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damaskus [11]
2 years ago
8

Which type of material uses the subtractive coloring process?

Physics
1 answer:
Triss [41]2 years ago
6 0
I think that the answer is A
You might be interested in
A pendulum swings back and forth 5 times in 10 seconds what is the period of the pendulum?
Ainat [17]
Period is T = 1/f.  The frequency, f, is 5cycles/10s = 0.5.  So the period T=1/0.2 = 2.
4 0
3 years ago
What is the Voltage of a circuit that has a resistance of 50 and a current of 2 A?
Nutka1998 [239]

Answer:

It's actually 100V

Explanation:

To answer this, you'd need to use Ohm's law, which states that Voltage is equivalent to the product of Current and Resistance (V=IR).

V = 50 × 2

V = 100

Therefore the voltage is a 100V.

None of the answers you have listed match this though, so I'm guessing it either must be typo or the question itself is wrong.

3 0
3 years ago
03: A mass with a 60 g vibrate at the end of a spring. The amplitude of the motion is 0.394 ft
Flauer [41]

Answer:

a) T = 1.69 s, b)  k = 0.825 N / m, c)  v = 1.46 feet/s, d) a = 5.41 ft / s²,

e)   v = - 1,319 ft / s,    a = - 2.70 ft / s², f) K = 4.8 10⁻³ J, U = 1.49 10⁻³ J

Explanation:

In a mass-spring system with simple harmonic motion, the angular velocity is

         w = \sqrt{\frac{k}{m} }

a) find the period

angular velocity, frequency, and period are related

         w = 2π f = 2π / T

          f = 1 / T

          T = 1 / f

           T = 1 / 0.59

           T = 1.69 s

b) the spring constant

         w = 2π f

         w = 2π 0.59

         w = 3.70 rad / s

         w² = k / m

          k = w² m

          k = 3.70² 0.060

          k = 0.825 N / m

c) the maximum speed

simple harmonic movement is described by the expression

          x = A cos (wt + Ф)

speed is defined by

         v =\frac{dx}{dt}

          v = -A w sin (wt + fi)

the speed is maximum when the cosine is ± 1

          v = A w

          v = 0.394 3.70

          v = 1.46 feet/s

d) maximum acceleration

            a = \frac{dv}{dt}

            a = - A w² cos wt + fi

the acceleration is maximum when the cosine is ±1

            a = A w²

            a = 0.394 3.70²

            a = 5.41 ft / s²

e) velocity and acceleration for x = 6 cm

let's reduce the cm to feet

            x = 6 cm (1 foot / 30.48 cm) = 0.1969 foot

Before doing this part we must find the phase angle (Ф), the most common way to start the movement is to move the spring a small distance and release it, so its initial speed is zero for t = 0 s

let's use the expression for the velocity

           v = -A w sin (0 + Фi)

           0 = - A w sin Ф

so sin Ф = 0 which implies that Фi = 0

the equation of motion is

            x = A cos wt

            x = 0.394 cos 3.70t

we substitute

           0.1969 = 0.394 cos 370t

           3.70 t = cos⁻¹ (0.1969 / 0.394)

let's not forget that the angle is in radians

           3.70, t = 1.047

           t = 1.047 / 3.70

           t = 0.2826 s

we substitute this time in the equation for velocity and acceleration

           v = - Aw sin wt

           v = - 0.394 3.70 sin 3.70 0.2826

           v = - 1,319 ft / s

           a = - A w² cos wt

           a = - 0.394 3.70² cos 3.70 0.2826

           a = - 2.70 ft / s²

f) the kinetic and potential energy at this point

           K = ½ m v²

let's slow down to the SI system

           v = 1.319 ft / s (1 m / 3.28 ft) = 0.402 m / s

           

           K = ½ 0.060 0.402²

           K = 4.8 10⁻³ J

           U = ½ k x²

           U = ½ 0.825 0.06²

           U = 1.49 10⁻³ J

5 0
2 years ago
A proton is moving at 425 m/s. (a) How much work must be done on it to stop it? (A proton has a mass of 1.67×10−27 kg.) (b) Assu
Tpy6a [65]

Answer:

a)1.51*10^-22joules b) 1.89*10^-7m

Explanation:

Work done to stop the proton = the kinetic energy of the proton = 1/2 mv^2 = 1/2* 1.67*10^-27* 425*425 = 1.51* 10 ^ -22 joules

b) net force acting to stop the proton = 8.01*10^-16

Work done needed to stop the proton = net force acting opposite the motion * distance

Distance covered = need work done/ net force

Distance = 1.51*10^-22/8.01*10^-16= 1.89*10^-7m

8 0
3 years ago
If your chunk of gold weighed 1 N in which case would you have the largest mass of gold?
kotykmax [81]
Ah ha !  Very interesting question.
Thought-provoking, even.

You have something that weighs 1 Newton, and you want to know 
the situation in which the object would have the greatest mass.

          Weight = (mass) x (local gravity)

          Mass  =  (weight) / (local gravity)

          Mass  =  (1 Newton) / (local gravity)

"Local gravity" is the denominator of the fraction, so the fraction
has its greatest value when 'local gravity' is smallest.  This is the
clue that gives it away.

If somebody offers you 1 chunk of gold that weighs 1 Newton,
you say to him:

   "Fine !  Great !  Golly gee, that's sure generous of you.  
But before you start weighing the chunk to give me, I want you
to take your gold and your scale to Pluto, and weigh my chunk
there.   And if you don't mind, be quick about it."

The local acceleration of gravity on Pluto is  0.62 m/s² ,
but on Earth, it's 9.81 m/s.

So if he weighs 1 Newton of gold for you on Pluto, its mass will be
1.613 kilograms, and it'll weigh 15.82 Newtons here on Earth. 

That's almost 3.6 pounds of gold, worth over $57,000 !


It would be even better if you could convince him to weigh it on
Halley's Comet, or on any asteroid.  Wherever he's willing to go
that has the smallest gravity.  That's the place where the largest
mass weighs 1 Newton.

3 0
3 years ago
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