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Alisiya [41]
3 years ago
11

Which circuit component is usually a device that uses electrical energy to to perform a certain function? A. wire B. power sourc

e C. load
Physics
2 answers:
gtnhenbr [62]3 years ago
4 0

Answer:

B

Explanation:

Just took the test

Ksenya-84 [330]3 years ago
3 0

Answer:

A

Explanation:

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A 100-g toy car is propelled by a compressed spring that starts it moving. the car follows a curved track. what is the final spe
wel

Answer:

0.687 m/s

Explanation:

Initial energy = final energy

1/2 mu² = mgh + 1/2 mv²

1/2 u² = gh + 1/2 v²

Given u = 2.00 m/s, g = 9.8 m/s², and h = 0.180 m:

1/2 (2.00 m/s)² = (9.8 m/s²) (0.180 m) + 1/2 v²

v = 0.687 m/s

5 0
3 years ago
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Which of the following simple machines is not properly identified? A. Scissors: lever B. Pizza cutter: wedge C. Screw: wheel and
GrogVix [38]

Pretty sure the answer is   C. Screw: wheel and axle

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3 years ago
Pls answer asap!! I need help on this question!
evablogger [386]

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8 0
2 years ago
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An ice skater has a moment of inertia of 5.0 kgm2 when her arms are outstretched. at this time she is spinning at 3.0 revolution
givi [52]
With arms outstretched,
Moment of inertia is I = 5.0 kg-m².
Rotational speed is ω = (3 rev/s)*(2π rad/rev) = 6π rad/s
The torque required is
T = Iω = (5.0 kg-m²)*(6π rad/s) = 30π 

Assume that the same torque drives the rotational motion at a moment of inertia of 2.0 kg-m².
If u = new rotational speed (rad/s), then
T = 2u = 30π
u = 15π rad/s
   = (15π rad/s)*(1 rev/2π rad)
   = 7.5 rev/s

Answer: 7.5  revolutions per second.

7 0
3 years ago
A meter stick is free to rotate about an axis through one of its end. Find the force F needed to balance this meter stick if the
Mumz [18]

Answer:

Explanation:

Component of force perpendicular to stick

= F Sin 60°

=√3 / 2 F.

Taking torque about the other end

= √3 / 2 F x 1 Nm

Weight of stick = 60 gm

= 60 x 10⁻³ kg

= 60 x 10⁻³ x 9.8 N

= .588 N

This weight will act from the middle point of stick so torque about the

other end

= .588 x 1 Nm

Balancing these two torques we have

.588 = √3 /2 F

F=\frac{2\times0.588}{\sqrt{3} }

F = 0.679 N

6 0
3 years ago
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