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solniwko [45]
2 years ago
13

Object A of mass M is moving east at speed v. It collides with object B of mass 2M that was initially at rest. The motion of the

objects before and after the collision is along the same line. After the collision, object A is moving west at a speed of v/3. What is the speed of object B immediately after the collision?
Physics
1 answer:
Firlakuza [10]2 years ago
3 0

Answer:

v_B=\frac{v}{3}

Explanation:

Given that:

mass of object A, m_A=M

mass of object B, m_B=2M

speed of object A, v_A=v

So, according to the conservation of momentum, the momentum before collision is equal to the momentum after conservation.

m_A.v+m_B\times 0=m_A\times v_A +m_B\times v_B

M\times v+0 = M\times \frac{v}{3}+2M\times v_B

2M\times v_B= \frac{2M\times v}{3}

v_B=\frac{v}{3}

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Answer:

a) 447.21m

b) -62.99 m/s

c)94.17 m/s

Explanation:

This situation we can divide in 2 parts:

⇒ Vertical : y =-200 m

y =1/2 at²

-200 = 1/2 *(-9.81)*t²

t= 6.388766 s

⇒Horizontal: Vx = Δx/Δt

Δx = 70 * 6.388766 = 447.21 m

b) ⇒ Horizontal

Vx = Δx/Δt ⇒ 70 = 400 /Δt

Δt= 5.7142857 s

⇒ Vertical:

y = v0t + 1/2 at²

-200 = v(5.7142857) + 1/2 *(-9.81) * 5.7142857²

v0= -7 m/s  ⇒ it's negative because it goes down.

v= v0 +at

v= -7 + (-9.81) * 5.7142857

v= -62.99 m/s

c) √(70² + 62.99²) = 94.17 m/s

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2 years ago
While you are studying for an upcoming physics exam, a lightning storm is brewing outside your window. Suddenly, you see a tree
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An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its
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Answer:

 Terminal velocity of object = 12.58 m/s

Explanation:

 We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.

Gravitational force = mg = 80 * 9.8 = 784 N

Drag force = 12.0v+4.00v^2

Equating both, we have

    784=12.0v+4.00v^2\\ \\ v^2+3v-196=0\\ \\ (v-12.58)(v+15.58)=0

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7 0
1 year ago
Given that the CFL bulb is rated at 10% efficiency, if the LED bulb consumes half the amount of electrical power, for the same a
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Answer and explanation:

The efficiency of the LED bulb is 20% while The efficiency of the Inc bulb is 1%.

The lightbulb efficiency can be defined as the quotient between the amount of usable light and the amount of electrical power output. For the efficiency of the CFL Bulb:

\epsilon_{CFL}=\frac{L_0}{P_0}=0.1

Therefore:

\epsilon_{led}=\frac{L_0}{0.5P_0}=2 \cdot0.1=0.2\\\epsilon_{INC}=\frac{L_0}{10P_0}=0.1 \cdot 0.1=0.01

4 0
1 year ago
Constants Periodic Table Suppose the top surface of the vessel in the figure (Figure 1) is subjected to an external gauge pressu
Gnom [1K]

Answer:

a)  v₁ = √ [2 (P₂-P₀) /ρ + 2 (y₂ -y₁)]

b) Water does not flow,

Explanation:

a) For this exercise we will use Bernoulli's equation, where index 1 is at the exit and index 2 on the surface of the water

            P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

This case does not indicate at the surface pressure is P₂, the pressure at the outlet is P₁ = P₀, the surface velocity is zero v₂ = 0

          P₀ + ½ ρ v₁² + ρ g y₁ = P₂ + 0 + ρ g y₂

           ½ ρ v₁² = P₂-P₀ + ρ (y₂ -y₁)

          v₁² = 2 (P₂-P₀) /ρ + 2 (y₂ -y₁)

          v₁ = √ [2 (P₂-P₀) /ρ + 2 (y₂ -y₁)]

b) Reduce the pressure to SI units

         P₂ = 0.86 atm (1.01 10⁵ Pa / 1 atm) = 0.8686 10⁵ Pa

         P₀ = 1.01 10⁵ Pa

         ρ = 1 10³ kg / m³

Let's calculate

         v₁ = √ [2 (0.8686 -1.01) 10⁵/10³ + 2 (2.6)]

         v₁ = √ [-0.2828 10² + 5.2] = Ra [-23]

Water does not flow, this is because the pressure of the inner part is less than atmospheric, so that the water flows the pressure P₂> = P₀

For example if the pressure P₂ = P₀

         v₁ = √ 5.2

          v₁ = 2.28 m / s

5 0
1 year ago
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