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2 years ago

9

A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of

the incline, and its speed after it has traveled 7.40 mm to the bottom of the incline is 3.80 m/sm/s.What is the speed of the block when it is 3.40 m from the top of the incline?

1 answer:

natita [175]2 years ago

6 0

Answer:

2.58 m/s

Explanation:

We can use the following equation of motion to find out the constant acceleration of the block:

$v^2 - v_0^2 = 2a\Delta s$

where v = 3.8 m/s is the final velocity after it has traveled 7.4 m, $v_0 = 0$ is the initial velocity of the block when it starts from rest, , and $\Delta s = 7.4$ is the distance traveled.

$3.8^2 - 0 = 2*a*7.4$

$14.44 = 14.8a$

$a = 14.44 / 14.8 = 0.98 m/s^2$

We can use the same motion equation to calculate block speed at the end of 3.4m track

$v_2^2 - 0 = 2*0.98*3.4$

$v_2^2 = 6.63$

$v_2 = \sqrt{6.63} = 2.58 m/s$

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The formula used for the radius of the Airy disk is given by,

$y_r=1.22\frac{\lambda f}{d}$

Where,

$y_r =$ Range of the radius

$\lambda =$ wavelength

f= focal length

Our values are given by,

State 1:

$d=2.00mm = 2*10^{-3}m$

$f= 25mm = 25*10^{-3}m$

$\lambda = 750nm = 750*10^{-9}m$

State 2:

$d=8.00mm = 8*10^{-3}m$

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$\lambda = 390nm = 390*10^{-9}$

Replacing in the first equation we have:

$y_{r1} = 1.22\frac{(750*10^{-9})(25*10^{-3})}{2*10^{-3}}$

$y_{r1}= 11.4\mu m$

And also for,

$y_{r2} =1.22\frac{(390*10^{-9})(25*10^{-3})}{8*10^{-3}}$

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2 years ago

An object with height h, mass M, and a uniform cross-sectional area A floats upright in a liquid with density ρ.

soldi70 [24.7K]

** Missing information: The vertical distance from surface of liquid to bottom of the object is sought in this question, with the condition that the object is at equilibrium **

Ans: The vertical distance = y = M/(ρA)

Explanation:

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Object's density = M/(A*h) (since A*h = volume)

By applying the condition,

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3 years ago

We need to find the launch velocity of our new marble launcher. we know that it will launch a 25g marble to a distance of 73 cm,

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The launch velocity of the marble launcher is 34.65 m/s

Given that the launch velocity of marble launcher, launches a 25g marble to a distance of 73 cm (0.73 m) and the marble roll up to 6.2 meters before stopping. The launch height is 20 cm (0.2 m).

The time for landing can be calculated by the second equation of motion formula:

h = ut + $\frac{1}{2}$ g $t^{2}$

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8 months ago

. Water is flowing at 12m/s in a horizontal pipe under a pressure of 600kpa radius 2cm. a. What is the speed of the water on the

lana66690 [7]

Answer:

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Explanation:

Givens:

Inlet Velocity, V₁ = 12 m/s

Inlet Pressure, P₁ = 600 kPa = 600,000 Pa

Inlet Radius r₁ = 0.5 cm

Outlet Velocity, V₂ = not given (we are asked to find this)

Outlet Pressure, P₂ = not given (we are asked to find this)

Outlet Radius, r₂ = 0.5 cm

From these, we can find the following:

Inlet Area, A₁ = π (r₁)² = π(2)² = 4π cm²

Outlet Area, A₂ = π (r₂)² = π(0.5)² = 0.25π cm²

<u>Part A :</u>

Assuming that water is incompressible, we can reason that within the same given time, the amount of volume of water entering the inlet must equal the volume of water exiting the outlet. Hence by the continuity equation (i.e. conservation of mass)

Inlet Volume flow rate = Outlet Volume flow rate

(recall that Volume flow rate in a pipe is given by Velocity x Cross Section Area), Hence the equation becomes

V₁ x A₁ = V₂ x A₂ (substituting the values that we know from above)

12 x 4π = V₂ x 0.25π (we don't have to change all to SI units because the conversion factors on the left will cancel out the conversion factors on the right).

V₂ = (12 x 4π) / (0.25π)

V₂ = 192 m/s (Answer)

<u>Part B:</u>

For Part B, if we assume a closed ideal system (control volume method), we can simply apply the energy equation (i.e Bernoulli's equation)

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Because the pipe is horizontal, there is no difference between h₁ and h₂, hence we can neglet this term:

P₁ + (1/2)ρV₁ = P₂ + (1/2)ρV₂ (rearranging)

P₂ = P₁ + (1/2)ρV₁ - (1/2)ρV₂

= P₁ + (1/2)ρ (V₁-V₂)

Assuming that the density of water is approx, ρ = 1000 kg/m³

P₂ = 600,000 + (1/2)(1000) (12-192)

= 600,000 + ( -90,000)

= 510,000 Pa

= 510 kPa (Answer)

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2 years ago

During takeoff, a plane goes from 0 to 50 m/s in 8s. What is its acceleration? Halos fast is it going after 5 s? How far had it

svlad2 [7]

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<span>distance= 400 m check that.</span>

4 0

2 years ago

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