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vladimir1956 [14]
3 years ago
5

Two horses are pulling a barge with a mass 2000 kg along a canal. The cable connected to the first horse makes an angle of θ1= 4

0 o with respect to the direction of the canal, while the cable connected to the second horse makes an angle of θ2= -20o . Find the acceleration of the barge (magnitude and direction), starting at rest, if each horse exerts a force of magnitude 600 N. ignore forces of resistance.
Physics
1 answer:
Shkiper50 [21]3 years ago
8 0

Answer:

a = 0.5195 m/s²

θ = 9.997º ≈ 10º

Explanation:

We apply Newton's 2nd Law as follows:

∑ Fx = m*ax

∑ Fy = m*ay

Then we have

∑ Fx = F₁x + F₂x = m*ax    ⇒ 600*Cos 40º + 600*Cos (-20º) = 2000*ax

⇒   ax = 0.5117 m/s²

∑ Fy = F₁y + F₂y = m*ay    ⇒ 600*Sin 40º + 600*Sin (-20º) = 2000*ay

⇒   ay = 0.0902 m/s²

the magnitude of the acceleration of the barge is

a = √(ax² + ay²) = √((0.5117 m/s²)² + (0.0902 m/s²))= 0.5196 m/s²

and the direction is

θ = Arctan (ay / ax) = Arctan (0.0902 / 0.5117) = 9.997º ≈ 10º

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3 years ago
A particle (charge = 40 μC) moves directly toward a second particle (charge = 80 μC) which is held in a fixed position. At an in
lisov135 [29]

Answer:

The distance separating the two particles when the moving particle is momentarily stopped is 0.947 m

Explanation:

Given that,

First charge = 40 μC

Second charge = 80 μC

Distance between the two particles = 2.0 m

Kinetic energy = 16 J

We need to calculate the distance separating the two particles when the moving particle is momentarily stopped

Using conservation of energy

K.E+\dfrac{kq_{1}q_{2}}{d}=\dfrac{kq_{1}q_{2}}{x}+K.E

Put the value into the formula

16+\dfrac{9\times10^{9}\times40\times10^{-6}\times80\times10^{-6}}{2}=\dfrac{9\times10^{9}\times40\times10^{-6}\times80\times10^{-6}}{x}+0

16+14.4=\dfrac{28.8}{x}

30.4x=28.8

x=\dfrac{28.8}{30.4}

x=0.947\ m

Hence, The distance separating the two particles when the moving particle is momentarily stopped is 0.947 m

8 0
3 years ago
What is the net force exerted by these two charges on a third charge q3 = 49.0nC placed between q1 and q2 at x3 = -1.085m ?Your
densk [106]

Answer:

The net force exerted by these two charges on a third charge is 5.468\times10^{-6}\ N

Explanation:

Given that,

Third charge q_{3}=49.0\ nC

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Suppose The magnitude of the force F between two particles with charges Q and Q' separated by a distance d. Consider two point charges located on the x axis one charge, q₁ = -12.5 nC , is located at x₁ = -1.650 m, the second charge, q₂ = 31.5 nC , is at the origin.

We need to calculate the total force will be the vector sum of two forces

Using Coulomb's law,

F_{13}=\dfrac{kq_{1}q_{3}}{(x_{1}-x_{3})^2}

Put the value into the formula

F_{13}=\dfrac{9\times10^{9}\times(-12.5\times10^{-9})\times49\times10^{-9}}{(-1.650-(-1.085))^2}

F_{13}=-17268.3\times10^{-9}\ N

We need to calculate the force will be to the negative charge with opposite charges

Using Coulomb's law,

F_{23}=\dfrac{kq_{2}q_{3}}{(x_{2}-x_{3})^2}

Put the value into the formula

F_{23}=\dfrac{9\times10^{9}\times(31.5\times10^{-9})\times49\times10^{-9}}{(-1.085)^2}

F_{23}=11800.2\times10^{-9}\ N

The force also will be to the negative side, charges with same charge sign

We need to calculate the net force exerted by these two charges on a third charge

Using formula of net force

F_{net}=F_{13}+F_{23}

F_{net}=-17268.3\times10^{-9}+11800.2\times10^{-9}

F_{net}=-0.0000054681\ N

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Negative sign shows the negative direction.

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That being said, you can simply divide the change in velocity by the change in time, giving you an answer of 9.8 m/s^2, which is the value of g. Even if they did not give you a time, the answer would still always be the value of g (that is if the question pertains to earth), as acceleration due to gravity is a constant.

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