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3 years ago

6

If you are building a clock by using a pendulum's oscillations to keep time and you find the period is a little too small (the c

lock counts seconds too quickly), what should you do to adjust the pendulum's period? O
a. Increase the length of the string
b. Increase the mass on the end of the string
c. Increase the amplitude of the swing
d. Decrease the length of the string
e. Decrease the mass on the end of the string

1 answer:

Leviafan [203]3 years ago

3 0

Time period of pendulum is given by

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

Here,

L is length

g is acceleration due to gravity

Therefore we can determine that the period of the pendulum is directly proportional to the square root of the length of the rope.

$T \propto \sqrt{L}$

Therefore, it is necessary to increase the length of the rope to increase the period so that it counts the seconds more slowly. The correct answer is A.

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Sliva [168]

The event in the life of a star that begins its expansion into a giant is its core that was hot enough for fusion reaction.

<h3>What is fusion reaction?</h3>

Nuclear fusion is a type of reaction in which two or more atomic nuclei are fuse to form one or more different atomic nuclei with the release or the absorption of energy.

So we can conclude that the event in the life of a star that begins its expansion into a giant is its core that was hot enough for fusion reaction.

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7 0

1 year ago

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo

djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for the force of static friction:

$f_s = \mu_s N$ --- (1)

where;

$f_s =$ static friction force

$\mu_s =$ coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

$N = ma_{min}$

From equation (1)

$f_s = \mu_s ma_{min}$

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

$F = f_s$

$mg = \mu_s ma_{min}$

$a_{min} = \dfrac{mg }{ \mu_s m}$

$a_{min} = \dfrac{g }{ \mu_s }$

where;

$\mu_s = 0.383$ and g = 9.8 m/s²

$a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }$

$\mathbf{a_{min}= 25.59 \ m/s^2}$

3 0

2 years ago

Which of these black holes exerts the weakest tidal force on an object near its event horizon? Which of these black holes exerts

ratelena [41]

Answer:

THA ANSWER IS B

Explanation:

YEH ITS B

6 0

2 years ago

An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr

Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

$F= mg$

Where, f = restoring force

$kx=mg$

$k=\dfrac{mg}{x}$

Put the value into the formula

$k=\dfrac{2.15\times9.8}{0.0895}$

$k=235.41\ N/m$

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Here, $v = A\omega$

$K.E=\dfrac{1}{2}m\times(A\omega)^2$

Here, $\omega=\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}kA^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2$

$K.E=0.06500\ J$

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0

2 years ago

A wrench 0.500 m long is applied to a nut with a force of 80.0 N. Because of the cramped space, the force must be exerted upward

riadik2000 [5.3K]

Answer:

Torque, $\tau=34.6\ N.m$

Explanation:

It is given that,

Length of the wrench, l = 0.5 m

Force acting on the wrench, F = 80 N

The force is acting upward at an angle of 60.0° with respect to a line from the bolt through the end of the wrench. We need to find the torque is applied to the nut. We know that torque acting on an object is equal to the cross product of force and distance. It is given by :

$\tau=Fr\ sin\theta$

$\tau=80\times 0.5\ sin(60)$

$\tau=34.6\ N.m$

So, the torque is applied to the nut is 34.6 N.m. Hence, this is the required solution.

7 0

2 years ago