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eduard
3 years ago
7

How does the angle of launch affect the kinetic energy of a rubber band?​

Physics
1 answer:
Lady_Fox [76]3 years ago
3 0

Answer:

The angle of launch of the rubber band affects the initial velocity. The more the rubber band is stretched the more force it applies to return to equilibrium and the more kinetic energy that results in.

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Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The
Galina-37 [17]

Answer:

220 A

Explanation:

The magnetic force on the floating rod due to the rod held close to the ground is F = BI₁L where B = magnetic field due to rod held close the ground = μ₀I₂/2πd where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, I₂ = current in rod close to ground and d = distance between both rods = 11 mm = 0.011 m. Also, I₁ = current in floating rod and L = length of rod = 1.1 m.

So, F = BI₁L

F = (μ₀I₂/2πd)I₁L

F = μ₀I₁I₂L/2πd

Given that the current in the rods are the same, I₁ = I₂ = I

So,

F = μ₀I²L/2πd

Now, the magnetic force on the floating rod equals its weight , W = mg where m = mass of rod = 0.10kg and g = acceleration due to gravity = 9.8 m/s²

So, F = W

μ₀I²L/2πd = mg

making I subject of the formula, we have

I² = 2πdmg/μ₀L

I = √(2πdmg/μ₀L)

substituting the values of the variables into the equation, we have

I = √(2π × 0.011 m × 0.1 kg × 9.8 m/s²/[4π × 10⁻⁷ H/m × 1.1 m])

I = √(0.01078 kgm²/s²/[2 × 10⁻⁷ H/m × 1.1 m])

I = √(0.01078 kgm²/s²/[2.2 × 10⁻⁷ H])

I = √(0.0049 × 10⁷kgm²/s²H)

I = √(0.049 × 10⁶kgm²/s²H)

I = 0.22 × 10³ A

I = 220 A

7 0
2 years ago
A golf ball strikes a hard, smooth floor at an angle of 28.2 ° and, as the drawing shows, rebounds at the same angle. The mass o
WITCHER [35]

Answer:

    I = 1.06886  N s

Explanation:

The expression for momentum is

          I = F t = Δp

therefore the momentum is a vector quantity, for which we define a reference system parallel to the floor

Let's find the components of the initial velocity

          sin 28.2 = v_y / v

          cos 28.2=  vₓ / v

          v_y = v sin 282

          vₓ = v cos 28.2

          v_y = 42.8 sin 28.2 = 20.225 m / s

          vₓ = 42.8 cos 28.2 = 37.72 m / s

since the ball is heading to the ground, the vertical velocity is negative and the horizontal velocity is positive, it can also be calculated by making

θ = -28.2

         v_y = -20.55 m / s

         v_x = 37.72 m / s

X axis

         Iₓ = Δpₓ = p_{fx} - p_{ox}

since the ball moves in the x-axis without changing the velocity, the change in moment must be zero

         Δpₓ = m v_{fx} - m v₀ₓ = 0

          v_{fx} = v₀ₓ

therefore

          Iₓ = 0

Y axis  

        I_y = Δp_y = p_{fy} -p_{oy}

when the ball reaches the floor its vertical speed is downwards and when it leaves the floor its speed has the same modulus but the direction is upwards

         v_{fy} = - v_{oy}

         Δp_y = 2 m v_{oy}

         Δp_y = 2 0.0260 (20.55)

         \Delta p_{y} = 1.0686 N s

the total impulse is

          I = Iₓ i ^ + I_y j ^

          I = 1.06886  j^ N s

7 0
2 years ago
A machine is classified as a compound machine when ___.
bearhunter [10]

Answer: it is made up two or more simple machines

Explanation: Brainliest Please!!!!

6 0
2 years ago
The position-time graph for two drivers (A and B) are shown in the figure below. Which of the following statements is correct?
elena-14-01-66 [18.8K]

Answer:

A

Explanation:

5 0
2 years ago
What happens in a change in temperature as it is heated
Andre45 [30]
It increases in temperature
5 0
2 years ago
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