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Yuri [45]
4 years ago
6

A dart is loaded into a spring-loaded toy dart gun by pushing the spring in by a distance d. For the next loading, the spring is

compressed a distance 8d. How much work is required to load the second dart compared to that required to load the first?
Physics
1 answer:
Aliun [14]4 years ago
8 0

Answer:

the work required for the loading of second dart is 64 times greater as work required for loading the first dart.

Explanation:

k = spring constant of the spring loaded toy dart gun

x₁ = compression of spring to load the first dart = d

x₂ = compression of spring to load the second dart = 8 d

E₁ = Work required to load the first dart

E₂ = Work required to load the second dart

Work required to load the first dart is given as

E₁ = (0.5) k x₁² = (0.5) k d²

Work required to load the second dart is given as

E₂ = (0.5) k x₂² = (0.5) k (8d)² = (64) (0.5) k d²

E₂ = 64 E₁

So the work required for the loading of second dart is 64 times greater as work required for loading the first dart

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in an experiment the following readings were observed volume of alcohol flowing per minute equals to 10 raise to power - 5 cube
tekilochka [14]

Answer:

The viscosity is 1.30 x 10^-3 deca poise.

Explanation:

Volume per minute, V = 10^-5 m^3

Volume per second, V = 1.67 x 10^-7 m^3

density, d = 800 kg/m^3

radius, r = 0.05 cm

Length, L = 0.5 m

Height, h = 60 cm

Pressure, P = h d g = 0.6 x 800 x 9.8 = 4704 Pa

Use the formula  of rate of flow

V = \frac{\pi p r^4}{8\eta L}\\\\1.67\times 10^{-7}\times8\times \eta\times 0.5 =  3.14\times 4707\times (0.05\times 10^{-2})^4\\\\\eta = 1.38\times 10^{-3} deca poise

6 0
3 years ago
A 170 g air-track glider is attached to a spring. The glider is pushed in 11.2 cm against the spring, then released. A student w
4vir4ik [10]

Answer:k = 10.83 N/m²

Explanation: The angular frequency (ω), spring constant (k) and mass is related by the formulae below

ω = √k/m

But ω = 2πf, where f = frequency.

f = number of oscillations /time taken

Number of oscillations = 14, time taken = 11s

f = 14/11 = 1.27Hz.

ω = 2×22/7×1.27

ω = 7.98 rad/s.

By substituting this parameters into ω = √k/m

Where ω = 7.98rad/s, m = 170g = 170/1000 = 0.17kg.

7.98 = √k/0.17

By squaring both sides

(7.98)² = k/ 0.17

k = (7.98)² × 0.17

k = 10.83 N/m²

7 0
4 years ago
What happens when an electron moving from the 3rd energy level to the 1st energy level?
Sholpan [36]

Answer:

A photon of wavelength 103 nm is released

Explanation:

When an electron in an atom jumps from a higher energy level to a lower energy level, it releases a photon whose energy is equal to the difference in energy between the two levels.

For example, if we are talking about a hydrogen atom, the energy of the levels are:

E_1 = -13.6 eV\\E_2 = -3.4 eV\\E_3 = -1.5 eV

So, the energy of the photon released when the electron jumps from the level n=3 to n=1 is

\Delta E = E_3 - E_1 = -1.5 -(-13.6)=12.1 eV

In Joules,

\Delta E =12.1\cdot 1.6\cdot 10^{-19} = 1.94\cdot 10^{-18} J

We can also find the wavelength of this photon, using the equation:

\Delta E = \frac{hc}{\lambda}\rightarrow  \lambda=\frac{hc}{\Delta E}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{1.94\cdot 10^{-18}}=1.03\cdot 10^{-7} m = 103 nm

7 0
4 years ago
Current what can do work in an electric device
Serggg [28]
It could help transport important information
5 0
3 years ago
If the bar magnet is flipped over and the south pole is brought near the hanging ball, the ball will be?
disa [49]

The ball may attracted to the magnet.

<h3>How can we understand that the hanging ball will be attracted to the magnet or not?</h3>
  • From the question, we understand that the ball is attracted by the north pole of the bar magnet, then the bar magnet flipped over and the south pole is brought near the hanging ball.
  • As we know, in this type of experiments of bar magnet most of the times the ball is made out of steel.
  • Steel is a magnetic material.
  • Magnetic materials gets attracted  to the magnet at both the North and South pole.
  • This can be compared to how neutral objects also gets attracted to the positively and negatively charged rods through the Polarization force.

So, If the bar magnet is flipped over and the south pole is brought near the hanging ball, The ball will be attracted to the magnet.

Learn more about the bar magnet:

brainly.com/question/27943723

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