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photoshop1234 [79]
3 years ago
5

A pendulum clock depends on the period of a pendulum to keep correct time. Suppose a pendulum clock is keeping correct time and

then Dennis the Menace slides the bob of the pendulum downward on the oscillating rod.
Does the clock run
(a) slow,
(b) fast, or
(c) correctly?
Physics
1 answer:
Mashutka [201]3 years ago
8 0

Answer:

time period is increased so the clock will become SLOW

Explanation:

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

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

here we know that

L = distance of the pendulum bob from the hinge

g = acceleration due to gravity

now here the bob slide down so that the length of the pendulum is being increased

so time period T of the pendulum is also increased

so here the pendulum will take more time to oscillate or to complete one oscillation

so clock will become SLOW

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A marble slides without friction in a vertical loop around the inside of a smooth, 28.6 cm diameter horizontal pipe. The marble'
Leviafan [203]

Answer:3.49 m/s

Explanation:

Given

Speed of marble at Bottom v=4.22 m/s

Diameter of loop d=28.6 cm

As Energy is conserved therefore Energy at top is equal to energy at bottom

E_T=E_B

\frac{mv^2}{2}+mgh=\frac{mv_0^2}{2}  ,where v_0 is the velocity at bottom

\frac{v^2}{2}+gh=\frac{v_0^2}{2}

v_0^2=v^2+2gh

v^2=v_0^2-2gh

v=\sqrt{v_0^2-2gh}

v=\sqrt{4.22^2-2\times 9.8\times 0.286}

v=\sqrt{17.8084-5.6056}

v=3.49 m/s

                       

7 0
3 years ago
A bungee cord can stretch, but it is never compressed. When the distance between the two ends of the cord is less than its unstr
Ksju [112]

Answer:

Explanation:

Given that

g=9.8m/s²

The spring constant is

k=50N/m

The length of the bungee cord is

Lo=32m

Height of bridge which one end of the bungee is tied is 91m

A steel ball of mass 92kg is attached to the other end of the bungee.

The potential energy(Us) of the steel ball before dropped from the bridge is given as

P.E= mgh

P.E= 92×9.8×91

P.E= 82045.6 J

Us= 82045.6 J

Potential energy)(Uc) of the cord is given as

Uc= ½ke²

Where 'e' is the extension

Then the extension is final height extended by cord minus height of cord

e=hf - hi

e=hf - 32

Uc= ½×50×(hf-32)²

Uc=25(hf-32)²

Using conservation of energy,

Then,

The potential energy of free fall equals the potential energy in string

Uc=Us

25(hf-32)²=82045.6

(hf-32)² = 82045.6/25

(hf-32)²=3281.825

Take square root of both sides

√(hf-32)²=√(3281.825)

hf-32=57.29

hf=57.29+32

hf=89.29m

We neglect the negative sign of the root because the string cannot compressed

3 0
3 years ago
If you fired a rifle straight upwards at 1000 m/s, how far up will the bullet get?
nadya68 [22]

Answer:

h = 51020.40 meters

Explanation:

Speed of the rifle, v = 1000 m/s

Let h is the height gained by the bullet. It can be calculated using the conservation of energy as :

\dfrac{1}{2}mv^2=mgh

h=\dfrac{v^2}{2g}

h=\dfrac{(1000\ m/s)^2}{2\times 9.8\ m/s^2}                    

h = 51020.40 meters

So, the bullet will get up to a height of 51020.40 meters. Hence, this is the required solution.          

4 0
3 years ago
Un satélite geoestacionario se encuentra a una distancia de 120.000 km sobre la superficie de Júpiter. Determine: a. El periodo
Lisa [10]

Answer:

a) a geostationary satellite is that it is always at the same point with respect to the planet,

b) f = 2.7777 10⁻⁵ Hz

c)                           d)   w = 1.745 10⁻⁴ rad / s

Explanation:

a) The definition of a geostationary satellite is that it is always at the same point with respect to the planet, that is, its period of revolutions is the same as the period of the planet

  •                T = 10 h (3600 s / 1h) = 3.6 104 s

b) the period the frequency are related

                T = 1 / f

                 f = 1 / T

                 f = 1 / 3.6 104

                 f = 2.7777 10⁻⁵ Hz

c) the distance traveled by the satellite in 1 day

The distance traveled is equal to the length of the circumference

                 d = 2pi (R + r)

                 d = 2pi (69 911 103 + 120 106)

                 d = 1193.24 m

d) the angular velocity is the angle traveled between the time used.

                 .w = 2pi /t

                  w = 2pi / 3.6 10⁴

                  w = 1.745 10⁻⁴ rad / s

how fast is

                  v = w r

                  v = 1.75 10-4 (69.911 106 + 120 106)

                  v = 190017 m / s

5 0
3 years ago
A loudspeaker diaphragm is vibrating in simple harmonic motion with a frequency of 760 Hz and a maximum displacement of 0.85 mm.
Alchen [17]

Answer:(a) 4775.2Hz (b) 4.06m/s (c) 19382.15m/s²

Explanation: Given that the frequency of oscilation f, is 760Hz and the maximum displacement x, is 0.85mm= 0.00085m

(a) Angular frequency w= 2πf

w= 2π × 760 = 4775.2Hz

(b) Maximum speed v is given as the product of angular frequency and maximum displacement

V=wx

V= 4775.2 × 0.00085

V= 4.06m/s

(c) The maximum acceleration a

= w²x

= (4775.2)² × (0.00085)

a= 19382.15m/s².

5 0
4 years ago
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