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Advocard [28]
4 years ago
14

Pendulum clock. Your friend is trying to construct a clock for a craft show and asks you for some advice. She has decided to con

struct the clock with a pendulum. The pendulum will be a very thin, very light wooden bar with a thin, but heavy, brass ring fastened to one end. The length of the rod is 80 cm and the diameter of the ring is 10 cm. She is planning to drill a hole in the bar to place the axis of rotation 15 cm from one end. She wants you to tell her the period of this pendulum.

Physics
1 answer:
levacccp [35]4 years ago
6 0

Answer:

The time period for this pendulum is 1.68 seconds

Explanation:

Solution

Given that:

The length of the pendulum is measured from the axis of rotation to the center of mass of the bob of the pendulum

Now,

In this case, the length becomes:

L= 80 - 15+5

L = 70 cm

The time period = T = 2π √L/g

T = 2* 3.14 *√0.7/9.8

= 1.68 seconds

Note: Kindly find an attached work to the part of the solution of the given question

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force goes as 1/d^2 ... (2d)^2 => 4d^2 ...

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3 years ago
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If a 15kg mass weighs 372 N on Jupiter, what is Jupiter's gravitational acceleration?
Dimas [21]

Answer:

Jupiter's gravitational acceleration is 24.8 m/s^2

Explanation:

Recall that the weight under the influence of a gravitational acceleration G is defined as:

Weight = m * G

Then, in our case we have

372 N = 15 kg * G

G = 372/15  m/s^2

G = 24.8 m/s^2

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3 years ago
ANSWER FAST FOR BRAINLIEST!
Ne4ueva [31]

Answer:

the last one

Explanation:

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3 years ago
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Can anyone help if u know it pls!!
LuckyWell [14K]

Answer:

(1)

H² = B² × L²

H² = (9.2)² × (5.2)²

H² = 84.64 × 27.04

H² = 2288.6656

H = √ 2288.6656

H = 47.84

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3 years ago
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A 1.20 kg solid ball of radius 40 cm rolls down a 5.20 m long incline of 25 degrees. Ignoring any loss due to friction, how fast
sladkih [1.3K]

Answer:

1.6s

Explanation:

Given that A 1.20 kg solid ball of radius 40 cm rolls down a 5.20 m long incline of 25 degrees. Ignoring any loss due to friction,

To know how fast the ball will roll when it reaches the bottom of the incline, we need to calculate the acceleration at which it is rolling.

Since the frictional force is negligible, at the top of the incline plane, the potential energy = mgh

Where h = 5.2sin25

h = 2.2 m

P.E = 1.2 × 9.8 × 2.2

P.E = 25.84 j

At the bottom, K.E = P.E

1/2mv^2 = 25.84

Substitutes mass into the formula

1.2 × V^2 = 51.69

V^2 = 51.69/1.2

V^2 = 43.07

V = 6.56 m/s

Using the third equation of motion

V^2 = U^2 + 2as

Since the object started from rest,

U = 0

6.56^2 = 2 × a × 5.2

43.07 = 10.4a

a = 43.07/10.4

a = 4.14 m/s^2

Using the first equation of motion,

V = U + at

Where U = 0

6.56 = 4.14t

t = 6.56/4.14

t = 1.58s

Therefore, the time the ball rolls when it reaches the bottom of the incline is approximately 1.6s

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3 years ago
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