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3 years ago

A merry-go-round on a playground consists of a horizontal solid disk with a weight

Physics

1 answer:

bezimeni [28]3 years ago

5 0

Answer:α = 1.00 rad/s², τ = 90.0 N•m, KEr = 1.12 kJ

Explanation:

m = 795/9.81 = 81.0 kg

ω₁ = 47.79 rev/min(2π rad/rev) / (60 s/min) = 5.00 rad/s

α = (ω₁ - ω₀)/τ = (5.00 - 0.00)/5.00 = 1.00 rad/s²

I = ½mR² = ½(81.0)(1.49²) = 90.0 kg•m²

τ = Iα = 90.0(1.00) = 90.0 N•m

KEr = ½Iω² = ½(90.0)5.00² = 1,124.477 ≈ 1.12 kJ

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What is the electric force acting between two charges of -0.0045 C and -0.0025 C that are 0.0060 m apart? Use Fe=kq1q2/r^2 and k

oksano4ka [1.4K]

Answer:

D. 2.8 × 10⁹ N

Explanation:

The force between two charges is directly proportional to the amount of charges at the two points and inversely proportional to the square of distance between the two points.

Fe= k Q₁Q₂/r²

Q₁= -0.0045 C

Q₂= -0.0025 C

r= 0.0060 m

k= 9.00 × 10 ⁹ Nm²/C²

Fe= (9.00 × 10 ⁹ Nm²/C²×-0.0045 C×-0.0025 C)/0.0060²

=2.8 × 10⁹ N

4 0

3 years ago

Read 2 more answers

Suppose a 48-N sled is resting on packed snow. The coefficient of kinetic friction is 0.10. If a person weighing 660 N sits on t

Annette [7]

Assume the snow is uniform, and horizontal.

Given:

coefficient of kinetic friction = 0.10 = muK

weight of sled = 48 N

weight of rider = 660 N

normal force on of sled with rider = 48+660 N = 708 N = N

Force required to maintain a uniform speed

= coefficient of kinetic friction * normal force

= muK * N

= 0.10 * 708 N

=70.8 N

Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.

8 0

3 years ago

A simple series circuit consists of a 120 Ω resistor, a 21.0 V battery, a switch, and a 3.50 pF parallel-plate capacitor (initia

slega [8]

Answer

Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)

Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.

ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)

Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC

30.7pC/εo = 3.47 V∙m <----- C)

ic(0.5ns) = 29.7ma <----- D)

8 0

3 years ago

Which is most likely the length of a student’s textbook?

UkoKoshka [18]

For this case, what we must do is to rewrite these measurements in the same unit in order to compare them.

By writing the measurements in meters we have:

$30 mm = (30) * (\frac{1}{1000}) = 0.030 m$

$30 cm = (30) * (\frac{1}{100}) = 0.30 m$

$30 dm = (30) * (\frac{1}{10}) = 3 m\\30 Hm = (30) * (100) = 3000 m$

Therefore, physically the correct measure is:

$0.30 m = 30 cm$

Answer:

the length of a student's textbook most likely is:

30 centimeters

3 0

3 years ago

Read 2 more answers

A wheel rotates about a fixed axis with an initial angular velocity of 13 rad/s. During a 8-s interval the angular velocity incr

____ [38]

Answer:

The number of revolutions is 44.6.

Explanation:

We can find the revolutions of the wheel with the following equation:

$\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2}$

Where:

$\omega_{0}$ : is the initial angular velocity = 13 rad/s

t: is the time = 8 s

α: is the angular acceleration

We can find the angular acceleration with the initial and final angular velocities:

$\omega_{f} = \omega_{0} + \alpha t$

Where:

$\omega_{f}$ : is the final angular velocity = 57 rad/s

$\alpha = \frac{\omega_{f} - \omega_{0}}{t} = \frac{57 rad/s - 13 rad/s}{8 s} = 5.5 rad/s^{2}$

Hence, the number of revolutions is:

$\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 13 rad/s*8 s + \frac{1}{2}*5.5 rad/s^{2}*(8 s)^{2} = 280 rad*\frac{1 rev}{2\pi rad} = 44.6 rev$

Therefore, the number of revolutions is 44.6.

I hope it helps you!

4 0

3 years ago