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3 years ago

12

¿Qué velocidad inicial debe impartirse a una masa de 5kg para elevarla a una altura de 10m? ¿Cuál es la energía total en cualqui

era de los puntos de su trayectoria?

1 answer:

Snowcat [4.5K]3 years ago

6 0

Cada 100 cm es de 1 m por lo que tomar los kilos a su vez a cm, a continuación, a su vez a metros

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When the temperature of a certain solid, rectangular object increases by AT, the length of one

Alex

Explanation:

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3 years ago

Which are ways to improve the design of this experiment? Check all that apply.

Arte-miy333 [17]

Answer:

* Experiment with a higher range of materials

* Use a galvanometer.

* Vary in number of coils of the electromagnet

Explanation:

This is an experiment of electricity and magnetism, in general the best way to improve the results are:

* Experiment with a higher range of materials

allowing to know the scope of the initial assumptions

* Use a galvanometer.

The more accurate the readings the error of the derived quantities is the less which will improve the precision of the experiment.

* Vary in number of coils of the electromagnet

Since it allows to have greater magnetic fields and therefore expand the range of measurements

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3 years ago

Read 2 more answers

According to the rayleigh criterion, what is the shortest object we could resolve at the 25.0 cm near point with light of wavele

Blizzard [7]

Answer:

$6.71 *10^{-5}$ rad

Explanation:

∅ = $\frac{1.22*wavelength}{D}$ = $\frac{1.22*550 * 10^{-9} }{25 * 10^{-2} }$

∅ = $6.71 *10^{-5}$ rad

The minimum resolvable angle = $6.71 *10^{-5}$ rad

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2 years ago

An object has a weight of 500 on earth what is the mass of this object

erastova [34]

550! OBVY! lol! ope this helps1

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3 years ago

An electric field of magnitude 2.35 V/m is oriented at an angle of 25.0° with respect to the positive z-direction. Determine the

zzz [600]

Answer:

The magnitude of the electric flux is $3.53\ N-m^2/C$

Explanation:

Given that,

Electric field = 2.35 V/m

Angle = 25.0°

Area $A= 1.65 m^2$

We need to calculate the flux

Using formula of the magnetic flux

$\phi=E\cdot A$

$\phi = EA\cos\theta$

Where,

A = area

E = electric field

Put the value into the formula

$\phi=2.35\times1.65\times\cos 25^{\circ}$

$\phi=2.35\times1.65\times0.91$

$\phi=3.53\ N-m^2/C$

Hence, The magnitude of the electric flux is $3.53\ N-m^2/C$

8 0

3 years ago