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ioda
3 years ago
15

Calculate a pendulum's frequency of oscillation (in Hz) if the pendulum completes one cycle in 0.5 s.

Physics
1 answer:
Marina86 [1]3 years ago
3 0
Time taken to complete one oscillation for a pendulum is Time Period, T = 0.5 s 
Frequency of the pendulum oscillation = 1 / Time Period => f = 1 / T = 1 / 0.5  
Frequency f = 2 Hz
You might be interested in
A doorframe is twice as tall as it is wide. There is a positive charge on the top left corner and an equal but negative charge i
Andrei [34K]

Answer:

α = 141.5°  (counterclockwise)

Explanation:

If

q₁ = +q

q₂ = -q

q₃ < 0

b = 2*a

We apply Coulomb's Law as follows

F₁₃ = K*q₁*q₃ / d₁₃² = + K*q*q₃ / (2*a)² = + K*q*q₃ / (4*a²)

F₂₃ = K*q₂*q₃ / d₂₃² = - K*q*q₃ / (5*a²)

(d₂₃² = a² + (2a)² = 5*a²)

Then

∅ = tan⁻¹(2a/a) = tan⁻¹(2) = 63.435°

we apply

F₃x = - F₂₃*Cos ∅ = - (K*q*q₃ / (5*a²))* Cos 63.435°

⇒  F₃x = - 0.0894*K*q*q₃ / a²

F₃y = - F₂₃*Sin ∅ + F₁₃

⇒  F₃y = - (K*q*q₃ / (5*a²))* Sin 63.435° + (K*q*q₃ / (4*a²))

⇒  F₃y = 0.0711*K*q*q₃ / a²

Now, we use the formula

α = tan⁻¹(F₃y / F₃x)

⇒  α = tan⁻¹((0.0711*K*q*q₃ / a²) / (- 0.0894*K*q*q₃ / a²)) = - 38.5°

The real angle is

α = 180° - 38.5° = 141.5°  (counterclockwise)

4 0
4 years ago
Dibuja la gráfica de calentamiento de un kilogramo de plomo que se encuentra inicialmente a 70ºC y pasa a una temperatura final
MariettaO [177]

Answer:

Q= m c_e ΔT and   Q = m L

Explanation:

For this graph of temperature vs energy (heating) we must use two relations

* for when there is no change of state

          Q= m c_e ΔT

* for using there is change of state

          Q = m L

the second expression is a consequence of the fact that all the energy supplied is used to change the state of the solid-liquid and liquid-gas system

the energy supplied is the sum of the energy in each interval

divide the system into intervals determined by the state change points

1) from T₀ = 70ºC to T_f = 327.4ºC, sample in solid-liquid state

           c_e = 128 J / kg ºC

           Q₁ = m c_e (T_f -To)

           Q₁=1  128 (327.4 -70)

           Q₁ = 3.29 10⁴ J

           Q = Q₁ = 3.29 10⁴ J

2) when is it changing from solid to liquid

            L = 2.45 10⁴ J / kg

            Q2 = 1 2.45 10⁴

            Q2 = 2.45 10⁴ J

            Q = Q₁ + Q₂

             Q = 5.74 10⁴ J

3) from to = 327.4ºC until T_f = 1725ºC, sample in liquid state

in the tables the specific heat of the solid and liquid state is the same

             Q3 = m c_e (T_f -To)

             Q3 = 1 128 (1725 -327.4)

             Q3 = 1.79 10⁵ J

              Q = Q₁ + Q₂ + Q₃

              Q = (3.29 +2.45 + 17.9) 10⁴ J

              Q = 23.64 10⁴ J

4) for when it is changing from the liquid state to the gaseous state

             L_v = 8.70 10⁵ J / kg

             Q₄ = m L_v

             Q₄ = 1 8.70 10⁵

             Q₄ = 8.70 10⁵ J

             Q = Q₁ + Q₂ + Q₃ + Q₄

              Q = (3.29 +5.74 + 17.9+ 87.0) 10⁴ J

               Q = 110.64 10⁴ J

5) from To = 1725ºC to T_f = 2000ºC, sample in gaseous state

             Q₅ = m c_e ΔT

             Q₅ = 1 128 (2000 -1725)

             Q₅ = 3.52 10⁴ J

             Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

              Q = 114.16 104 J

the following table shows the points to be plotted

         Energy (10⁴ J)  Temperature (ºC)

                  0                     70

                 3.29             327.4

                 5.74             327.4

               23.64           1725

               110.64          1725

                114.16         2000

In the attachment we can see a graph of Temperature versus energy supplied

8 0
3 years ago
Why is it important to have a balanced centrifuge?
ratelena [41]

It is important to properly balance a centrifuge because an unbalanced machine can damage the rotor, cause catastrophic damage to the machine itself, or even injure or kill lab personnel working in the room. Balancing a centrifuge involves spreading the weight of the samples across the entire rotor.

