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GREYUIT [131]
2 years ago
13

A soccer player with a mass of 60 kg is traveling at 8 m/s when he completes a corner kick on a 0.45 kg soccer ball. The soccer

ball travels toward the goal at a speed of 35 m/s. In this elastic collision, what is the speed of the soccer player immediately after kicking the ball?
A) 5.4 m/s
B) 7 m/s
C) 7.7 m/s
D) 8 m/s
Physics
2 answers:
trasher [3.6K]2 years ago
6 0
It must be noted that during an elastic collision both the momentum and kinetic energy are conserved. For the kinetic energy, it can be solved through the equation,
                          KE = 0.5mv²
Equating the kinetic energies before and after collision,
            0.5(60)(8 m/s)² = (0.5)(60)(x²) + (0.5)(0.45)(35 m/s)²
The value of x from the equation is approximately 7.40 m/s

azamat2 years ago
6 0

Answer:

v = 7.73 m/s

Explanation:

It is given that,

Mass of the player, m₁ = 60 kg

Initial speed of the player, u₁ = 8 m/s

Mass of the soccer ball, m₂ = 0.45 kg

Initial speed of the player, u₂ = 0 (at rest)

After the collision,

Final speed of the player, v₁

Final speed of the ball, v₂ = 35 m/s

Using the conservation of linear momentum to find the final speed of the player. Mathematically, it is given by :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

m_1u_1=m_1v_1+m_2v_2

60\times 8=60\times v_1+0.45\times 35          

v_1=7.73\ m/s

So, the speed of the soccer player immediately after kicking the ball is 7.73 m/s. Hence, this is the required solution.

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Cindy exerts a force of 40 newtons and moves a chair 6 meters. Her brother Andy pushes a different chair for 6 meters while exer
sesenic [268]
Work formula:
W = F * d
F 1 = 40 N, d 1 = 6 m;
F 2 = 30 N; d 2 = 6 m.
W ( Cindy ) = 40 * 6 = 240 Nm
W ( Andy ) = 30 * 6 = 180 Nm
The difference of their amounts if work:
240 Nm - 180  Nm = 60 nm

hope it helps!
3 0
2 years ago
A rocket moves upward, starting from rest with an acceleration of +29.4 for 3.98 s. it runs out of fuel at the end of the 3.98 s
topjm [15]
U = 0, initial upward speed
a = 29.4 m/s², acceleration up to 3.98 s
a = -9.8 m/s², acceleration after 3.98s

Let h₁ =  the height at time t, for t ≤ 3.98 s
Let h₂ =  the height at time t > 3.98 s

Motion for  t ≤ 3.98 s:
h₁ = (1/2)*(29.4 m/s²)*(3.98 s)² = 232.854 m
Calculate the upward velocity at t = 3.98 s
v₁ = (29.4 m/s²)*(3.98 s) = 117.012 m/s

Motion for t  > 3.98 s
At maximum height, the upward velocity is zero.
Calculate the extra distance traveled before the velocity is zero.
(117.012 m/s)² + 2*(-9.8 m/s²)*(h₂ m) = 0
h₂ = 698.562 m

The total height is
h₁ + h₂ = 232.854 + 698.562 = 931.416 m

Answer: 931.4 m (nearest tenth)

6 0
2 years ago
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You are working as an assistant to an air-traffic controller at the local airport, from which small airplanes take off and land.
Alika [10]

Answer:

d = 2021.6 km

Explanation:

We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them

Airplane 1

Height   y₁ = 800m

Angle θ = 25°

           cos 25 = x / r

           sin 25 = z / r

           x₁ = r cos 20

           z₁ = r sin 25

          x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

          z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m

2 plane

Height   y₂ = 1100 m

Angle θ = 20°

          x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

          z₂ = 20 103 without 25 = 8.452 103 m = 8452 m

The distance between the planes using the Pythagorean Theorem is

         d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Let's calculate

        d² = (18126-16314)²  + (1100-800)² + (8452-7607)²

        d² = 3,283 106 +9 104 + 7,140 105

        d² = (328.3 + 9 + 71.40) 10⁴

        d = √(408.7 10⁴)

        d = 20,216 10² m

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7 0
2 years ago
At the normal boiling temperature of iron, TB = 3330 K, the rate of change of the vapor pressure of liquid iron with temperature
Margaret [11]

The molar latent enthalpy of boiling of iron at 3330 K is  ΔH = 342 \times 10^3 J.

<u>Explanation:</u>

Molar enthalpy of fusion is the amount of energy needed to change one mole of a substance from the solid phase to the liquid phase at constant temperature and pressure.

                      d ln p = (ΔH / RT^2) dt

                   (1/p) dp = (ΔH / RT^2) dt

                    dp / dt = p (ΔH / RT^2) = 3.72 \times 10^-3

                  (p) (ΔH) / (8.31) (3330)^2 = 3.72 \times 10^-3

                          ΔH = 342 \times 10^3 J.

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2 years ago
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ki77a [65]

Answer:

Explanation:

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