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Liula [17]
3 years ago
10

Does distance from modem affect internet speed

Physics
1 answer:
7nadin3 [17]3 years ago
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 massless spring with force constant ????=200N/m hangs from the ceiling. A 2.0-kg block is attached to the free end of the spri
Makovka662 [10]

Answer:

-0.4454 Joules

Explanation:

m = Mass of block = 2 kg

h = Height of extension = 17 cm = x

g = Acceleration due to gravity = 9.81 m/s²

Potential energy of the spring

P=mgh\\\Rightarrow P=2\times 9.81\times 0.17\\\Rightarrow P=3.3354\ J

The kinetic energy of the spring

K=\frac{1}{2}mx^2\\\Rightarrow K=\frac{1}{2}\times 200\times 0.17^2\\\Rightarrow K=2.89\ J

In this system as the potential and kinetic energy is conserved from work energy equivalence we get

W=P-K\\\Rightarrow W=2.89-3.3354\\\Rightarrow W=-0.4454\ J

The work done by friction is -0.4454 Joules

8 0
3 years ago
At what time was the person at a position of 0m?
Anni [7]

Answer: The person was not at a position of "0" at any time. The person started at 10 metres from the starting line. The explanation below shows how to use the standard formula for position when the initial position is not "0". It is noteworthy that the standard expression of the formula for distance travelled does not include a variable (e.g. "d") for distance at the start (when t(time) = 0)

Explanation: At time = 0, the start, the person was at 10m distance from the starting line. Therefore, to use the standard equation, "s + ut + 1/2att (t squared, that is), distance from starting line = 10 + s, that is, total distance from starting line  equals initial position, 10 metres, plus "s" (distance travelled from t = 0 to t = 1) in metres.

for the section of the graph from "0" seconds (t = 0) to 1 second (t = 1):

s = ut + 1/2att

the initial position is 10 metres.

s = 10

the distance is constant from t = 0 to t = 1, therefore the velocity for the whole of that section of graph must be 0.

u = 0

there is no change in the velocity from t = 0 to t= 1, therefore the acceleration for the first section of the graph must be 0.

a = 0

s = ut + 1/2att

  = (0 x 1) + 1/2 (0 x 1 x 1)

  = (0) + 1/2 (0)

  = 0

total distance from starting line (position) equals initial position plus change in position (distance travelled).

at t = 1,

position = 10 + 0

 = 10 metres

The whole of the graph can be analysed using this process for each straight section of the graph separately, adding "s" for each section to the previous total of distance from starting line.

using "d" for initial distance from starting line ( position ), d1 for distance from starting line at t = 1, d2 for distance from starting line at t = 2, etcetera:

section 1, t = 0 to t = 1:

d1 (t=0 to t=1)  =  10 + s (t=0 to t=1).

section 2, t= 1 to t = 2:

d2 (t=0 to t=2) = 10 + s (t=0 to t=1) + s (t=1 to t=2).

etcetera.

8 0
3 years ago
A thin, flat washer is a disk with an outer diameter of 14 cm and a hole in the center with a diameter of 7 cm. The washer has a
Kryger [21]

Answer:

2583.9 N/C

Explanation:

Parameters given:

Outer diameter = 14 cm

Outer radius, R = 7cm = 0.07m

Inner diameter = 7 cm

Inner radius, r = 3.5 cm = 0.035m

Charge of washer = 8 nC = 8 * 10^(-9)C

Distance from washer, z = 33 cm = 0.33m

The electric field due to a washer (hollow disk) is given as:

E = k * σ * 2π [ 1 - z/(√(z² + R²)]

Where σ = charge per unit area

σ = q/π(R² - r²)

σ = 8 * 10^(-9) /(π*(0.07 - 0.035)²)

σ = 2.077 * 10^(-6) C/m²

=> E = 9 * 10^9 * 2.077 * 10^(-6) * 2π * [1 - 0.33/(√(0.33² + 0.07²)]

E = 117.467 * 10^3 * (1 - 0.978)

E = 117.467 * 10^3 * 0.022

E = 2583.9 N/C

6 0
3 years ago
When the starter motor on a car is engaged, there is a 320 A current in the wires between the battery and the motor. Suppose the
Salsk061 [2.6K]

Answer:

0.5

Explanation:

because it is V uwbsusvegwjosnfvehdbuxbdusndgdghqbwbwbbeuehdbdhdhdhdudhdhdudhdnskowoqllqlqlqnebe

8 0
2 years ago
The flywheel of an engine has moment of inertia 2.50 kg m2 about its rotation axis. What constant torque is required to bring it
MrRissso [65]

Answer:

Explanation:

From the question we are told that

   The moment of inertia is  I = 2.50 \ kg \cdot m^2

    The final  angular speed is w_f =  400 rev/min  =  \frac{400 * 2\pi}{60}  = 41.89 \ rad/s

     The time taken is  t =  8.0 s

      The initial angular speed is  w_i  =  0\ rad/s

Generally the average angular acceleration is mathematically represented as

        \alpha  =  \frac{w_f - w_i }{t}

=>     \alpha  =  \frac{41.89}{8}

=>      \alpha  = 5.24 \ rad/s^2

Generally the torque is mathematically represented as

   \tau  =  I  *  \alpha

=>    \tau   =  5.24 *  2.50

=>     \tau   =  13.09 \  N \cdot m

5 0
3 years ago
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