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balu736 [363]
3 years ago
7

A simple harmonic oscillator consists of a spring of constant k = 100 N/m and a block of mass 2.0 kg. When t = 1.0 s, the positi

on and velocity of the block are x = 0.129 m and v = 3.415 m/s. (a) Findthe amplitude of the oscillations? (b) Determinethe location (x) of the block at t = 0? (c) What is the velocity v(t) of the block at t = 0?

Physics
1 answer:
ludmilkaskok [199]3 years ago
5 0

Answer:

a) 0,18 m b) 0,18 m c) 0 m/s

Explanation:

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andre [41]
Opening the valve allows more of the pressurized material out. If the area is decreased, less of the pressurized material is released, and its force ends up more spread out, reducing the pressure. Opening the valve will increase volume of transfer of your liquid.
8 0
3 years ago
A 10 Ω resistor is connected to a 120-V ac power supply. What is the peak current through the resistor?
11111nata11111 [884]

Answer:

Peak current = 16.9 A

Explanation:

Given that

RMS voltage = 120 Volts

V_{rms} = 120 V

AC is connected across resistance

R = 10 ohm

now by ohm's law

V = i R

120 = i (10)

i_{rms} = \frac{120}{10} = 12 A

now peak value of current will be given as

i_{peak} = \sqrt{2} i_{rms}

i_{peak} = \sqrt2 (12) = 16.9 A

8 0
3 years ago
The driver of a train moving at 23m/s applies the breaks when it pases an amber signal. The next signal is 1km down the track an
pochemuha

Answer:

3.2 m/s

Explanation:

Given:

Δx = 1000 m

v₀ = 23 m/s

a = -0.26 m/s²

t = 76 s

Find: v

This problem is over-defined.  We only need 3 pieces of information, and we're given 4.  There are several equations we can use.  For example:

v = at + v₀

v = (-0.26 m/s²) (76 s) + (23 m/s)

v = 3.2 m/s

Or:

Δx = ½ (v + v₀) t

(1000 m) = ½ (v + 23 m/s) (76 s)

v = 3.3 m/s

Or:

v² = v₀² + 2aΔx

v² = (23 m/s)² + 2(-0.26 m/s²)(1000 m)

v = 3.0 m/s

Or:

Δx = vt − ½ at²

(1000 m) = v (76 s) − ½ (-0.26 m/s²) (76 s)²

v = 3.3 m/s

As you can see, you get slightly different answers depending on which variables you use.  Since 1000 m has 1 significant figure, compared to the other variables which have 2 significant figures, I recommend using the first equation.

8 0
3 years ago
Light of wavelength O is passed through a diffraction grating with N lines/meter and then lands on a screen a distance L from th
rodikova [14]

Condition for diffraction

dsin\theta = m\lambda

Where

a = Distance between slits

m = Order of the fringes

\lambda = Wavelength

\theta = At the angle between the ray of light and the projected distance perpendicular between the two objects

For small angles

sin\theta = \approx tan\theta

Where

tan\theta = \frac{Y}{L}

Where L is the distance between the slits and Y the length of the light.

Replacing we have

d\frac{Y}{L} = \lambda m

Y = \frac{m\lambda L}{d}

The distance between slits d can be expressed also as d= \frac{L}{N} Where N is the number of the fringes, then

Y_n = mN\lambda L

Similarly when there is added a new Fringe we have the change of the distance would be :

Y_{n+1} = (m+1)N\lambda L

Linear distance between fringes is

\Delta Y = \Delta Y_{m+1}-Y_m

\Delta Y = (m+1)N\lambda L - mN\lambda L

Therefore the answer is

\Delta Y = N\lambda L

8 0
3 years ago
A 58-kg boy swings a baseball bat, which causes a 0.140-kg baseball to move toward 3rd base with a velocity of 38.0 m/s.
Sophie [7]

Answer:
101 J
Explanation:
The kinetic energy of an object is given by

where
m is the mass of the object
v is its speed
For the baseball in this problem,
m = 0.140 kg
v = 38.0 m/s
Substituting into the equation, we find the kinetic energy of the ball:

4
answers left
3 0
3 years ago
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