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Kitty [74]
3 years ago
14

Why is it important to be able to trace the pole connection on a meter back to the same type of pole at the electrical source?

Physics
1 answer:
lozanna [386]3 years ago
6 0

Answer:

Explanation:

A grounded wire is sometimes strung along the tops of the towers to provide lightning protection.

In areas where the neutral is grounded or earthed, it is essential to endure that the neutral and the live or hot wires are not confused for each other.

When this happens, the fuses on the transformer will not operate unless the fault is very close to the transformer. The fuses in the consumer's intake box, will not operate.

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A juggler throws two balls to the same height so that one is at the halfway point going up when the other is at the halfway poin
horrorfan [7]

We know that in a projectile motion the acceleration always points down and it has magnitude of 9.8 m/s^2 (the acceleration due to gravity). In this kind of motion we know that at equal heights the velocty has the same magnitude but opposite direction.

Since the juggler throws the ball to the same height this means that the balls will follow the same projectile motion then the properties mentioned above are satisfied.

Therefore we conclude that at that point:

Their accelerations are equal but their velocities are equal and opposite.

4 0
1 year ago
What is the term that describes<br> the material through which a wave travels?
Genrish500 [490]

Answer:

A WAVE IS ANY DISTURBANCE THAT TRANSMITS ENERGY THROUGH MATTER OR SPACE. ... HOWEVER, THE MATERIAL THROUGH WHICH THE WAVE TRAVELS DOES NOT MOVE WITH THE ENERGY. A MEDIUM IS A SUBSTANCE THROUGH WHICH A WAVE CAN TRAVEL. A MEDIUM CAN BE A SOLID, A LIQUID, OR A GAS.

i hope this is helpful :)

Explanation:

3 0
3 years ago
A car speeds up from 14 meters per second to 21 meters per second in 6 seconds. Whats the acceleration and the distance passed w
Solnce55 [7]

Answer:

a = 1.666... m/s²

Explanation:

a = v2 - v1 / t2 - t1

a = 21m/s - 14m/s / 6s - 0s

a = 7m/s / 6s

a = 1.666... m/s²

7 0
2 years ago
A delivery truck leaves a warehouse and travels 2.60 km north. The truck makes a left turn and travels 1.25 km west before makin
yulyashka [42]

Answer:

4.19 km and 107.35 degrees north of east

Explanation:

So in the end, the truck is (2.6 + 1.4 = 4km) north and 1.25 km west from the warehouse. We can use the Pythagorean formula to calculate the magnitude and direction α of the truck displacement from the warehouse:

s = \sqrt{s_n^2 + s_w^2} = \sqrt{4^2 + 1.25^2} = \sqrt{16 + 1.5625} = \sqrt{17.5625} = 4.19 km

tan\alpha = \frac{s_n}{s_w} = \frac{4}{1.25} = 3.2

\alpha = tan^{-1}3.2 = 1.27 rad \approx 72.65 degrees north or west or (180 - 72.65) = 107.35 degrees north of east

3 0
3 years ago
Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m
Phoenix [80]

Answer:

a. t_1=12.5\ s

b. a_2=-13.61\ m.s^{-2}  must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. t_2=2.5714\ s is the time taken to stop after braking

Explanation:

Given:

  • speed of leading car, u_1=25\ m.s^{-1}
  • speed of lagging car, u_{2}=35\ m.s^{-1}
  • distance between the cars, \Delta s=45\ m
  • deceleration of the leading car after braking, a_1=-2\ m.s^{-2}

a.

Time taken by the car to stop:

v_1=u_1+a_1.t_1

where:

v_1=0 , final velocity after braking

t_1= time taken

0=25-2\times t_1

t_1=12.5\ s

b.

using the eq. of motion for the given condition:

v_2^2=u_2^2+2.a_2.\Delta s

where:

v_2= final velocity of the chasing car after braking = 0

a_2= acceleration of the chasing car after braking

0^2=35^2+2\times a_2\times 45

a_2=-13.61\ m.s^{-2} must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

v_2=u_2+a_2.t_2

0=35-13.61\times t_2

t_2=2.5714\ s  is the time taken to stop after braking

7 0
3 years ago
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