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KonstantinChe [14]
3 years ago
7

A car with a mass of 1.1 x10 to the third power killograms hits a stationary truck with a mass of 2.3kg x10 to the third power f

rom the rear ean. The initial velocity of the car is +22.0 meters/second. After the collision the velocity of the car is -11.0 meters/second. What is the velocity of the truck after this elastic collision.
Physics
1 answer:
Anton [14]3 years ago
5 0

Answer:

v2f = +15.8 m/s

Explanation:

Conservation law of linear momentum:

m1v1i + m2vi2 = m1v1f + m2v2f

Given:

m1 = 1.1 × 10^3 kg

m2 = 2.3 × 10^3 kg

v1i = +22.0 m/s

v2i = 0

v1f = -11.0 m/s

v2f = ?

Re-arranging the conservation law, we get

m1v1i = m1v1f + m2v2f

Solving for v2f,

m2v2f =m1(v1i - v1f)

or

v2f = m1(v1i - v1f)/m2

= (1.1 × 10^3 kg)(22.0 m/s - (-11.0 m/s))/(2.3 m/s)

= (1.1 × 10^3 kg)(33.0 m/s)/(2.3 × 10^3 kg)

= +15.8 m/s

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Water flows straight down from an open faucet. The cross-sectional area of the faucet is 2.4 × 10-4m2 and the speed of the water
Ksenya-84 [330]

To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.

By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,

v_2^2 = v_1^2 + 2gh

Where,

V_i = Velocity in each state

g= Gravity

h = Height

Our values are given as,

A_1 = 2.4*10^{-4} m^2

v_1 = 0.8 m/s

h = 0.11m

Replacing at the kinetic equation to find V_2 we have,

v_2 = \sqrt{v_1^2 + 2gh}

v_2 = \sqrt{(0.8 m/s)^2 + 2(9.80 m/s2)(0.11 m)}

v_2= 1.67 m/s

Applying the concepts of continuity,

A_1v_1 = A_2v_2

We need to find A_2 then,

A_2= \frac{A_1v_1 }{v_2}

So the cross sectional area of the water stream at a point 0.11 m below the faucet is

A_2= \frac{A_1v_1 }{v_2}

A_2= \frac{(2.4*10^{-4})(0.8)}{(1.67)}

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8 0
3 years ago
A flat sheet of ice has a thickness of 1.4 cm. It is on top of a flat sheet of crown glass that has a thickness of 3.0 cm. Light
MAXImum [283]

Answer:

t = 2.13 10-10 s , d = 6.39 cm

Explanation:

For this exercise we use the definition of refractive index

        n = c / v

Where n is the refraction index, c the speed of light and v the speed in the material medium.

The refractive indices of ice and crown glass are 1.13 and 1.52, respectively, therefore the speed of the beam in the material medium is

        v = c / n

As the beam strikes perpendicularly, the beam path is equal to the distance of the leaves, there is no refraction, so we can use the uniform motion relationships

        v = d / t

        t = d / v

        t = d n / c

Let's look for the times on each sheet

Ice

        t₁ = 1.4 10⁻² 1.31 / 3 10⁸

        t₁ = 0.6113 10⁻¹⁰ s

Crown glass (BK7)

        t₂ = 3.0 10⁻² 1.52 / 3.0 10⁸

        t₂ = 1.52 10⁻¹⁰ s

Time is a scalar therefore it is additive

         t = t₁ + t₂

         t = (0.6113 + 1.52) 10⁻¹⁰

         t = 2.13 10-10 s

The distance traveled by this time in a vacuum would be

        d = c t

       d = 3 10⁸ 2.13 10⁻¹⁰

       d = 6.39 10⁻² m

       d = 6.39 cm

3 0
3 years ago
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