respectively.

The initial momentum will be:

$p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}$

But, as they are at rest, initially

$p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0$

$p_{horizontal_i} = 0$

So, this means:

$p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0$

We know that the have an combined mass of 195 kg:

$m_{total} = m_{Al} + m_{Jo} = 195 \ kg$ .

so:

$m_{Jo} = 195 \ kg - m_{Al}$ .

$m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0$

$m_{Al} \ v_{Al_f} - m_{Al} \ v_{Jo_f}= - 195 \ kg \ v_{Jo_f}$

$m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}$

$m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } { v_{Al_f} - v_{Jo_f} }$

$m_{Al} = \frac{195 \ kg \ v_{Jo_f} } { v_{Jo_f} - v_{Al_f} }$

Now, we can use the values:

$v_{Al_f}= 10.2 \frac{m}{s}$

$v_{Jo_f}= - 11.4 \frac{m}{s}$

where the minus sign appears as they are moving at opposite directions

$m_{Al} = \frac{195 \ kg ( - 11.4 \frac{m}{s} ) } { (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }$

$m_{Al} = 102.92 \ kg$

and this is the Al's mass.

5 0

2 years ago

A 0.50-kg red cart is moving rightward with a speed of 40 cm/s when it collides

lukranit [14]

The momentum of the red cart before the collision is 0.2 kgm/s and the blue cart is 0.

The momentum of the red cart after the collision is 0.05 kgm/s and the blue cart is 0.15 kgm/s.

The change in momentum of the system of the carts is 0.

<h3>Initial momentum of the carts before collision</h3>

The momentum of the carts before the collision is calculated as follows;

P(red) = 0.5 kg x 0.4 m/s = 0.2 kgm/s

P(blue) = 1.5 x 0 = 0

<h3>Momentum of the carts after collision</h3>

The momentum of the carts after the collision is calculated as follows;

P(red) = 0.5 x 0.1 = 0.05 kgm/s

P(blue) = 1.5 0.1 = 0.15 kgm/s

<h3>Change in momentum of the carts</h3>

$\Delta P = P_f - P_i$

ΔP = (0.05 + 0.15) - (0.2)

ΔP = 0

Learn more about momentum here: brainly.com/question/7538238

8 0

1 year ago