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3 years ago

Does distance from modem affect internet speed

1 answer:

5 0

It shouldn't, but to be honest, it kind of does.
The further you are away from the modem,
the less connectivity you have.
Because you can't connect really well, your internet slows down.

So... yes?

You might be interested in

A moving small car has a head-on collision with a large stationary truck 7.3 times the mass of the car. Which statement is true

Trava [24]

The car would speed off twice as fast as the speed of the heavy truck provided the the collision is an elastic collision where there's no or little friction occurring within the scenario. Newton's law proves that an object with a greater mass can move objects of lesser mass at greater distances and speed.

4 0

3 years ago

Read 2 more answers

a cart is loaded with a brick and pulled up at a constant speed along an inclined plane to the height of a seattop. if the mass

Fudgin [204]

Answer:

6 J

Explanation:

Gravitational Potential Energy can be calculated using the formula:

PE = mgh

where m is mass of object in kg

g is the gravitational acceleration of 10 m/s^2

h is the height at which object is thrown in m.

Hence, by substituting in the values, we get:

GPE = (4.0)(10)(0.15)

6 J

Hence, the potential energy is 6 J.

8 0

3 years ago

Read 2 more answers

What is heroism??? plz answer

AnnyKZ [126]

Answer:

heroism is great bravery..

hope it helps !

6 0

3 years ago

if 130N centripetal force is needed to keep a 0.45kg ball that is attached to a string that is 0.7m long to complete 5 full rota

andrezito [222]

Answer:

centripetal acceleration of the ball is 6.9 m/s/s

tangential speed of the ball is 2.2 m/s

Explanation:

As we know that ball complete 5 rotations in 10 seconds

so frequency of rotation of ball is given as

$f = \frac{5}{10} = 0.5 Hz$

now we know that angular frequency is given as

$\omega = 2\pi f$

$\omega = 2(\pi)(0.5)$

$\omega = \pi rad/s$

Now centripetal acceleration is given as

$a_c = \omega^2 R$

$a_c = \pi^2 (0.7)$

$a_c = 6.9 m/s^2$

now the velocity of the ball at this angular frequency is given as

$v = R\omega$

$v = 0.7 (\pi)$

$v = 2.2 m/s$

4 0

3 years ago

How much work did the movers do (horizontally) pushing a 41.0-kg crate 10.6 m across a rough floor without acceleration, if the

VLD [36.1K]

Answer:

The required work done is $2555.448~J$

Explanation:

Consider 'F' is the applied force on the crate and 'f' be the force created by friction. According to the figure if ' $\mu_{k}$ ' be the coefficient of friction, then

$f = \mu_{k} \times N = \mu_{k} \times Mg$

where 'M', 'N' and 'g' are the mass of the crate, the normal force aced upon the block and the acceleration due to gravity respectively.

Since the application of force by the movers does not create any acceleration to the block, we can write

$F = f = \mu_{k} \times M \times g = 0.6 \times 41~Kg~ \times 9.8~m~s^{-2} = 241.08~N$

So the work done (W) in moving the crate by a distance s = 10.6 m is

$W = F \times s = 241.08~N \times 10.6~m = 2555.448 J$

5 0

3 years ago