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pychu [463]
3 years ago
10

You throw a snowball horizontally. It hits a target 2 meters away. You notice that it hit o.9m below where you aimed. With what

horizontal velocity did you throw the snowball?
Physics
1 answer:
ludmilkaskok [199]3 years ago
3 0

The snowball travels a horizontal distance <em>x</em> and has height <em>y</em> at time <em>t</em> according to

<em>x</em> = <em>vt</em>

<em>y</em> = 0.9 m - 1/2 <em>gt</em> ²

where <em>g</em> = 9.8 m/s² is the magnitude of the acceleration due to gravity.

Set <em>y</em> = 0 and solve for <em>t</em> :

0 = 0.9 m - 1/2 <em>gt</em> ²

<em>t</em> ² = (1.8 m) / <em>g</em>

<em>t</em> ≈ 0.43 s

The target is 2 m away from where the snowball is thrown, so it was thrown with speed <em>v</em> such that

2 m = <em>v </em>(0.429 s)

<em>v</em> = (2 m) / (0.43 s)

<em>v</em> ≈ 4.7 m/s

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What is the buoyant force on an object that weighs 340 N and is floating on a lake? if you could explain the answer that would b
Kryger [21]

You just said that the object is "floating".  

(As soon as you said that, a picture of a duck flashed through my mind.  But then I knew right away that the duck could not be an accurate representation of the situation you're describing.  340 N would be <u><em>some duck</em></u> ... about 76 pounds ... and that duck would have been caught and eaten a long time ago. I mean ... what could a 76-pound duck do ?  Could it fly away ?  Could it run away ? ?  Not likely.)

So it's not a duck, but whatever it is, it's just sitting there on the water, floating.  What's important is that it's <u><em>not accelerating</em></u> up or down.  THAT tells us that the vertical forces on it are balanced so that there's NO NET vertical force on it at all.

What are the vertical forces on it ?  There's gravity, pulling it DOWN with a force of 340 N, and there's buoyancy, pushing it UP.  The SUM of those two forces must be <em>zero</em> ... otherwise the object would be accelerating up or down.

It's not.  So (gravity) + (buoyancy) must add up to zero.

The buoyant force on the object is <em>340 N UPward.</em>

5 0
3 years ago
In case A below, a 1 kg solid sphere is released from rest at point S. It rolls without slipping down the ramp shown, and is lau
mestny [16]

Answer:

the block reaches higher than the sphere

\frac{y_{sphere}} {y_block} = 5/7

Explanation:

We are going to solve this interesting problem

A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp

Let's use the concept of conservation of energy

starting point. At the top of the ramp

         Em₀ = U = m g y₁

final point. At the exit of the ramp

         Em_f = K + U = ½ m v² + ½ I w² + m g y₂

notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂

energy is conserved

          Em₀ = Em_f

         m g y₁ = ½ m v² + ½ I w² + m g y₂

angular and linear velocity are related

        v = w r

the moment of inertia of a sphere is

         I = \frac{2}{5} m r²

we substitute

         m g (y₁ - y₂) = ½ m v² + ½ (\frac{2}{5} m r²) (\frac{v}{r})²

         m g h = ½ m v² (1 + \frac{2}{5})

where h is the difference in height between the two sides of the ramp

h = y₂ -y₁

         mg h = \frac{7}{5} (\frac{1}{2} m v²)

         v = √5/7  √2gh

This is the exit velocity of the vertical movement of the sphere

         v_sphere = 0.845 √2gh

B) is the same case, but for a box without friction

   starting point

          Em₀ = U = mg y₁

   final point

          Em_f = K + U = ½ m v² + m g y₂

          Em₀ = Em_f

          mg y₁ = ½ m v² + m g y₂

          m g (y₁ -y₂) = ½ m v²

          v = √2gh

this is the speed of the box

          v_box = √2gh

to know which body reaches higher in the air we can use the kinematic relations

          v² = v₀² - 2 g y

at the highest point v = 0

           y = vo₀²/ 2g

for the sphere

           y_sphere = 5/7 2gh / 2g

           y_esfera = 5/7 h

for the block

           y_block = 2gh / 2g

            y_block = h

       

therefore the block reaches higher than the sphere

         \frac{y_{sphere}} {y_bolck} = 5/7

3 0
3 years ago
72.0-kg person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface of the door. The doorknob is locat
expeople1 [14]

Answer:I=2 kg-m^2

Explanation:

Given

mass m=72 kg

Force F=5 N

door knob is located at a distance of r=0.8 m from axis

Angular acceleration of door \alpha =2 rad/s^2

Torque T=I\alpha =F\times r

where I=moment of inertia

5\times 0.8=I\times 2

I=2 kg-m^2

4 0
3 years ago
A bag of sugar weighs 5.00 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration is one-sixt
gizmo_the_mogwai [7]

Answer:

Earth: 22.246 N

Moon: 3.71 N

Jupiter: 58.72 N

Explanation:

The mass of an object will remain constant in any location, its weight however, can fluctuate depending on its location. For example, a golf ball will weigh less on the moon, but its mass will not be different if it was on earth.

To calculate anything, we need to convert to standard measurements.

5.00 lbs = 2.27 kg

On earth, gravity is measured to be 9.8 m/s², so the weight in Newtons on Earth would be: (2.27 kg) x (9.8 m/s²) = 22.246 N

Repeated on the moon where gravity is (9.8 m/s²) x (1/6) = 1.633 m/s², so the weight in Newtons on the moon would be: (2.27 kg) x (1.633 m/s²) = 3.71 N

Repeated on Jupiter where gravity is (9.8 m/s²) x (2.64) = 25.87 m/s², so the wight in Newtons on Jupiter would be: (2.27 kg) x (25.87 m/s²) = 58.72 N

3 0
3 years ago
A balloon is filled with helium at a pressure of 2.4 x 105 Pa. The balloon is at a
garri49 [273]
1)
p = 2.4 * 10^5 Pa
T = 18° C + 273.15 = 291.15 k
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Use ideal gas equation: pV = nRT => n = pV / RT = [2.4*10^5 Pa * 0.06545 m^3] / [8.31 J/k*mol * 291.15k] = 6.492 mol

Avogadro number = 1 mol = 6.022 * 10^23 atoms

Number of atoms = 6.492 mol * 6.022 *10^23 atom/mol = 39.097 * 10^23 atoms = 3.91 * 10^24 atoms

2) Double atoms => double volume

V2 / V1 = r2 ^3 / r1/3

2 = r2 ^3 / r1 ^3 => r2 ^3 = 2* r1 ^3

r2 = [∛2]r1

The factor is ∛2
5 0
3 years ago
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