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3 years ago

9

What is the force in Newtons when the mass of a truck is 1000Kg and the acceleration of the truck is 3m/s2?

1 answer:

olga nikolaevna [1]3 years ago

5 0

Answer:

thank u for the happiest year of my life life

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3 years ago

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2 years ago

La unidad de calor es la caloría; es posible expresarla en el sistema internacional de medidas??

d1i1m1o1n [39]

Answer:

Yes, the calorie can be expressed in SI units

Explanation:

1 calorie (1 cal) is defined as the amount of heat energy that must be supplied to 1 gram of water in order to raise its temperature by 1 degree Celsius ( $1^{\circ}C$ .

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However, it is possible to convert energy from calories to Joules, and viceversa. In fact, the conversion factor between the two units is:

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So, to convert from calories to Joules we simply multiply by 4.184, while if we want to convert from Joules to calories, we just divide by 4.184.

8 0

3 years ago

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N

Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

$r=\sqrt{(r_{1})^2+(r_{2})^2}$

Put the value into the formula

$r=\sqrt{(9.8)^2+(2.1)^2}$

$r=10.02\ cm$

We need to calculate the force

Using formula of force

$F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}$

Force F₁₂,

$F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}$

$F_{12}=13.33\ N$

$F_{21}=-13.33\ N$

Force F₂₃,

$F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}$

We need to calculate the value of third charge

$F_{net}=F_{21}+F_{23}$

$4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}$

$q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}$

$q_{3}=7.99\times10^{-7}\ C$

$q_{3}=0.8\times10^{-6}\ C$

Hence, The value of third charge is 0.8μC.

4 0

3 years ago