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Arlecino [84]
2 years ago
7

Unlike acceleration and velocity, speed does not need to specify

Physics
1 answer:
frosja888 [35]2 years ago
4 0

Unlike acceleration and velocity, speed does not need to specify the direction of motion. Speed is a scalar quality.

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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
2 years ago
A scientist carries out an experiment. How could she help other scientists judge the validity of her results?
NNADVOKAT [17]
One way to do it is she could right down the data that she got
4 0
3 years ago
Read 2 more answers
As a battery is used to charge a capacitor, does the overall charge inside the battery get smaller, greater, or stay the same?
irga5000 [103]
The Charge Gets Smaller.
5 0
2 years ago
A 5.0-kg centrifuge takes 95 s to spin up from rest to its final angular speed with constant angular acceleration. A point locat
stellarik [79]

Answer:

(a) 17.37 rad/s^2

(b) 12479

Explanation:

t = 95 s, r = 6 cm = 0.06 m, v = 99 m/s, w0 = 0

w = v / r = 99 / 0.06 = 1650 rad/s

(a) Use first equation of motion for rotational motion

w = w0 + α t

1650 = 0 + α x 95

α = 17.37 rad/s^2

(b) Let θ be the angular displacement

Use third equation of motion for rotational motion

w^2 = w0^2 + 2 α θ

1650^2 = 0 + 2 x 17.37 x θ

θ = 78367.87 rad

number of revolutions, n = θ / 2 π

n = 78367.87 / ( 2 x 3.14)

n = 12478.9 ≈ 12479

4 0
3 years ago
A physics professor is pushed up a ramp inclined upward at 30.0° above the horizontal as she sits in her desk chair, which slide
11111nata11111 [884]

Answer:

V = 3.17 m/s

Explanation:

Given

Mass of the professor m = 85.0 kg

Angle of the ramp θ = 30.0°

Length travelled L = 2.50 m

Force applied F = 600 N

Initial Speed  u = 2.00 m/s

Solution

Work = Change in kinetic energy

F_{net}d = \frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}\\\frac{2F_{net}d }{m} = v^{2} -u^{2}\\ v^{2} =\frac{2F_{net}d }{m} +u^{2}\\ v^{2} =\frac{2(600cos30 - 85\times 9.8 \times sin30) \times 2.5 }{85} +2.00^{2}\\ v^{2} = 10.066\\v = 3..17m/s

7 0
3 years ago
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