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alexandr1967 [171]

3 years ago

8

In an experiment, you determined the density of the wood to be 0.45g/cc, whereas the standard value was 0.47g/cc. Determine the

percentage difference. [Hint: Look at the procedure section of Part Al[1 Point] 5. 6. How do you determine y-intercept from a graph? [1 Point]

1 answer:

Zielflug [23.3K]3 years ago

7 0

Answer:

Percentage difference is 4.25 %.

Explanation:

Standard value of the density of wood, $\rho_s=0.47\ g/cc$

Experimental value of the density of wood, $\rho_e=0.45\ g/cc$

We need to find the percentage difference on the density of wood. It is given by :

$\%=|\dfrac{\rho_s-\rho_e}{\rho_s}|\times 100$

$\%=|\dfrac{0.47-0.45}{0.47}|\times 100$

Percentage difference in the density of the wood is 4.25 %. Hence, this is the required solution.

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In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0m/s at a 40.0? angle from the horizontal.

HACTEHA [7]

A) Horizontal range: 16.34 m

B) Horizontal range: 16.38 m

C) Horizontal range: 16.34 m

D) Horizontal range: 16.07 m

E) The angle that gives the maximum range is $41.9^{\circ}$

Explanation:

A)

The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:

$s=u_y t + \frac{1}{2}at^2$ (1)

where

s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)

$u_y = u sin \theta$ is the initial vertical velocity, where

u = 12.0 m/s is the initial speed

$\theta=40.0^{\circ}$ is the angle of projection

So

$u_y=(12.0)(sin 40.0^{\circ})=7.7 m/s$

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (downward)

Substituting the numbers, we get

$-1.80 = 7.7t -4.9t^2\\4.9t^2-7.7t-1.80=0$

which has two solutions:

t = -0.21 s (negative, we ignore it)

t = 1.778 s (this is the time of flight)

The horizontal motion is instead uniform, so the horizontal range is given by

$d=u_x t$

where

$u_x = u cos \theta=(12.0)(cos 40^{\circ})=9.19 m/s$ is the horizontal velocity

t = 1.778 s is the time of flight

Solving, we find

$d=(9.19)(1.778)=16.34 m$

B)

In this second case,

$\theta=42.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 42.5^{\circ})=8.1 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.1t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.851 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 42.5^{\circ})=8.85 m/s$

So, the range of the shot is

$d=u_x t = (8.85)(1.851)=16.38 m$

C)

In this third case,

$\theta=45^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 45^{\circ})=8.5 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.5t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.925 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 45^{\circ})=8.49 m/s$

So, the range of the shot is

$d=u_x t = (8.49)(1.925)=16.34$ m

D)

In this 4th case,

$\theta=47.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 47.5^{\circ})=8.8 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.8t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.981 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 47.5^{\circ})=8.11 m/s$

So, the range of the shot is

$d=u_x t = (8.11)(1.981)=16.07 m$

E)

From the previous parts, we see that the maximum range is obtained when the angle of releases is $\theta=42.5^{\circ}$ .

The actual angle of release which corresponds to the maximum range can be obtained as follows:

The equation for the vertical motion can be rewritten as

$s-u sin \theta t + \frac{1}{2}gt^2=0$

The solutions of this quadratic equation are

$t=\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g}$

This is the time of flight: so, the horizontal range is

$d=u_x t = u cos \theta (\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g})=\\=\frac{u^2}{-2g}(1+\sqrt{1+\frac{2gs}{u^2 sin^2 \theta}})sin 2\theta$

It can be found that the maximum of this function is obtained when the angle is

$\theta=cos^{-1}(\sqrt{\frac{2gs+u^2}{2gs+2u^2}})$

Therefore in this problem, the angle which leads to the maximum range is

$\theta=cos^{-1}(\sqrt{\frac{2(-9.8)(-1.80)+(12.0)^2}{2(-9.8)(-1.80)+2(12.0)^2}})=41.9^{\circ}$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

8 0

3 years ago

Two parallel conducting plates are separated by 9.2 cm, and one of them is taken to be at a potential of zero volts.What is the

True [87]

Answer:

$E=54V/cm$

$\Delta V=496.8V$ between the plates.

Explanation:

The equation for change of voltage between two points separated a distance d inside parallel conducting plates (<em>which have between them constant electric field</em>) is:

$\Delta V=Ed$

So to calculate our electric field strength we use the fact that the potential 8.8 cm from the zero volt plate is 475 V:

$E=\frac{\Delta V}{d}=\frac{475 V}{8.8cm}=54V/cm$

And we use the fact that the plates are 9.2cm apart to calculate the voltage between them:

$\Delta V=Ed=(54V/cm)(9.2cm)=496.8V$

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15.One girl was involved in a bus crash. She told the police that she was seated in the last line of seats and that when the bus

Alona [7]

Answer: It is not likely.

Explanation:

When the bus is moving forward, all the objects inside of it also are moving forward.

Now, as the objects inside the buss are not fixed to the bus, if the bus suddenly stops the objects inside of it will keep moving forward, because of the conservation of the momentum, defined as the quantity of motion (Similar to when you are in a car and it suddenly stops, you can feel the forward impulse).

Then is not likely that, in a case where the bus stops suddenly, an object inside the bus flies backward in opposite direction to the previous movement of the bus.

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nikdorinn [45]

b. The sweater has a tendency to attract protons.

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