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xz_007 [3.2K]
3 years ago
7

A grandfather clock depends on the period of a pendulum to keep correct time. Suppose a grandfather clock is calibrated correctl

y and then a mischievous child slides the bob of the pendulum downward on the oscillating rod. The grandfather clock will run:
Physics
2 answers:
sweet-ann [11.9K]3 years ago
7 0

The grandfather clock will run: slower

<h3>Explanation: </h3>

A grandfather clock depends on the period of a pendulum to keep correct time. Suppose a grandfather clock is calibrated correctly and then a mischievous child slides the bob of the pendulum downward on the oscillating rod.

A pendulum is a weight that get suspended from a pivot, therefore it can swing freely. A pendulum clock is the clock using a pendulum, a swinging weight, as its timekeeping element.

Calibrate in physics means to ascertain the caliber of and to determine, rectify, or mark the graduations of something, such as a thermometer tube.

A bob is the weight located on the end of a pendulum found in pendulum clocks.

The time period of simple pendulum is given by:

T = 2 \pi \sqrt{\frac{L}{g} }

The time period of a simple pendulum is defined as the time taken by the pendulum to finish one full oscillation.

Where l is the effective length of the pendulum, and g is the acceleration due to gravity

The grandfather clock will run: Slower. Because as the length of the pendulum increase, the tie period of the pendulum also increase.  It means the clock run slow.

Learn more about   pendulum brainly.com/question/8411900

#LearnWithBrainly

Setler [38]3 years ago
6 0

Answer:

Slow

Explanation:

As we know that the time period of simple pendulum is given by

T = 2\pi \sqrt{\frac{L}{g}}

Where, l be the effective length of the pendulum and g be the acceleration due to gravity

As the length of the pendulum increase, then the tie period of the pendulum will also increase.

It means the clock run slow.

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Two identical 9.10-g metal spheres (small enough to be treated as particles) are hung from separate 300-mm strings attached to t
Musya8 [376]

Answer:

n = 1.266\times 10^{12}

Explanation:

Given data:

mass of sphere is 10 g

Angle between string and vertical axis is \theta = 13 degree

thickness of string  300 mm = 0.3 m

sin\theta =\frac{2}{0.3 m}

r =0.3 sin 13 = 0.067 m

Fe = \frac{ kq_1 q-2}{d^2}

Fe = \frac{kq^2}{(2r)^2} = mg tan\theta

q^2 =  mg tan\theta \frac{(2r)^2}{k}

    = 0.0091 \times 9.8 tan13 \times \frac{(2\times 0.067)^2}{9\times 10^9}

q^2 = 4.10\times 10^{-14}

q = 2.026 \times 10^{-7} C

q = ne

n = \frac{1.6\times 10^{-19}}{2.02\times 10^{-7}}

n = 1.266\times 10^{12}

3 0
3 years ago
uphill at a rate of 2.5 mi/h from the base of a 6-mi trail. At the same time, Edwin walks downhill at a rate of 3.5 mi/h from th
uranmaximum [27]
<h2>After 1 hour they meet.</h2>

Explanation:

Distance between them = 6 miles

Speed of uphill person = 2.5 miles per hour

Speed of downhill person = 3.5 miles per hour

Relative velocity = 2.5 - ( -3.5 ) = 6 miles per hour

We know

         Displacement = Velocity x Time

         6 = 6 x Time taken

Time taken = 1 hour

After 1 hour they meet.

4 0
3 years ago
A 4.00-g bullet, traveling horizontally with a velocity of magnitude 400 m/s, is fired into a wooden block with mass 0.650 kg ,
Maru [420]

Answer:

a) Coefficient of kinetic friction between block and surface = 0.12

b) Decrease in kinetic energy of the bullet = 247.8 J

c) Kinetic energy of the block at the instant after the bullet passes through it = 0.541 J

Explanation:

Given,

Mass of bullet = 4.00 g = 0.004 kg

Initial velocity of the bullet = 400 m/s

Mass of wooden block = 0.65 kg

Initial velocity of the wooden block = 0 m/s (since it was initially at rest)

Final velocity of the bullet = 190 m/s

Distance slid through by the block after the collision = d = 72.0 cm = 0.72 m

Let the velocity of the wooden block after collision be v

According to the law of conservation of momentum,

Momentum before collision = Momentum after collision

Momentum before collision = (Momentum of bullet before collision) + (Momentum of wooden block before collision)

Momentum of bullet before collision = (0.004×400) = 1.6 kgm/s

Momentum of wooden block before collision = (0.65)(0) = 0 kgm/s

Momentum after collision = (Momentum of bullet after collision) + (Momentum of wooden block after collision)

Momentum of bullet after collision = (0.004×190) = 0.76 kgm/s

Momentum of wooden block after collision = (0.65)(v) = (0.65v) kgm/s

Momentum balance gives

1.6 + 0 = 0.76 + 0.65v

0.65v = 1.6 - 0.76 = 0.84

v = (0.84/0.65)

v = 1.29 m/s

The velocity of the wooden block after collision = 1.29 m/s

To obtain the coefficient of kinetic friction between block and surface, we will apply the work-energy theorem.

The work-energy theorem states that the work done in moving the block from one point to another is equal to the change in kinetic energy of the block between these two points.

The points to consider are the point when the block starts moving (immediately after collision) and when it stops as a result of frictional force.

Mathematically,

W = ΔK.E

W = workdone by the frictional force in stopping the wooden block (since there is no other horizontal force acting on the block)

W = -F.d (minus sign because the frictional force opposes motion)

d = Distance slid through by the block after the collision = 0.72 m

F = Frictional force = μN

where N = normal reaction of the surface on the wooden block and it is equal to the weight of the block.

N = W = mg

F = μmg

W = - μmg × d = (-μ)(0.65)(9.8) × 0.72 = (-4.59μ) J

ΔK.E = (final kinetic energy of the block) - (initial kinetic energy of the block)

Final kinetic energy of the block = 0 J (since the block comes to a rest)

(Initial kinetic energy of the block) = (1/2)(0.65)(1.29²) = 0.541 J

ΔK.E = 0 - 0.541 = - 0.541 J

W = ΔK.E

-4.59μ = -0.541

μ = (0.541/4.59)

μ = 0.12

b) The decrease in kinetic energy of the bullet

(Decrease in kinetic energy of the bullet) = (Kinetic energy of the bullet before collision) - (Kinetic energy of the bullet after collision)

Kinetic energy of the bullet before collision = (1/2)(0.004)(400²) = 320 J

Kinetic energy of the bullet after collision = (1/2)(0.004)(190²) = 72.2 J

Decrease in kinetic energy of the bullet = 320 - 72.2 = 247.8 J

c) Kinetic energy of the block at the instant after the bullet passes through it = (1/2)(0.65)(1.29²) = 0.541 J

Hope this Helps!!!

4 0
3 years ago
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bogdanovich [222]

Answer:

im not really good at explaining, but i found this website url:

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same question just with the explanation

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A block of wood has a mass of 120 g and a volume of 200cm what is the density of the word
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