If the volume of a hot air balloon remains constant, what happens as the temperature of the air inside the balloon increases?

You wad up a piece of paper and throw it into the wastebasket. How far will

astraxan [27]

Answer:

Since the paper is wadded up tight, and if there's any

air resistance left we assume there isn't any, it might

just as well be a stone that's tossed. This is just a

stripped down projectile situation.

You said "an angle of 36 degrees", but you didn't say relative

to what. I'll assume that it's 36 degrees above horizontal, and

now I'll proceed to answer the question with the information that

I just gave myself.

-- The vertical component of the velocity is 1.4 sin(36)

= 0.823 m/s up.

-- The projectile rises for (0.823/9.8) second, runs out of gas,

and then falls for another (0.823/9.8) second to its original height.

So it's in the air for

2 (0.823/9.8) = 0.168 second

(not very long at all)

-- The horizontal component of the velocity is 1.4 cos(36)

= 1.133 m/s

and it doesn't change.

-- During the 0.168 second that it's in the air,

the wad travels horizontally

(0.168 s) x (1.133 m/s)

= 0.19 meter

(19 cm, ~ 7.5 inches)

If you find my mistake on this one, please please tell me.

As of now, it looks like with that velocity at that angle, your

paper wad only makes it 7.5 inches from your hand into the can.

Explanation:

6 0

3 years ago

Read 2 more answers

A box is pulled to the right with a force of 65 N at an angle of 58 degrees to the horizontal. The surface is frictionless. The

Citrus2011 [14]

The free-body diagram is missing, but I assume the only forces acting on the box are the force F pushing the box, the weight of the object and the normal reaction of the surface.

Since the weight and the normal reaction acts in the vertical (y) direction, the only force acting on the box in the horizontal (x) direction is the horizontal component of the force F, which is given by
$F_x = F \cos 58^{\circ} = (65 N)(\cos 58^{\circ} )=34.4 N$
And so this is the net force in the x-direction.

5 0

3 years ago

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The valu

natulia [17]

Answer:

$\% Error = 2.6\%$

Explanation:

Given

$x: 1.54, 1.53, 1.44, 1.54, 1.56, 1.45$

Required

Determine the percentage error

First, we calculate the mean

$\bar x = \frac{\sum x}{n}$

This gives:

$\bar x = \frac{1.54+ 1.53+ 1.44+ 1.54+ 1.56+ 1.45}{6}$

$\bar x = \frac{9.06}{6}$

$\bar x = 1.51$

Next, calculate the mean absolute error (E)

$|E| = \sqrt{\frac{1}{6}\sum(x - \bar x)^2}$

This gives:

$|E| = \sqrt{\frac{1}{6}*[(1.54 - 1.51)^2 +(1.53- 1.51)^2 +.... +(1.45- 1.51)^2]}$

$|E| = \sqrt{\frac{1}{6}*0.0132}$

$|E| = \sqrt{0.0022}$

$|E| = 0.04$

Next, calculate the relative error (R)

$R = \frac{|E|}{\bar x}$

$R = \frac{0.04}{1.51}$

$R = 0.026$

Lastly, the percentage error is calculated as:

$\% Error = R * 100\%$

$\% Error = 0.026 * 100\%$

$\% Error = 2.6\%$

4 0

3 years ago

A 25 MGD surface water drinking plant has four circular clarifiers (aka sedimentation basin), operated in parallel with each bas

dangina [55]

Answer:

can y

Explanation:

jj

8 0

3 years ago

Starting from rest, a boulder rolls down a hill with constant acceleration and travels 3.00m during the first second.

Alexus [3.1K]

Answer:

a) 9.00 m b) 6.00 m/s c) 12.00 m/s

Explanation:

a) If the acceleration is constant, and we know that the displacement during the first second was 3.00 m, as the boulder (assumed that we can treat it as a point mass) started from rest, we can say the following:

Δx = $\frac{1}{2}*a*t^{2}$ = 3.00 m

As t = 1 s, replacing in the expression above, and solving for a, we have:

$a = \frac{2*3.00m}{1s2}$ = 6.00 m/s²

In order to know how far it travels during the second second, we need to know the value of the speed after the first second, as it is the initial velocity when the second second begins:

vf = v₀ + a*t ⇒ vf = 0 + 6 m/s²*1s = 6.00 m/s

The total displacement, during the second second, will be as follows:

Δx = v₀*t + $\frac{1}{2}*a*t^{2}$ = 6.00m/s*1s + $\frac{1}{2}*6.00 m/s2*1s^{2} = 9.00 m$

⇒ Δx = 9.00 m

b) At the end of the first second, the final speed can be obtained as follows:

vf = v₀ + a*t ⇒ vf = 0 + 6 m/s²*1s = 6.00 m/s

c) At the end of the second second, the final speed can be obtained as follows:

vf = v₀ + a*t ⇒ vf = 0 + 6 m/s²*2s = 12.00 m/s

3 0

3 years ago