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Orlov [11]
3 years ago
5

An object of mass m is attached to a horizontal spring, stretched to a displacement A from equilibrium and released, undergoing

harmonic oscillations on a frictionless surface with period T0. The experiment is then repeated with a mass of 4m. The subsequent total mechanical energy of the object with mass 4m is which of the following when compared to the original total mechanical energy?
a. greater than
b. less than
c. equal to
Physics
1 answer:
astraxan [27]3 years ago
6 0

Answer:

a

Explanation:

filler filler filler filler

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Find the slit separation of a double-slit arrangement that will produce interference fringes 0.018 rad apart on a distant screen
inessss [21]

Answer:

1.64 * 10^(-5) m

Explanation:

Parameters given:

Angular separation, θ = 0.018 rad

Wavelength, λ = 589 nm = 5.89 * 10^(-7) m

The angular separation when there are 2 slots is given as

θ = λ/2d

where d = separation between slits

d = λ/2θ

d = (589 * 10^(-9))/(2 * 0.018)

d = 1.64 * 10^(-5) m

5 0
3 years ago
SCIENCE HELP! PLEASE HELP URGENT!
Nonamiya [84]
It is also tripled, there is a rule to everything, whatever you do to one thing, you do the exact thing to the other. Hope this solves it :)
6 0
3 years ago
A circuit consists of one 330 ohm resistor in series with two other resistors (470 ohms and 220 ohms) connected in parallel. Wha
umka21 [38]

Answer:

power drain on an ideal battery, P = 0.017 W

Given:

R_{1} = 330\ohm

R_{2} = 470\ohm

R_{3} = 220\ohm

Since, R_{2} = 470\ohm and R_{3} = 220\ohm are in parallel and this combination is in series with R_{1} = 330\ohm, so,

Equivalent resistance of the circuit is given by:

R_{eq} = \frac{R_{2}R_{3}}{R_{2} + R_{3}} + R_{1}

R_{eq} = \frac{470\times 220}{470 + 220} + 330

R_{eq} = 149.85 + 330 = 479.85 \ohm

power drain on an ideal battery, P = \frac{V^{2}}{R_{eq}}

                                                      P = \frac{2.9^{2}}{479.85}

                                                      P = 0.017 W

4 0
4 years ago
Calculate the work done by a force of 30 N in lifting a load of 2 kg to a height of 10 m (g = 10 m/s2)
Mars2501 [29]

Answer:

<u>300 J</u>

Explanation:

Given :

  1. Applied force = 30 N
  2. Mass = 2 kg
  3. Height = 10 m
  4. g = 10 m/s²

Work done by the force :

  • Work done = Force x Displacement
  • Work done = mg x h
  • Work done = 30 N x 10 m
  • Work done = <u>300 J</u>

<u></u>

<u><em>Note</em></u> :

  • What you have calculated is the work done by gravitational force on the object (that too, incorrectly)
  • But in the end, it asks for work done by the force of 30N
  • Hence, the given answer ~
3 0
3 years ago
A system consists of a disk rotating on a frictionless axle and a piece of clay moving toward it, as shown in the figure above.
cestrela7 [59]

Explanation:

Forgive me If I am wrong, it's been a while since I've studied Torque.

The formula for the angular momentum is

momentum= I*w.

We can also write I*W as 1/2MR^2 * W so the extra mass coming from the block of clay would most likely cause the angular momentum to increase from the amount it was before.

6 0
2 years ago
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