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3 years ago

5

A technician has verified that a recent loss of network connectivity to multiple workstations is due

1 answer:

Natasha_Volkova [10]3 years ago

8 0

Answer:

D. Toner probe

Explanation:

The tone probe allows you to apply a tone signal to a pair of wires or a single conductor, while the trace is done with an amplified probe. When used in conjunction with the amplified probe, the tone generator allows technicians to identify a wire within any beam, at cross connections or at remote ends. You can use twisted pair, coaxial and disconnected AC wiring.

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Any 1 know aboyt atoms plz help atoms question in picture

alina1380 [7]

The first choice is the answer.

5 0

3 years ago

The following objects have velocity and mass as given: Object A: m=5 kg, ⃗v=−11 ^j Object B: m=6 kg, ⃗v=5^ i +8.7 ^ j Object C:

Fofino [41]

Answer:

For object A

m = 5 kg , v= -11 j

For B object

m = 6 kg , v= 5 i +8.7 j

For object C

m = 10 kg , v= -10 i

We know that

Linear momentum P= m v kg.m/s

a) A and C

Momentum in y direction

Py=- 5 x 11 j= - 55 j kg.m/s

Momentum in x direction

Px=- 10 x 10 j= - 100 i kg.m/s

b) B and C

Momentum in y direction

Py=6 x 8.7 j= 52.2 j kg.m/s

Momentum in x direction

Px=( 6 x 5 - 10 x 10 ) i = - 70 i kg.m/s

c) A ,B and C

By using data of a and b

Momentum in y direction

Py= 6 x 8.7 - 5 x 11 j= -2.8 j kg.m/s

Momentum in x direction

Px= 6 x 5 -10 x 10 i = -70 i kg.m/s

4 0

3 years ago

SOMEONE HELP QUICK! (question in the picture)

kondor19780726 [428]

1) 4.8
2) 5.6
3)0.8
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8 0

3 years ago

Consider that a clay model of a tiger has a mass of 0.195 kg and glides on ice at a speed of 0.75 m/s. It hits another clay mode

juin [17]

Answer:

The final velocity after the collision is 0.27 m/s.

Explanation:

Given that,

Mass of tiger, m = 0.195 kg

Initial speed of tiger model, v = 0.75 m/s

Mass of another clay model, m' = 0.335 kg

Initially, second model is at rest, v' = 0

We need to find the final velocity after the collision. It is a case of inelastic collision. Using the conservation of linear momentum as :

$mv+m'v'=(m+m')V\\\\V=\dfrac{mv+m'v'}{(m+m')}\\\\V=\dfrac{0.195\times 0.75}{(0.195 +0.335 )}\\\\V=0.27\ m/s$

So, the final velocity after the collision is 0.27 m/s.

6 0

4 years ago

A 77.0−kg short-track ice skater is racing at a speed of 12.6 m/s when he falls down and slides across the ice into a padded wal

sukhopar [10]

Answer:

-6112.26 J

Explanation:

The initial kinetic energy, $KE_i$ is given by

$KE_i=0.5mv_1^{2$ } where m is the mass of a body and $v_i$ is the initial velocity

The final kinetic energy, $KE_f$ is given by

$KE_f=0.5mv_f^{2}$ where $v_f$ is the final velocity

Change in kinetic energy, $\triangle KE$ is given by

$\triangle KE=KE_f-KE_i=0.5mv_f^{2}-0.5mv_1^{2}=0.5m(v_f^{2}-v_i^{2})$

Since the skater finally comes to rest, the final velocity is zero. Substituting 0 for $v_f$ and 12.6 m/s for $v_i$ and 77 Kg for m we obtain

$\triangle KE=0.5*77*0^{2}-0.5*77*(0^{2}-12.6^{2})=-6112.26 J$

From work energy theorem, work done by a force is equal to the change in kinetic energy hence for this case work done equals <u>-6112.26 J</u>

3 0

3 years ago