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VashaNatasha [74]

2 years ago

13

un cuerpo solido de cierto material,se midio su masa y se encontro un valor de 450 gramos, al medir su volumen este fue de 2890c

cm³. calcular la densidad en el sistema internacional

1 answer:

vaieri [72.5K]2 years ago

7 0

Answer:

my bad i wish i could ansrew

Explanation:im only six

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Two tiny particles having charges 20.0 μC and 8.00 μC are separated by a distance of 20.0 cm What are the magnitude and directio

Alecsey [184]

Answer:

The magnitude and direction of electric field midway between these two charges is $10.8\times10^{5}\ N/C$ along AB.

Explanation:

Given that,

First charge $q_{1}= 20\mu C$

second charge $q_{2}= 8\mu C$

Distance = 20 cm

We need to calculate the electric field

For first charge,

Using formula of electric field

$E_{1}= \dfrac{kq_{1}}{r^2}$

Put the valueinto the formula

$E_{1}=\dfrac{9\times10^{9}\times20\times10^{-6}}{10\times10^{-2}}$

$E_{1}=18\times10^{5}\ N/C$

Direction of electric field along AB

We need to calculate the electric field

For second charge,

Using formula of electric field

$E_{2}= \dfrac{kq_{2}}{r^2}$

Put the valueinto the formula

$E_{2}=\dfrac{9\times10^{9}\times8\times10^{-6}}{10\times10^{-2}}$

$E_{2}=7.2\times10^{5}\ N/C$

Direction of electric field along AO

We need to calculate the net electric field at midpoint

$E_{net}=E_{1}-E_{2}$

$E_{net}=(18-7.2)\times10^{5}\ N/C$

$E_{net}=10.8\times10^{5}\ N/C$

Direction of net electric field along AB

Hence, The magnitude and direction of electric field midway between these two charges is $10.8\times10^{5}\ N/C$ along AB.

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How much can having a lighted candle increase the temperature inside the shelter?

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3 years ago

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3 years ago

A(n) 0.2 kg object is swung in a vertical circular path on a string 0.1 m long. The acceleration of gravity is 9.8 m/s2 . If a c

Leya [2.2K]

Answer:

$T=83.37N$

Explanation:

Since the object is under a circular motion, according to Newton's second law, when the object is at the top of the circle we have:

$\sum F_y: T-mg=F_c$

Where $F_c$ is the centripetal force and is given by:

$F_c=ma_c=m\frac{v^2}{r}$

Replacing and solving for T:

$T=m\frac{v^2}{r}+mg\\T=0.2kg\frac{(6.38\frac{m}{s})^2}{0.1m}+0.2kg(9.8\frac{m}{s^2})\\T=83.37N$

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