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2 years ago

If the electric field is 100N/C at a distance of 50 cm from a point charge , what is the value of ?.

Physics

1 answer:

White raven [17]2 years ago

5 0

Answer:

$e \: = \: \frac{kq}{{r}^{2} }$

then q = e r² / k

then q = 100 * (50 ×10^-2)² / 9×10^9 = 2.777777778×10^-9 C

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What mass has a rest energy of 100J?

alexandr1967 [171]

Answer:

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Explanation:

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2 years ago

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A 60 kg person sits in a chair. How much does the earth pull on the person? (Acceleration due to gravity is -10m/s2) F = mxa → F

MA_775_DIABLO [31]

Answer:

<em>600N(downwards)</em>

Explanations

<em>600N(downwards)</em>

Mas of the person = 60kg

Acceleration due to gravity = -10m/s²

To get the earths pull on the person, we will use the Newton second law of motion;

Force = mass * acceleration;

Force = 60 * -10

Force- -600N

<em>Hence the earth gravitational pull on the person is 600N(downwards). It is downwards due to the negative sign.</em>

7 0

2 years ago

A 5-g lead bullet traveling in 20°C air at 300 m/s strikes a flat steel plate and stops.

densk [106]

To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:

$KE = \frac{1}{2} mv^2$

Where,

m = Mass

v = Velocity

Replacing we have that the Total Kinetic Energy is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (5*10^{-3})(300)^2$

$KE = 225J$

On the other hand the required Energy to heat up t melting point is

$Q_1 = mC_p \Delta T$

$Q_2 = L_f m$

Where,

m = Mass

$C_p =$ Specific Heat

$\Delta T =$ Change at temperature

$L_f =$ Latent heat of fussion

Heat required to heat up to melting point,

$Q = Q_1+Q_2$

$Q = mC_p \Delta T+L_f m$

$Q = 5*0.128*(327-20) + 5*24.7$

$Q = 310J$

The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.

4 0

3 years ago