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Scorpion4ik [409]
3 years ago
12

As a 5.0 × 10^2-newton basketball player jumps

Physics
1 answer:
AysviL [449]3 years ago
4 0
     This player is initially at rest, then the Force Weight (W) and the Normal Force (N) have a same module. When applies a Extra Force (E) on the floor, this reacts with this increase, causing the player to be released upwards.
     Using the Newton's Secound Law, we have:

F_{R}=ma \\ W+E-N=ma \\ N =W+E \\ N=5*10^2+10*10^2 \\ \boxed {N=1.5*10^3N}

Number 3

If you notice any mistake in my english, please let me know, because i am not native.
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April does 1800 J of work to get a canoe moving from rest. April and the canoe have a mass of 70 kg. a. What is the kinetic ener
Katyanochek1 [597]

E = (1/2)mv²

1800 = (1/2)(70)v²

Solving for v²

1800(2)/70 = v²

v² = 55.714

√v² = √55.714

v = ~7.46 m/s

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A track consists spring launcher on one end. A spring which is compressed 0.5 m has a
patriot [66]

(a) The velocity of the first ball before the collision with the second ball is 11.18 m/s.

(b) The final velocity of the two balls after the collision is determined as 5.59 m/s.

<h3>Speed of the block when pushed by the spring</h3>

The speed of the block when pushed by the spring is calculated as follows;

K.E = Ux

¹/₂mv² = ¹/₂kx²

mv² = kx²

v² = kx²/m

v² = (25 x 0.5²)/0.05

v² = 125

v = 11.18 m/s

<h3>Final velocity of the two balls after the collision</h3>

The velocity of the two balls after the collision is calculated as follows;

Pi = Pf

where;

  • Pi is initial momentum
  • Pf is final momentum

m1u1 + m2u2 = v(m1 + m2)

0.05(11.18) + 0.05(0) = v(0.05 + 0.05)

0.559 = 0.1v

v = 5.59 m/s

Learn more about velocity here: brainly.com/question/4931057

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4 0
2 years ago
What waves travel through a medium? Transverse, longitudinal, surface, electromagnetic, and mechanical are the choices
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Hoped this helped!
6 0
3 years ago
If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her
Marysya12 [62]

Answer:

 μ = 0.336

Explanation:

We will work on this exercise with the expressions of transactional and rotational equilibrium.

Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation

          fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ  = 0

          fr sin θ  - cos θ  (W / 2 + 0,3 W_painter) = 0

          fr = cotan θ  (W / 2 + 0,3 W_painter)

Now let's write the equilibrium translation equation

     

X axis

        F1 - fr = 0

        F1 = fr

the friction force has the expression

       fr = μ N

Y Axis

       N - W - W_painter = 0

       N = W + W_painter

       

we substitute

      fr = μ (W + W_painter)

we substitute in the endowment equilibrium equation

     μ (W + W_painter) = cotan θ  (W / 2 + 0,3 W_painter)

      μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)

we substitute the values ​​they give

      μ = cotan θ  (12/2 + 0.3 55) / (12 + 55)

      μ = cotan θ  (22.5 / 67)

      μ = cotan tea (0.336)

To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45

       cotan 45 = 1 / tan 45 = 1

the result is

    μ = 0.336

5 0
3 years ago
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