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maxonik [38]
3 years ago
10

CAN YOU HELP, JUST NUMBER 3​

Physics
2 answers:
bonufazy [111]3 years ago
6 0

Answer:

its b option balance hi your last question

Volgvan3 years ago
6 0

Answer:

c. thermometer ....................

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To reduce inductive reactance, what devices are normally placed on transmission or distribution lines?
UkoKoshka [18]

Answer : Capacitors

Explanation : Capacitors are normally  placed on transmission or distribution lines when to reduce inductive reactance.

This is because it enhances electromechanical and voltage stability , limit voltage dips at network nodes and reduces the power loss.

So, we can say that inductive reactance normally replace by the capacitors.



4 0
3 years ago
A projectile fired from a gun has initial horizontal and vertical components of velocity equal to 60 m/s and 80 m/s, respectivel
DerKrebs [107]

This question involves the concepts of projectile motion and launch speed.

(a) The initial launch speed of the projectile is "100 m/s".

(b) The launch angle of the projectile is "53.13°".

<h3>(a) LAUNCH SPEED</h3>

A projectile motion is a motion that takes place on both x and y axes, simultaneously. In this motion the initial launch speed is given by the following formula:

v_o=\sqrt{v_{ox}^2+v_{oy}^2}

where,

  • v_o = initial launch speed = ?
  • v_{ox} = horizontal component of initial launch speed = 60 m/s
  • v_{oy} = vertical component of initial launch speed = 80 m/s

Therefore,

v_o = \sqrt{(60\ m/s)^2+(80\ m/s)^2}\\\\v_o = 100 m/s

<h3>(b) LAUNCH ANGLE</h3>

Launch angle is given by th following formula:

\theta = tan^{-1}(\frac{v_{oy}}{v_{ox}})=tan^{-1}(\frac{80\ m/s}{60\ m/s})\\\\\theta=53.13^o

Learn more about the projectile motion here:

brainly.com/question/11049671

6 0
3 years ago
Which of these charges is experiencing the electric field with the largest magnitude? A 2C charge acted on by a 4 N electric for
Pavlova-9 [17]

Answer:

The highest electric field is experienced by a 2 C charge acted on by a 6 N electric force. Its magnitude is 3 N.

Explanation:

The formula for electric field is given as:

E = F/q

where,

E = Electric field

F = Electric Force

q = Charge Experiencing Force

Now, we apply this formula to all the cases given in question.

A) <u>A 2C charge acted on by a 4 N electric force</u>

F = 4 N

q = 2 C

Therefore,

E = 4 N/2 C = 2 N/C

B) <u>A 3 C charge acted on by a 5 N electric force</u>

F = 5 N

q = 3 C

Therefore,

E = 5 N/3 C = 1.67 N/C

C) <u>A 4 C charge acted on by a 6 N electric force</u>

F = 6 N

q = 4 C

Therefore,

E = 6 N/4 C = 1.5 N/C

D) <u>A 2 C charge acted on by a 6 N electric force</u>

F = 6 N

q = 2 C

Therefore,

E = 6 N/2 C = 3 N/C

E) <u>A 3 C charge acted on by a 3 N electric force</u>

F = 3 N

q = 3 C

Therefore,

E = 3 N/3 C = 1 N/C

F) <u>A 4 C charge acted on by a 2 N electric force</u>

F = 2 N

q = 4 C

Therefore,

E = 2 N/4 C = 0.5 N/C

The highest field is 3 N, which is found in part D.

<u>A 2 C charge acted on by a 6 N electric force</u>

3 0
4 years ago
How to show your work for 0.04% of 245
Harlamova29_29 [7]
0.04*245 is the work
7 0
3 years ago
Read 2 more answers
5. Read the final paragraph on the opposite page. What<br> is the speed of the car in km/h?
Vera_Pavlovna [14]

Answer:

yhjkhjjoknhtyhhjhhynn

jjjjjtsstujbfryjbctybggn

8 0
3 years ago
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