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Fed [463]
2 years ago
7

answer A vertical spring stretches 3.4 cm when a 8-g object is hung from it. The object is replaced with a block of mass 26 g th

at oscillates in simple harmonic motion. Calculate the period of motion.
Physics
2 answers:
victus00 [196]2 years ago
8 0

Answer:

0.695s

Explanation:

From Hooke's law, the restoring force is given has

F = -ky .......1

Where F is the force, y is the spring displacement and k force constant of the spring.

Also recall,

F=mg ............ 2

Where m is the mass of object, g is the acceleration due to gravity.

Equating 1 and 2

Ky = mg

Given that g=9.8m/s2 , y is 3.4cm and g is 8g

K×3.4/100m =8/1000kg × 9.8m/s2

K= ( 0.008kg × 9.8m/s2 ) ÷ 0.034

K= 0.0784÷0.035

K=2.24N/m

Mass ofvthe second object is 25g =0.025kg

Period of oscillation T

T=2π√m/k

T=2×3.142√0.025/2.24

T=6.284√0.0111

T=0.659seconds

notsponge [240]2 years ago
8 0

Answer:

df,jgukilhugyftdrswjdfggfjgjfjgfthf

Explanation:

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8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s
allsm [11]

Answer:

a) \omega = 50\,\frac{rad}{s}, b) \omega = 0\,\frac{rad}{s}

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

\tau = F \cdot r

\tau = m\cdot a \cdot r

\tau = m \cdot \alpha \cdot r^{2}

Where \alpha is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}

\alpha = -3\,\frac{rad}{s^{2}}

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)

\omega = 50\,\frac{rad}{s}

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}

t = 66.667\,s

Since t > 66.667\,s, then the angular velocity is equal to zero. Therefore:

\omega = 0\,\frac{rad}{s}

7 0
3 years ago
Light of wavelength 633 nm from a He-Ne laser passes through a circular aperture and is observed on a screen 4.0 m behind the ap
Verizon [17]

Answer:

The answer is "1.144 \times 10^{-4} \ m".

Explanation:

w=\frac{2.44 \lambda L}{D}\\\\D=\frac{2.44 \lambda L}{w}\\\\

   =\frac{2.44 \times 633 \times 10^{-9}\times 4 }{0.054}\\\\=\frac{6178.08\times 10^{-9}}{0.054}\\\\=1.144 \times 10^{-4} \ m

8 0
3 years ago
A proton moves perpendicular to a uniform magnetic field B with arrow at a speed of 2.50 107 m/s and experiences an acceleration
KIM [24]

Answer:

A) B = 0.009185 T

B) Drection is negative y-direction

Explanation:

A) We are given;

Speed(v) = 2.5 x 10^(7) m/s

Acceleration (a) = 2.2 x 10^(13) m/s²

We also know that charge of proton(q) = 1.6 x 10^(-19)

Mass of proton(m) = 1.67 x 10^(-27)

Now, Since the proton is moving by circular motion, this force is equal to the centripetal force which is given as;

F = qvBsinθ = ma

Since perpendicular, θ = 90°

And so, sinθ = sin 90 = 1

Thus, qvB = ma

Making B the subject gives;

B = ma/qv

B = (1.67 X 10^(-27) X 2.2 X 10^13)) / (1.6 X 10^(-19) X 2.5 X 10^(7))

= 0.009185 T

B) By use of Flemings right hand rule, we can see that the middle finger points toward negative y-direction, so the magnetic field is in the negative y-direction

4 0
3 years ago
Read 2 more answers
I need help for the first 3 plz.
madam [21]
1) hypothesis
2) data
3) method

I think these are correct.
7 0
3 years ago
Read 2 more answers
The current that flows from and back to the power supply in a parallel circuit is called __________ current.
blagie [28]

Answer:

mainline current

Explanation:

Current that flows from and back to the power supply in a parallel circuit. Fuse. A type of circuit protection device.

6 0
2 years ago
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