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3 years ago

What is the net displacement of the particle between 0 seconds and 80 seconds?

1 answer:

7 0

After a thorough research, there exists the same question that has choices and the link of the graph (http://i37.servimg.com/u/f37/16/73/53/52/graph410.png)

<span>Choices:
A. 160 meters
B. 80 meters
C. 40 meters
D. 20 meters
E. 0 meters
</span>
The correct answer is letter E. 0 meters.

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A swimming pool has the shape of a right circular cylinder with radius 21 feet and height 10 feet. Suppose that the pool is full

AysviL [449]

Answer:

The water required to pump all the water to a platform 2 feet above the top of the pool is is 6061310.32 foot-pound.

Explanation:

Given that,

Radius = 21 feet

Height = 10 feet

Weighing = 62.5 pounds/cubic

Work = 4329507.37572

Height = 2 feet

Let's look at a horizontal slice of water at a height of h from bottom of pool

We need to calculate the area of slice

Using formula of area

$A=\pi r^2$

Put the value into the formula

$A=\pi\times21^2$

$A=441\pi\ feet^2$

Thickness of slice $t=\Delta h\ ft$

The volume is,

$V=(441\pi\times\Delta h)\ ft^3$

We need to calculate the force

Using formula of force

$F=W\times V$

Where, W = water weight

V = volume

Put the value into the formula

$F=62.5\times(441\pi\times\Delta h)$

$F=27562.5\pi\times\Delta h\ lbs$

We need to calculate the work done

Using formula of work done

$W=F\times d$

Put the value into the formula

$W=27562.5\pi\times\Delta h\times(10-h)\ ft\ lbs$

We do this by integrating from h = 0 to h = 10

We need to find the total work,

Using formula of work done

$W=\int_{0}^{h}{W}$

Put the value into the formula

$W=\int_{0}^{10}{27562.5\pi\\times(10-h)}dh$

$W=27562.5\pi(10h-\dfrac{h^2}{2})_{0}^{10}$

$W=27562.5\pi(10\times10-\dfrac{100}{2}-0)$

$W=4329507.37572$

To pump 2 feet above platform, then each slice has to be lifted an extra 2 feet,

So, the total distance to lift slice is (12-h) instead of of 10-h

We need to calculate the water required to pump all the water to a platform 2 feet above the top of the pool

Using formula of work done

$W=\int_{0}^{h}{W}$

Put the value into the formula

$W=\int_{0}^{10}{27562.5\pi\\times(12-h)}dh$

$W=27562.5\pi(12h-\dfrac{h^2}{2})_{0}^{10}$

$W=27562.5\pi(12\times10-\dfrac{100}{2}-0)$

$W=1929375\pi$

$W=6061310.32\ foot- pound$

Hence, The water required to pump all the water to a platform 2 feet above the top of the pool is is 6061310.32 foot-pound.

8 0

3 years ago

REMARKS The speed found in part (a) is the same as if the woman fell vertically through a distance of 21.9 m. The result of part

sasho [114]

Answer:

Yes, if the system has friction, the final result is affected by the loss of energy.

Explanation:

The result that you are showing is the conservation of mechanical energy between two points in the upper one, the energy is only potential and the lower one is only kinetic.

In the case of some type of friction, the change in energy between the same points is equal to the work of the friction forces

$W_{fr}$ = ΔEm

$W_{fr}$ = $Em_{f}$ -Em₀

As we can see now there is another quantity and for which the final energy is lower and therefore the final speed would be less than what you found in the case without friction.

$Em_{f}$ = $W_{fr}$ + Em₀

Remember that the work of the rubbing force is negative, let's write the work of the rubbing force explicitly, to make it clearer

½ m v² = -fr d + mgh

v = √(-fr d 2/m + 2 gh)

v = √ (2gh - 2fr d/m)

Now it is clear that there is a decrease in the final body speed.

Consequently, if the system has friction, the final result is affected by the loss of energy.

5 0

2 years ago

choose the correct answer 14: which of the following force follows the inverse square law of distance A: gravitaional force B: e

Delicious77 [7]

Answer:

C. both A and B

Explanation:

Sana nakatulong

8 0

3 years ago

What kind of model is this?

fredd [130]

Answer:

I think it's Experimental

Explanation:

it has atoms on it.

5 0

2 years ago

Read 2 more answers

A 3.9 g dart is fired into a block of wood with a mass of 24.6 g. The wood block is initially at rest on a 1.5 m tall post. Afte

Galina-37 [17]

Answer:

46.48m/s

Explanation:

The problem is a combination of the principle of conservation of linear momentum and projectile motion.

The principle of conservation of linear momentum states that in a closed system, the total momentum of colliding bodies before impact is equal to the total momentum after impact. The masses stated in the problem experienced an inelastic collision. In an inelastic collision, the bodies involved stick together after the collision and move with a common velocity.

For two bodies of masses $m_1$ and $m_2$ moving with velocities $u_1$ and $u_2$ before impact, if they experience inelastic collision, the conservation of their momenta is as stated in equation (1);

$m_1u_1+m_2u_2=(m_1+m_2)v..................(1)$

were v is their common velocity after impact. If the second mass $m_2$ was at rest before the impact, then its initial velocity $u_2=0m/s$ . therefore $m_2u_2=0$ . Equation (1) then becomes;

$m_1u_1=(m_1+m_2)v..............(2)$

In the problem stated, the second mass taken as the mass of the wooden block was at rest before the impact and the collision was inelastic since both the wood and the dart stuck together and moved with a common velocity after the impact. Therefore we can use equation (2) for the problem.

Given;

$m_1=3.9g=0.0039kg\\u_1=?\\m_2=24.6g=0.0246kg\\v=?$

Substituting these values into (2), we get the following;

$0.0039*u_1=(0.0039+0.024)v\\0.0039u_1=0.0285v.........(3)$

Their common v velocity after impact now makes both the wooden block and the dart (as a single body) to fall vertically through a height h of 1.5m over a range R of 3.5m as stated by the problem; hence by the principle of projectile motion for a body projected horizontally, the following relationship holds;

$R= vt............(4)$