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Sergeu [11.5K]
3 years ago
10

base your answer to this question on the information below and on your knowledge of physics. A toy launcher that is used to laun

ch small plastic spheres horizontally contains a spring with a spring constant of 50. Newtons per meter. The spring is compressed a distance of 0.10 meter when the launcher is ready to launch a plastic sphere. The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the plastic sphere will be launched.
Physics
1 answer:
Nadya [2.5K]3 years ago
3 0

Answer: v = 2.24 m/s

Explanation: The <u>Law</u> <u>of</u> <u>Conservation</u> <u>of</u> <u>Energy</u> states that total energy is constant in any process and, it cannot be created nor destroyed, only transformed.

So, in the toy launcher, the energy of the compressed spring, called <u>Elastic</u> <u>Potential</u> <u>Energy (PE)</u>, transforms into the movement of the plastic sphere, called <u>Kinetic</u> <u>Energy (KE)</u>. Since total energy must be constant:

KE_{i}+PE_{i}=KE_{f}+PE_{f}

where the terms with subscript i are related to the initial of the process and the terms with subscript f relates to the final process.

The equation is calculated as:

\frac{1}{2}kx^{2}+0=0+\frac{1}{2}mv^{2}

\frac{1}{2}kx^{2}=\frac{1}{2}mv^{2}

\frac{1}{2}50(0.1)^{2}=\frac{1}{2}(0.1)v^{2}

v^{2}=\frac{50(0.1)^{2}}{0.1}

v=\sqrt{50(0.1)}

v=\sqrt{5}

v = 2.24

The maximum speed the plastic sphere will be launched is 2.24 m/s.

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Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if
klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:  

"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

F_{E}= K\frac{q_{1}.q_{2}}{d^{2}} (1)

Where:

F_{E}=60 N  is the electrostatic force

K=8.99(10)^{9} Nm^{2}/C^{2} is the Coulomb's constant  

q_{1} and q_{2} are the electric charges

d=3 m is the separation distance between the charges  

Then:

60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}} (2)

Isolating q_{1} and q_{2}:

q_{1}q_{2}=6(10)^{-8} C^{2} (3)

Now, if we keep the same charges but we decrease the distance to d_{1}=1 m, (1) is rewritten as:

F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}} (4)

Then, the new electrostatic force will be:

F_{E}= 539.4 N (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0
3 years ago
An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +44 ft/s2. After some time t1,
mash [69]

Answer:

a) t₁ = 4.76 s, t₂ = 85.2 s

b) v = 209 ft/s

Explanation:

Constant acceleration equations:

x = x₀ + v₀ t + ½ at²

v = at + v₀

where x is final position,

x₀ is initial position,

v₀ is initial velocity,

a is acceleration,

and t is time.

When the engine is on and the sled is accelerating:

x₀ = 0 ft

v₀ = 0 ft/s

a = 44 ft/s²

t = t₁

So:

x = 22 t₁²

v = 44 t₁

When the engine is off and the sled is coasting:

x = 18350 ft

x₀ = 22 t₁²

v₀ = 44 t₁

a = 0 ft/s²

t = t₂

So:

18350 = 22 t₁² + (44 t₁) t₂

Given that t₁ + t₂ = 90:

18350 = 22 t₁² + (44 t₁) (90 − t₁)

Now we can solve for t₁:

18350 = 22 t₁² + 3960 t₁ − 44 t₁²

18350 = 3960 t₁ − 22 t₁²

9175 = 1980 t₁ − 11 t₁²

11 t₁² − 1980 t₁ + 9175 = 0

Using quadratic formula:

t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22

t₁ = 4.76, 175

Since t₁ can't be greater than 90, t₁ = 4.76 s.

Therefore, t₂ = 85.2 s.

And v = 44 t₁ = 209 ft/s.

