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2 years ago

- Una carga de -5x10-6 C está situada a 20cm delante de otra

Physics

1 answer:

saul85 [17]2 years ago

3 0

Answer:

$F = 5.625\,N$ (Fuerza de repulsión/Repulsive force)

Explanation:

La fuerza ejercida entre las dos partículas se calcula por la Ley de Coulomb (The force exerted between the two particles is determined by Coulomb's Law):

$F = k \cdot \frac{q_{1}\cdot q_{2}}{r^{2}}$

Donde (Where):

$k$ - Constante electrostática, medido en $\frac{N\cdot m^{2}}{C^{2}}$ (Electrostatic constant, measured in $\frac{N\cdot m^{2}}{C^{2}}$ ).

$q_{1}$ , $q_{2}$ - Magnitudes de las cargas de cada partícula, medidos en Coulombs. (Magnitudes of charges from each particle, measured in Coulombs).

$r$ - Distancia entre las partículas, medida en metros.

La fuerza electrostática es (Electrostatic force is):

$F = \left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot \frac{(5\times 10^{-6}\,C)\cdot (5\times 10^{-6}\,C)}{(0.20\,m)^{2}}$

$F = 5.625\,N$

Dado que las partículas tienen el mismo signo de carga, la fuerza es de repulsión. (Given that both particles have the same charge sign, the force is of repulsive nature)

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An 8.5 kg crate is pulled 5.1 m up a 30 degree incline by a rope angled 17 degrees above the incline. The tension in the rope is

Ulleksa [173]

Answer:

1. a $W_t=746.63 J$

b $E_p=212.415 J$

c $W_n=183.96J$

2. $T_e=99.71J$

Explanation:

a). The work done by the tension is:

$W_t=T*dt$

$dt=\frac{5.1}{cos(17)}$

$W_t=140N*\frac{5.1}{cos(17)}$

$W_t=746.63 J$

b). The work done potential of gravity

$E_p=m*g*h$

$h=5.1*sin(17)$

$E_p=8.5kh*9.8*5.1*sin(30)$

$E_p=212.415 J$

c). The work done by the normal force

$W_n=N*d_n$

$d_n=5.1*sin(30)=2.55$

$W_n=8.5kg*9.8*cos(30)*2.55$

$W_n=183.96J$

2. The increase in thermal energy is:

$T_e=F*d$

$F_k=u_k*m*g=0.271*8.5kg*9.8*cos(30)$

$F_k=19.5N$

$T_e=19.55*5.1m$

$T_e=99.71J$

4 0

3 years ago

If two objects A and B have the same kinetic energy but A has three times the momentum of B, what is the ratio of their inertias

Thepotemich [5.8K]

Answer:

$\frac{inertia_B}{inertia_A}=9$

Explanation:

First of all, let's remind that:

- The kinetic energy of an object is given by $K=\frac{1}{2}mv^2$ , where m is the mass and v is the speed

- The momentum of an object is given by $p=mv$

- The inertia of an object is proportional to its mass, so we can write $I=km$ , where k just indicates a constant of proportionality

In this problem, we have:

- $K_A = K_B$ (the two objects have same kinetic energy)

- $p_A = 3 p_B$ (A has three times the momentum of B)

Re-writing both equation we have:

$\frac{1}{2}m_A v_A^2 = \frac{1}{2}m_B v_B^2\\m_A v_A = 3 m_B v_B$

If we divide first equation by second one we get

$v_A = 3 v_B$

And if we substitute it into the first equation we get

$m_A (3 v_B)^2 = m_B v_B^2\\9 m_A v_B^2 = m_B v_B^2\\m_B = 9 m_A$

So, B has 9 times more mass than A, and so B has 9 times more inertia than A, and their ratio is:

$\frac{I_B}{I_A}=\frac{km_B}{km_A}=\frac{9m_A}{m_A}=9$

7 0

2 years ago

Question 1 (1 point)

Dmitry [639]

1). trajectory

2). person sitting in a chair

3). 490 meters

4). 65 m/s

5). False. The projectile's displacement, velocity, and acceleration have vertical and horizontal components, but the projectile doesn't.

6). False

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The vertical components do NOT equal the horizontal components.

8). Decreasing if you include the effects of air resistance. Constant if you don't. Gravity has no effect on horizontal velocity.

9). We can't see the simulation. But if the projectile doesn't have jets on it, then as it travels upward, its vertical velocity must decrease, because gravity is trying to not let it get away.

10). We can't see the simulation. But if the projectile is traveling downward, we would call that "falling", and its vertical velocity must increase, because gravity is pulling it downward.

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cestrela7 [59]

ITS C

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