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lbvjy [14]
3 years ago
7

2) Two railway tracks are parallel to west east direction. Along one track, train A moves with a speed of 45 m/s from

Physics
1 answer:
OLEGan [10]3 years ago
4 0

Answer:

Relative\ Velocity = 105m/s

Explanation:

Given

V_A = 45m/s

V_B = 60m/s

Required

Determine the speed of B w.r.t A

The question implies that, we determine the relative velocity of B w.r.t A

Because both trains are moving towards one another, the required velocity is a sum\ of\ velocities\ of both trains:

This is shown below:

Relative\ Velocity = V_A + V_B

Relative\ Velocity = 45m/s + 60m/s

Relative\ Velocity = 105m/s

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Water of density 1000 kg/m3 falls without splashing at a rate of 0.373 L/s from a height of 40.5 m into a 0.64 kg bucket on a sc
Sphinxa [80]

Answer:

       F_scale = 20.18 N

Explanation:

The scale reading corresponds to two factors, the first the weight of the water in the container and the second the force of the liquid that is falling at the moment of reading.

* Let's find the amount of liquid in the container for a time of t = 2.93 s

Let's use a direct proportion rule. If 0.373 l falls in one second at t = 2.93 s, how many liters are there

        V_{water} = 2.93 s (0.373 l / 1s) = 1.09 l

        V_{water} = 1.09 10⁻³ m³

the amount of water is

       ρ = m / V

       m = ρ V

       m = 1000 1.09 10⁻³

       m = 1.09 kg

so the weight of the liquid in the container for this time is

       W = mg

       W = 1.09 9.8

       W = 10.68 N

* Let's look for the force of the falling jet

Let's use Bernoulli's equation, where the subscript 1 is for the container and the subscript 2 is for the water at a height h

        P₁ + 1/2 ρ g v₁² + ρ g y₁ = P₂ + 1/2  ρ g v₂² + ρ g y₂

In this case, the water falls freely, so the external pressure is atmospheric.

         P₂ = P_{atm}

since they indicate that the water falls, we assume that its initial velocity is zero v₂ = 0

let's use kinematics to find the speed of a drop when it reaches the container y = 0

         v² = v₀² - 2 g (y-y₀)

         v = \sqrt{0 -2 g ( 0-y_o)}

let's calculate

         v = √(2 9.8 40.5)

         v = 28.17 m / s

this is the speed in the container v₁ = 28.17 m / s

the height from where it falls is y₂ = 40.5 and reaches the container y₁ = 0

we substitute in Bernoulli's equation

         P₁ +1/2 ρ g v₁² + 0 = P_{atm} + 0 + ρ g y₂

         P₁ + ½ ρ g v₁² = P_{atm} + ρ g y₂

         P₁ = P_{atm} + ρ g y₂ - ½ ρ g v₁²

         P₁ = 1 10⁵ + 1000 9.8 40.5 - ½ 1000 28.17²

         P₁ = 1 10⁵ + 3.97 10⁵ - 3.69 10⁵

         P₁ = 1.28 10⁵ Pa

The definition of Pressure is

         P = F / A

         F = P A

We must suppose a time to carry out the reading suppose an average time of the modern equipment t = 0.1 s, in this time how much is now arriving

          m₂ = 0.373 0.2 = 0.0746 l = 0.0746 10⁻³ m³

the volume is V = A l

if the length of l = 1 m

A = 0.0746 10⁻³ m³ = 7.45 10⁻⁵ m²

the force of this jet is

            F = P A

            F = 1.28 10⁵  7.46 10⁻⁵

            F = 9.5 N

with these data let's use the equilibrium equation

           F_ scale -W - F = 0

           F_scale = W + F

           F_scale = 10.682 + 9.5

           F_scale = 20.18 N

4 0
3 years ago
Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma
AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

m_1 = Mass of first car = 120 kg

m_2 = Mass of second car

u_1 = Initial Velocity of first car = 14 m/s

u_2 = Initial Velocity of second car = 0 m/s

v_1 = Final Velocity of first car = -2 m/s

v_2 = Final Velocity of second car

For perfectly elastic collision

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}

Applying in the next equation

v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s

m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg

Mass of second car = 160 kg

Velocity of second car = 12 m/s

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svet-max [94.6K]

Answer:

By Gaining Electrons

Explanation:

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