An electron with speed v0 = 27.5 X 106 m/s is traveling parallel to a uniform electric field of magnitude E = 11.4 X 103 N/C. Ho

Answer:

5.7 m

Explanation:

AD = length of the ladder = L = 8 m

AB = the position of the ladder's center of mass = (0.5) L = (0.5) 8 = 4 m

AC = distance of the climber from the bottom of the ladder = x

W = weight of the ladder = 240 N

= weight of the climber = 710 N

F = force exerted by the wall on the ladder

N = normal force acting on the ladder from the ground =?

By applying force equilibrium in the vertical direction

N = + W

N = 710 + 240

N = 950 N

μ = Coefficient of static friction = 0.55

f = static friction force on the ladder

Static friction force can be expressed as

f = μ N

f = (0.55) (950)

f = 522.5 N

The equation for force along the horizontal axis reads

F = f

F = 522.5 N

using torque equilibrium around point A

F Sin50 (AD) = W Cos50 (AB) + ( Cos50 (AC))

(522.5) Sin50 (8) = (240) Cos50 (4) + (710) Cos50 (x)

x = 5.7 m

The kangaroo reaches a maximum vertical altitude of 2.8 m, which can be calculated using the formula 2.8 = 1/2 * 9.8 * t^2. Thus, applying the equation s = ut + 1/2at^2.

Answer:

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Explanation:

Answer:

F = 0.535 N

Explanation:

We will apply energy concepts, considering both the peak and the bottom of the path.

Top

   Em₀ = U = mg y

Bottom

    = K = ½ m v²

    Emo =

    mg y = ½ m v²

    v = √ (2gy)

   y = L - L cos θ

  v = √ (2g L (1 - cos θ))

Next, we will employ Newton's second law at the lowest point where the acceleration is centripetal.

     F = ma

     a = v² / r

For the turning radius, the cable length is r = L.

    F = m 2g (1 - cos θ)

Now, let's find the result.

    F = 2  1.25  9.8 (1 - cos 12)

    F = 0.535 N