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3 years ago

what is the mass of an object that is experiencing a net force of 200N and an acceleration of 500m/s2

Physics

1 answer:

Bas_tet [7]3 years ago

4 0

Answer:

F=200N

a=500m/s2

Mass=?

Explanation:

F=ma

200=m*500

200/500=m

Mass=0.4kg

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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 = 14.413

natima [27]

Answer:

The value of charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force $F=14.413\ N$

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

$r=\sqrt{x_{2}^2+y_{2}^2}$

Put the value into the formula

$r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}$

$r=0.0876\ m$

We need to calculate the magnitude of the charge q₃

Using formula of net force

$F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})$

Put the value into the formula

$14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})$

$(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}$

$\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}$

$q_{3}=0.0210909\times(0.0438)^2$

$q_{3}=40.46\times10^{-6}\ C$

$q_{3}=40.46\ \mu C$

Hence, The value of charge q₃ is 40.46 μC.

5 0

2 years ago

The speed that a tsunami can travel is modeled by the equation , where s is the speed in kilometers per hour and d is the averag

CaHeK987 [17]

The speed of tsunami is a.0.32 km.

Steps involved :

The equation s = 356d models the maximum speed that a tsunami can move at. It reads as follows: s = 200 km/h d =?

Let's now change s to s in the equation to determine d: s = 356√d 200 = 356√d √d = 200 ÷ 356 √d = 0.562 Let's square the equation now by squaring both sides: (√d)² = (0.562) ² d = (0.562)² = 0.316 ≈ 0.32

As a result, 0.32 km is roughly the depth (d) of water for a tsunami moving at 200 km/h.

To learn more about tsunami refer : brainly.com/question/11687903

#SPJ4

6 0

1 year ago

A convection cell is BEST described as ___.

Helga [31]

D, a loop of moving air

6 0

2 years ago

Speed equals distance divided by? a. time b. velocity c. size d. motion

lukranit [14]

I believe it's A. time.

7 0

2 years ago

A ray of light exits from a material with a refractive index of 1.75, traveling into air. The angle of refraction is 25°. What w

Viktor [21]

Answer:

θi = 47.7°

Explanation:

The formula for the refractive index is as follows:

$n = \frac{Sin\theta_i}{Sin\theta_r}$

where,

n = refractive index = 1.75

θi = angle of incidence = ?

θr = angle of refraction = 25°

Therefore,

$1.75 = \frac{Sin\ \theta_i}{Sin\ 25^o} \\\\(1.75)(Sin\ \ 25^o) = Sin\ \theta_i\\\\\theta_i = Sin^{-1}(0.739)$

θi = 47.7°

3 0

2 years ago

what is the mass of an object that is experiencing a net force of 200N and an acceleration of 500m/s2​

what is the mass of an object that is experiencing a net force of 200N and an acceleration of 500m/s2