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alukav5142 [94]
3 years ago
5

Alguien que sepa de electromecánica porfavor

Physics
1 answer:
kkurt [141]3 years ago
8 0
LAPA HDIDOSHSUWJWVWIHDHDOSSHSVWIME
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High blood levels of LDL cholesterol reduces your risk of developing heart
SVEN [57.7K]

Answer:

false

Explanation:

3 0
3 years ago
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Por una resistencia de 10 Ω fluyen 5A. ¿Cuál será la diferencia de potencial que se le debe aplicar a la resistencia?
ra1l [238]

Answer:

V = 50 volts

Explanation:

Given that,

Resistance, R = 10 ohms

Current, I = 5 A

We need to find the potential difference across the circuit. We know that,

V = IR

Put all the values,

V = 5 × 10

V = 50 volts

Hence, the potential difference is equal to 50 volts.

8 0
3 years ago
Help please I have to turn this in tonight!!
inna [77]

Answer:

True

Explanation:

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3 years ago
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How does the input distance of a single fixed pulley compare to the out- put distance?
ololo11 [35]

A pulley is another sort of basic machine in the lever family. We may have utilized a pulley to lift things, for example, a banner on a flagpole.

<u>Explanation:</u>

The point in a fixed pulley resembles the support of a lever. The remainder of the pulley behaves like the fixed arm of a first-class lever, since it rotates around a point. The distance from the fulcrum is the equivalent on the two sides of a fixed pulley. A fixed pulley has a mechanical advantage of one. Hence, a fixed pulley doesn't increase the force.

It essentially alters the direction of the force. A moveable pulley or a mix of pulleys can deliver a mechanical advantage of more than one. Moveable pulleys are appended to the item being moved. Fixed and moveable pulleys can be consolidated into a solitary unit to create a greater mechanical advantage.

4 0
3 years ago
A convex security mirror has a radius of curvature of 12.0 cm. What is the magnification of a pare 3.0 m from the mirror?
Makovka662 [10]

Answer:

magnification will be -0.025

Explanation:

We have given the radius of curvature = 12 cm

And object distance = 3 m

So focal length f=\frac{R}{2}=\frac{12}{2}=6cm

Now for mirror we know that \frac{1}{f}=\frac{1}{u}+\frac{1}{v}

So \frac{1}{0.06}=\frac{1}{3}+\frac{1}{v}

16.66-0.333=\frac{1}{v}

v = 0.750 m

Now magnification of the mirror is m=\frac{-v}{u}=\frac{-0.750}{3}=-0.025

5 0
3 years ago
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