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EastWind [94]
2 years ago
9

Write the formula of Lever, Pulleys, wheel and axle and inclined plane.​

Physics
1 answer:
maxonik [38]2 years ago
6 0

Answer:

Lever => d_{e} = d_{r}

Pulley => G = M x n (gravitational acceleration)

Wheel and axle => M.A = Radius of the wheel/radius of the axle = R/r

Inclined plane => It can be divided into two components: Fi = Fg * sinθ - parallel to inclined plane. Fn = Fg * cosθ - perpendicular one.

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Two identical tiny spheres of mass m =2g and charge q hang from a non-conducting strings, each of length L = 10cm. At equilibriu
Citrus2011 [14]

Answer:

0.247 μC

Explanation:

As both sphere will be at the same level at wquilibrium, the direction of the electric force will be on the x axis. As you can see in the picture below, the x component of the tension of the string of any of the spheres should be equal to the electric force of repulsion. And its y component will be equal to the weight of one sphere. We can use trigonometry to find the components of the tensions:

F_y:  T_y - W = 0\\T_y = m*g = 0.002 kg *9.81m/s^2 = 0.01962 N

T_y = T_*cos(50)\\T = \frac{T_y}{cos(50)} = 0.0305 N

T_x = T*sin(50) = 0.0234 N

The electric force is given by the expression:

F = k*\frac{q_1*q_2}{r^2}

In equilibrium, the distance between the spheres will be equal to 2 times the length of the string times sin(50):

r = 2*L*sin(50) = 2 * 0.1m * sin(50) 0.1532 m

And k is the coulomb constan equal to 9 *10^9 N*m^2/C^2. q1 y q2 is the charge of each particle, in this case, they are equal.

F_x = T_x - F_e = 0\\T_x = F_e = k*\frac{q^2}{r^2}

q = \sqrt{T_x *\frac{r^2}{k}} = \sqrt{0.0234 N * \frac{(0.1532m)^2}{9*10^9 N*m^2/C^2} } = 2.4704 * 10^-7 C

O 0.247 μC

8 0
2 years ago
A two-stage rocket moves in space at a constant velocity of +4300 m/s. The two stages are then separated by a small explosive ch
ololo11 [35]

Answer:

 v_{1f} = +3,394 103 m / s

Explanation:

We will solve this problem with the concept of the moment. Let's start by defining the system that is formed by the complete rocket before and after the explosions, bone with the two stages, for this system the moment is conserved.

The data they give is the mass of the first stage m1 = 2100 kg, the mass of the second stage m2 = 1160 kg and its final velocity v2f = +5940 m / s and the speed of the rocket before the explosion vo = +4300 m / s

The moment before the explosion

      p₀ = (m₁ + m₂) v₀

After the explosion

      pf = m₁ v_{1f} + m₂ v_{2f}

     p₀ = [texpv_{f}[/tex]

     (m₁ + m₂) v₀ = m₁ v_{1f} + m₂ v_{2f}

Let's calculate the final speed (v1f) of the first stage

     v_{1f} = ((m₁ + m₂) v₀ - m₂ v_{2f}) / m₁

     

     v_{1f} = ((2100 +1160) 4300 - 1160 5940) / 2100

     v_{1f} = (14,018 10 6 - 6,890 106) / 2100

     v_{1f} = 7,128 106/2100

     v_{1f} = +3,394 103 m / s

come the same direction of the final stage, but more slowly

4 0
3 years ago
A cart loaded with bricks has a total mass of 22.2 kg and is pulled at constant speed by a rope. The rope is inclined at 27.5 ◦
musickatia [10]

Answer:

W = 1.432 KJ

Explanation:

given,

mass = 22.2 Kg

angle of the rope = 27.5°

distance on the ground = 24 m

kinetic friction= μ = 0.32

acceleration due to gravity, g = 9.8 m/s²

Work done = ?

W = F d cosθ

a = 0 because it is moving with constant speed

equating all the forces acting  in x direction

F cosθ = F friction = μN  

equating all the forces acting  in y direction

F sinθ + N -mg =0

now,

N = mg - F sinθ

putting value of N

F cosθ = μ mg -μ F sinθ

F (cosθ + μsinθ ) = μ mg

F = \dfrac{\mu mg}{cos\theta + \mu sin\theta}

F = \dfrac{0.32 \times 22.2 \times 9.8}{cos 27.5^0+0.32 \times sin27.5^0}

F =67.28 N

now,

W=F d cosθ

W =67.28 x 24 x cos(27.5)

W =1432.27 J

W = 1.432 KJ

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2 years ago
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An electron drops from the n=6 to the n=4 level of an infinite square well that is 3.10-11 m wide. What is the wavelength of the
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The answer would be C !
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3 years ago
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