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3 years ago

PLEASE HELP WILL GIVE BRAINLIEST!!!

Physics

1 answer:

Elan Coil [88]3 years ago

6 0

The spring constant is 181.0 N/m

Explanation:

We can solve the problem by applying the law of conservation of energy. In fact, the elastic potential energy initially stored in the compressed spring is completely converted into gravitational potential energy of the dart when the dart is at its maximum height. Therefore, we can write:

$\frac{1}{2}kx^2 = mgh$

where the term on the left represents the elastic potential energy of the spring while the term on the right is the gravitational potential energy of the dart at maximum height, and where

k is the spring constant of the spring

x = 2.08 cm = 0.0208 m is the compression of the spring

m = 12.3 g = 0.00123 kg is the mass of the dart

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 3.25 m is the maximum height of the dart

Solving for k, we find:

$k=\frac{2mgh}{x^2}=\frac{2(0.00123)(9.8)(3.25)}{(0.0208)^2}=181.0 N/m$

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A farmer had 15 sheep, and all but 8 died. How many are left?

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A 0.290 cm diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 30.0 pC charge on it

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Answer:

Therefore,

The potential (in V) near its surface is 186.13 Volt.

Explanation:

Given:

Diameter of sphere,

d= 0.29 cm

$radius=\dfrac{d}{2}=\dfrac{0.29}{2}=0.145\ cm$

$r = 0.145\ cm = 0.145\times 10^{-2}\ m$

Charge ,

$Q = 30.0\ pC=30\times 10^{-12}$

To Find:

Electric potential , V = ?

Solution:

Electric Potential at point surface is given as,

$V=\dfrac{1}{4\pi\epsilon_{0}}\times \dfrac{Q}{r}$

Where,

V= Electric potential,

ε0 = permeability free space = 8.85 × 10–12 F/m

Q = Charge

r = Radius

Substituting the values we get

$V=\dfrac{1}{4\times 3.14\times 8.85\times 10^{-12}}\times \dfrac{30\times 10^{-12}}{0.145\times 10^{-2}}$

$V=\dfrac{30}{16.117\times 10^{-2}}=186.13\ Volt$

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The potential (in V) near its surface is 186.13 Volt.

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3 years ago

A sheet of glass is coated with a 500-nm-thick layer of oil (n = 1.42).

tamaranim1 [39]

Answer:

a) 473.33 nm 

b) 568 nm and 406 nm

c) bluish green and blue

Explanation:

a) As the light traverses the layer of oil it first reflects at the front surface of the oil. Here the index of refraction increases from that of air to that of the oil , so a phase change occurs. The light then reflects from the rear surface of oil. The index of refraction increases from that of the oil to that of the glass , so again a phase change occurs. Thus two phase changes occur.

In thin-film interference with 0 or 2 phase changes, condition for constructive interference is:

2t=mλ/n

So:

λ= 2tn/m

For m=1

λ=1420 nm

For m=2

λ=710 nm

For m=3

λ=473.33 nm

For m=4

λ=355 nm

Thus the only wavelength in the visible spectrum (400 - 700 nm) that will give constructive interference is 473.33 nm 

In thin-film interference with 0 or 2 phase changes, condition for destructive interference is:

2t=(m+1/2)λ/n=(2m+1)*λ/2n

so;

λ=4tn/(2m+1)

For m=1

λ=946.667 nm

For m=2

λ=568 nm

For m=3

λ=405.33 nm

For m=4

λ=315.56 nm

Thus the wavelengths in the visible spectrum (400 to 700 nm) that will give destructive interference are 568 nm and 406 nm

c) The color of reflected light is bluish green since the wavelength is 473.3 nm . We know that the colors of reflected and transmitted light are complimentary to each other.Thus the color of transmitted light is blue (due to the combination of wavelengths 568 nm (green) and 406 nm (deep violet).

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3 years ago

Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Expl

professor190 [17]

Complete question:

A force F is applied to the block as shown (check attached image). With an applied force of 1.5 N, the block moves with a constant velocity.

Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain

Answer:

The applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N

Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the initially applied force.

Explanation:

Given;

magnitude of applied force, F = 1.5 N

Apply Newton's second law of motion;

F = ma

$F = m(\frac{v}{t} )\\\\F = \frac{m}{t} v\\\\Let \ \frac{m}{t} \ be \ constant = k\\F = kv\\\\k = \frac{F}{v} \\\\\frac{F_1}{v_1} = \frac{F_2}{v_2}$

The applied force needed to keep the box moving with a constant velocity that is twice as fast as before;

$\frac{F_1}{v_1} = \frac{F_2}{v_2} \\\\(v_2 = 2v_1, \ and \ F_1 = 1.5N)\\\\\frac{1.5}{v_1} = \frac{F_2}{2v_1} \\\\1.5 = \frac{F_2}{2}\\\\F_2 = 2*1.5\\\\F_2 = 3 N$

Therefore, the applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N

Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the applied force.

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3 years ago