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3 years ago

A pendulum oscillates 50 times in 6 seconds. Find its time period and frequency?

Physics

1 answer:

iren [92.7K]3 years ago

4 0

Explanation:

time taken fir 50 oscillations is 6 seconds

time taken for 1 oscillation is 6/50

convert it into a decimal

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A subway train is traveling at 22.2 m/s when it approaches a slower train 50m ahead traveling in the same direction at 6.94 m/s.

Amiraneli [1.4K]

Answer:

Time that they collide = 4.99s

Relative speed of the trains when they collide: The relative speed of The first train relative to the second, slower train at collision = 4.781 m/s

Explanation:

We will use the equations of motion to obtain the solution required

At time t = 0

speed of first train = 22.2 m/s

Initial space between the two trains = 50 m

Speed of second train = 6.94 m/s

For the first car, distance covered by the first train = y

y = distance covered between the beginning of the deceleration and the point where the the two trains hit one another.

u = initial velocity = 22.2 m/s

t = time taken for all this to happen

a = deceleration = - 2.1 m/s²

y = ut + (1/2)at²

y = 22.2t - 1.05t² (eqn 1)

For the second train,

At t = 0, y = 50 m

Let the new distance moved by the second train before collision = (y - 50)

u = initial velocity = 6.94 m/s

t = time taken = t

a = acceleration of the second train = 0 m/s² (constant velocity)

(y - 50) = ut + (1/2)at²

y - 50 = 6.94t

y = 6.94t + 50 (eqn 2)

substituting for y in eqn 2 using the expression obtained in eqn 1

y = 22.2t - 1.05t²

y = 6.94t + 50

22.2t - 1.05t² = 6.94t + 50

1.05t² - 15.26t + 50 = 0

Solving this quadratic equation

t = 4.99 s or 9.54 s

The position of the two trains are the same at those two times, but the first time is when they hit each other.

t = 4.99 s

At 4.99 s, the the velocity of the first train

v = u + at

v = 22.2 + (-2.1×4.99) = 11.721 m/s in the same direction as the second train.

Relative velocity at this point will be

= 11.721 - 6.94 = 4.781 m/s

Relative speed of the trains when they collide: The relative speed of The first train relative to the second, slower train at collision = 4.781 m/s

Hope this Helps!!!

4 0

3 years ago

An Iowa class warship holds the record for the fastest warship. If the ship accelerates uniformly from rest at 0.15 m/s² for 2 m

Mumz [18]

The equation to be used is the derived formulas for rectilinear motion at a constant acceleration. The formula for acceleration is

a = (v - v₀)/t
where
v and v₀ are the initial and final velocities, respectively
t is the time
a is the acceleration

Since it started from rest, v₀ = 0. Using the formula:

0.15 m/s² = (v - 0)/[2 minutes*(60 s/1 min)]
Solving for v,
v = 18 m/s

3 0

3 years ago

Steel blocks A and B, which have equal masses, are at TA = 300 oC and T8 = 400 oC. Block C, with mc - 2mA, is at TC = 350 oC. Bl

shepuryov [24]

Answer:

b) TA = TB = TC

Explanation:

When put in contact each other, and isolated, both blocks will exchange heat till they reach to thermal equilibrium.
During this process, the body at a higher temperature, will loss heat, tat it will be gained by the other body.
The equilibrium condition will be reached when the following equation be met:

$\Delta Q = c_{st}* m_{A} * (T_{fin} - T_{0A} ) = c_{st}* m_{B} * (T_{0B} - T_{fin} )$

Replacing by the values of T₀A = 300º C, and T₀B = 400ºC, and simplifying common terms as mA = mB, we can solve for Tfin, as follows:

$(400 \ºC - T_{fin}) = (T_{fin} - 300 \ºC) \\ \\ 2* T_{fin} = 700\ºC\\ \\ T_{fin} = 350\ºC$

So, when both blocks reach to equilibrium, they will be at a common final temperature, 350ºC.
When put in contact with block C, at the same temperature, at that instant, the three blocks will have the same common temperature of 350 ºC.
So, option b) is the right one.

8 0

3 years ago

The atmospheric pressure on the top of the Engineering Sciences Building (ESB) is 97.6 kPa, while that in Room G39-ESB (ground f

Stells [14]

Answer:

Δ h = 52.78 m

Explanation:

given,

Atmospheric pressure at the top of building = 97.6 kPa

Atmospheric pressure at the bottom of building = 98.2 kPa

Density of air = 1.16 kg/m³

acceleration due to gravity, g = 9.8 m/s²

height of the building = ?

We know,

Δ P = ρ g Δ h

(98.2-97.6) x 10³ = 1.16 x 9.8 x Δ h

11.368 Δ h = 600

Δ h = 52.78 m

Hence, the height of the building is equal to 52.78 m.

6 0

3 years ago

SOMEONE PLEASE HELP PLEASE PLEASE

Black_prince [1.1K]

Ais the correct answer

4 0

2 years ago

A pendulum oscillates 50 times in 6 seconds. Find its time period and frequency? ​

A pendulum oscillates 50 times in 6 seconds. Find its time period and frequency?