8 0
4 years ago
A roller coaster starts at the top of a hill of height h, goes down the hill, and does a circular loop of radius r before contin
jeka94

a) See free-body diagram in attachment

b) Net force in the y-direction: F_y=mg+N[/tex]

c) The velocity at which the roller coaster will fall is [tex]v=\sqrt{gr}[/tex]

d) The speed of the roller coaster must be 17.1 m/s

e) The roller coaster should start from a height of 90 m

f) The roller coaster should start from a height of 100 m

Explanation:

a)

See the free-body diagram in attachment. There are only two forces acting on the roller coaster at the top of the loop:

  • The weight of the roller coaster, acting downward, indicated by mg (where m is the mass of the roller coaster and g is the acceleration of gravity)
  • The normal reaction exerted by the track on the roller coaster, acting downward, and indicated with N

The two forces are represented in the diagram as two downward arrows (the length is not proportional to their magnitude, in this case)

b)

Since there are only two forces acting on the roller coaster at the top of the loop, and both forces are acting downward, then we can write the vertical net force as follows (we take downward as positive direction):

F_y = mg + N

where

mg is the weight

N is the normal reaction

Since the roller coaster is in circular motion, this net force must be equal to the centripetal force, therefore

m\frac{v^2}{r}=mg+N

where v is the speed of the car at the top of the loop and r is the radius of the loop.

c)

For this part of the problem, we start from the equation written in part b)

m\frac{v^2}{r}=mg+N

where the term on the left represents the centripetal force, and the terms on the right are the weight and the normal reaction.

We now re-arrange the equation making v (the speed) as the subject:

v=\sqrt{gr+\frac{Nr}{m}}

However, the velocity at which the roller coaster will fall is the velocity at which the normal reaction becomes zero (the roller coaster loses contact with the track), so when

N = 0

And as a result, the minimum velocity of the cart is

v=\sqrt{gr}

d)

In this part, we are told that the radius of the loop is

r = 30 m

And the mass of the cart is

m = 50 kg

Moreover, the acceleration of gravity is

g=9.8 m/s^2

We said that the minimum velocity that the cart must have in order not to fall at the top is

v=\sqrt{gr}

And substituting, we find

v=\sqrt{(9.8)(30)}=17.1 m/s

e)

According to the law of conservation of energy, the initial gravitational energy of the roller coaster at the starting point must be equal to the sum of the kinetic energy + gravitational potential energy at the top of the loop, therefore:

mgh = \frac{1}{2}mv^2 + mg(2r)

where

h is the initial height at the starting point

(2r) is the height of the roller coaster at the top of the loop

We can re-arrange the equation making h the subject,

h=\frac{v^2}{2g}+2r

And substituting the minimum speed of the cart,

v=\sqrt{gr}

this becomes

h=r+2r=3r

And since r = 30 m, we find

h=3(30)=90 m

f)

In this case, 10% of the initial energy is lost during the motion of the roller coaster. We can rewrite the equation of the previous part as

0.90mgh = \frac{1}{2}mv^2 + mg(2r)

Because only 90% (0.90) of the initial energy is converted into useful energy (kinetic+potential) when the cart reaches the top of the loop.

Re-arranging the equation, this time we get

h=\frac{\frac{v^2}{2g}+2r}{0.90}

Again, by substituting v=\sqrt{gr}, we get

h=\frac{3r}{0.90}

And therefore, the new initial height must be

h=\frac{3(30)}{0.9}=100 m

Learn more about kinetic and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

4 0
3 years ago
John is gliding along on his scooter at a speed of 2.8 m/s when
Anna71 [15]

Answer: His acceleration is -18.66m/s^2

Explanation:

Ok, the initial speed is 2.8m/s. (we can define the initial direction as the positive direction).

And he wants to stop, so he must accelerate in the opposite direction as the initial movement, then we would have:

a(t) = -A.

So we have a constant, and negative acceleration.

Now, if we want to find the velocity we must integrate over time, and we will get:

v(t) = -A*t + V0

where V0 is the initial velocity, we know that it is 2.8m/s, and t is the time in seconds.

Then the velocity is:

v(t) = -A*t + 2.8m/s.

Now we know that John is brought to rest in 0.15 seconds after he starts slowing down, this means that at t = 0.15 seconds, his velocity is equal to zero.

v(0.15s) = 0m/s = -A*0.15s + 2.8m/s

2.8m/s = A*0.15s

2.8m/s/0.15s = 18.66m/s^2 = A.

So his acceleration is -A, then we have that:

His acceleration is -18.66m/s^2

8 0
4 years ago
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