3 0
3 years ago
(1) A positive charge +3 C is separated from another positive charge of +5 C by a distance of 7m. What is the magnitude of the e
Aneli [31]

1. The magnitude of the electric force between the two charges is 2.8×10⁹ N (Option B)

2. The net charge on the molecule is -8×10⁻¹⁹ C (Option D)

3. The magnitude of the force between the charges is 16000 N (Option C)

4. The correct statement is: A neutral object has equal numbers of protons and electrons. (Option C)

<h3>1. How to determine the force</h3>
  • Charge 1 (q₁) = +3 C
  • Charge 2 (q₂) = +5 C
  • Electric constant (K) = 9×10⁹ Nm²/C²
  • Distance apart (r) = 7 m
  • Force (F) =?

F = Kq₁q₂ / r²

F = (9×10⁹ × 3 × 5) / (7)²

F = 2.8×10⁹ N

<h3>2. How to determine the net charge on the molecule</h3>
  • Electron = 223 electrons
  • Proton = 218 protons
  • Net Charge =?

Charge = Proton - Electron

Charge = 218 - 223

Charge = -5 electrons

But

1 electron = 1.6×10⁻¹⁹ C

Thus,

Net Charge = -5 × 1.6×10⁻¹⁹ C

Net Charge = -8×10⁻¹⁹ C

<h3>3. How to determine the force</h3>
  • Charge 1 (q₁) = 2×10⁻⁴ C
  • Charge 2 (q₂) = 8×10⁻⁴ C
  • Electric constant (K) = 9×10⁹ Nm²/C²
  • Distance apart (r) = 0.3 m
  • Force (F) =?

F = Kq₁q₂ / r²

F = (9×10⁹ × 2×10⁻⁴ × 8×10⁻⁴) / (0.3)²

F = 16000 N

<h3>4. What is a neutral object?</h3>

A neutral object is an object having equal numbers of protons and electrons. For example, an object with 4 protons and 4 electrons is said to be neutral as illustrated below

  • Electron = 4 electrons
  • Proton = 4 protons
  • Net Charge =?

Charge = Proton - Electron

Charge = 4 - 4

Charge = 0 (neutral)

Thus, the correct statement about neutral object, given in the question is: A neutral object has equal numbers of protons and electrons (Option C)

Learn more about Coulomb's law:

brainly.com/question/506926

#SPJ1

6 0
2 years ago
A circular pipe of 25-mm outside diameter is placed in an airstream at 25C and 1-atm pressure. The air moves in cross flow over
kifflom [539]

Answer:

f_D = =3.24 N/m

Explanation:

data given

properties of air\nu\ of air =19.31*10^{-6} m2/s

\rho = 1.048 kg/m3

k = 0.0288 W/m.K

WE KNOW THAT

Reynold's number is given as

Re =\frac{VD}{\nu}

      = \frac{ 15*0.025}{19.31*10^{-6}}

      = 1.941 *10^4

drag coffecient is given as

C_D = \frac{f_D}{A_f\frac{\rho v^2}{2}}

solving for f_D

f_D = C_D A_f*\frac{\rho v^2}{2}

     =C_D D*\frac{\rho v^2}{2}

Drag coffecient for smooth circular cylinder is 1.1

therefore Drag force is

f_D = 1.1*0.025 *\frac{1.048*15^2}{2}

f_D = =3.24 N/m

4 0
3 years ago
At what point in its path does a baseball that is hit to the outfield have its minimum vertical speed?
ioda

Answer:

At the point when it reaches its maximum height (answer C)

Explanation:

As soon as the ball leaves the bat is has not only horizontal component of the velocity, but also vertical, which allows it travel upwards in its trajectory.

When it reaches its maximum elevation, and the trajectory starts curving down, is when the velocity of the baseball is just horizontal (the vertical component has been reduced to zero due to the constant action of the acceleration of gravity which was slowing down it ascending pattern.

After that point, the ball starts gaining now vertical velocity being accelerated towards the center of the Earth due to gravity. At the maximum height of its path is when the vertical component of the velocity switches from positive (pointing up) to negative (pointing down) going through zero magnitude.

5 0
3 years ago
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