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2 years ago

9

A 20000 kg railroad car is rolling at 1.00 m/s when a 1000 kg load of gravel is suddenly dropped in. part a what is the car's sp

eed just after the gravel is loaded?

1 answer:

Talja [164]2 years ago

8 0

Using conservation of energy and momentum we get m1*v1=(m1+m2)*v2 so rearranging for v2 and plugging the given values in we get:
(200000kg*1.00m/s)/(21000kg)=.952m/s

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One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a s

DedPeter [7]

Complete Question

One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a small patch of light on the far wall. Having recently studied optics in your physics class, you're not too surprised to see that the patch of light seems to be a circular diffraction pattern. It appears that the central maximum is about 2 cm across, and you estimate that the distance from the window shade to the wall is about 5 m.

Required:

Estimate the diameter of the pinhole.

Answer:

The diameter is $d =0.000336 m$

Explanation:

From the question we are told that

The central maxima is $D= 2cm = \frac{2}{100} = 0.02m$

The distance from the window shade is $L = 5m$

The average wavelength of the sun is mathematically evaluated as

$\lambda_{ave } = \frac{\lambda_i + \lambda_f}{2}$

Generally the visible light spectrum has a wavelength range between 400 nm to 700 nm

So the initial wavelength of the sun is $\lambda _i = 400nm$

and the final wavelength is $\lambda_f = 700nm$

Substituting this into the above equation

$\lambda_{sun} = \frac{400nm +700nm}{2}$

$= 550nm$

The diameter is evaluated as

$d = \frac{2.44 \lambda_{sun} L}{D}$

substituting values

$d = \frac{2.44 * 550*10^{-9} * 5 }{0.02}$

$d =0.000336 m$

5 0

3 years ago

PLEASE I NEED HELP I AM REALLY STUCK!!!

AlekseyPX

Answer: Force = Mass X Acceleration

F = 5 x 2

F = 10 N

8 0

3 years ago

3. A football is kicked with a speed of 35 m/s at an angle of 40°.

jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

$u_y = u sin \theta$

where

u = 35 m/s is the initial speed

$\theta=40^{\circ}$ is the angle of projection of the ball

Substituting into the equation, we find

$u_y = (35)(sin 40)=22.5 m/s$

b) 26.8 m/s

The initial horizontal velocity is given by:

$u_x = u cos \theta$

where

u = 35 m/s is the initial speed

$\theta=40^{\circ}$ is the angle of projection of the ball

Substituting into the equation, we find

$u_x = (35)(cos 40)=26.8 m/s$

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

$v_y = u_y + at$

where

$v_y$ is the vertical velocity at time t

$u_y = 22.5 m/s$

$a=g=-9.8 m/s^2$ is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, $v_y =0$ ; substituting, we find the time t at which this happens:

$0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s$

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

$s=u_y t + \frac{1}{2}gt^2$

where

s is the vertical displacement at time t

$u_y = 22.5 m/s$

$g=-9.8 m/s^2$

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

$s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m$

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

$t=2(2.30 s)=4.60 s$

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

$v_x = 26.8 m/s$

Therefore, the horizontal distance travelled during the whole motion is

$d=v_x t = (26.8)(4.60)=123.3 m$

So, the ball lands 123.3 m far from the initial point.

4 0

3 years ago

The volume flow rate in an artery supplying the brain is 4.00 10-6 m3/s. (a) If the radius of the artery is 4.50 10-3 m, determi

mr_godi [17]

Answer:

(a) 0.063 m/s

(b) 1.01 m/s

Explanation:

rate of volume flow, V = 4 x 10^-6 m^3/s

(a) radius, r = 4.5 x 10^-3 m

Let the speed of blood is v.

So, V = A x v

where A be the area of crossection of artery

4 x 10^-6 = 3.14 x 4.5 x 10^-3 x 4.5 x 10^-3 x v

v = 0.063 m/s

Thus, the speed of flow of blood is 0.063 m/s .

(b) Now r' = r / 4 = 4.5 /4 x 10^-3 m = 1.125 x 10^-3 m

Let the speed is v'.

So, V = A' x v'

4 x 10^-6 = 3.14 x 1.125 x 10^-3 x 1.125 x 10^-3 x v'

v' = 1.01 m/s

Thus, the speed of flow of blood is 1.01 m/s .

8 0

3 years ago

How would decreasing the volume of the reaction vessel affect each of the following equilibria?2NOBr(g)⇌2NO(g)+Br2(g)

mafiozo [28]

Answer:

The equilibrium position will shift towards the left hand side or reactants side

Explanation:

Decreasing the volume (increasing the pressure) of the system will shift the equilibrium position towards the lefthand side or reactants side. This is because, decreasing the volume (increasing the pressure) implies shifting the equilibrium position towards the side having the least number of moles.

There are two moles of reactants and a total of three moles of products(total). Hence decreasing the volume and increasing the pressure of the gas phase reaction will shift the equilibrium position towards the lefthand side.

5 0

3 years ago