Physics Middle School

A simple pendulum has a period of 3.45 second, when the length of the pendulum is shortened by 1.0m, the period is 2.81 second calculate, the original length of the pendulum, the value of accelerations due to gravity​

Answers

Answer 1

Answer:

Original length = 2.97 m

Explanation:

Let the original length of the pendulum be 'L' m

Given:

Acceleration due to gravity (g) = 9.8 m/s²

Original time period of the pendulum (T) = 3.45 s

Now, the length is shortened by 1.0 m. So, the new length is 1 m less than the original length.

New length of the pendulum is, [tex]L_1=L-1[/tex]

New time period of the pendulum is, [tex]T_1=2.81\ s[/tex]

We know that, the time period of a simple pendulum of length 'L' is given as:

[tex]T=2\pi\sqrt{\frac{L}{g}}[/tex]-------------- (1)

So, for the new length, the time period is given as:

[tex]T_1=2\pi\sqrt{\frac{L_1}{g}}[/tex]------------ (2)

Squaring both the equations and then dividing them, we get:

[tex]\dfrac{T^2}{T_1^2}=\dfrac{(2\pi)^2\frac{L}{g}}{(2\pi)^2\frac{L_1}{g}}\\\\\\\dfrac{T^2}{T_1^2}=\dfrac{L}{L_1}\\\\\\L=\dfrac{T^2}{T_1^2}\times L_1[/tex]

Now, plug in the given values and calculate 'L'. This gives,

[tex]L=\frac{3.45^2}{2.81^2}\times (L-1)\\\\L=1.507L-1.507\\\\L-1.507L=-1.507\\\\-0.507L=-1.507\\\\L=\frac{-1.507}{-0.507}=2.97\ m[/tex]

Therefore, the original length of the simple pendulum is 2.97 m

Answer 2
Final answer:

The period of a simple pendulum can be calculated with the formula T = 2π√(L/g). When the length of the pendulum is shortened by 1.0m, the period becomes 2.81 seconds. By solving a system of equations related to the period and the length of the pendulum, we can get the original length and the acceleration.

Explanation:

The subject matter relates to physics, specifically kinematics, and the topic is the period of a simple pendulum. The formula to calculate the period of a simple pendulum is T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity. We know that when the length of the pendulum was shortened by 1.0m, its period became 2.81 seconds. We can plug these values into our formula and solve for the original length (L).

To find the original length, we first let L0 be the original length and L1 be the shortened length. Therefore, L1 = L0 - 1m. Now we have two equations, T0 = 2π√(L0/g) and T1 = 2π√(L1/g). Plug in the known values and solve the system of equations, you can get the original length and the acceleration. Note, that the typical value for acceleration due to gravity on Earth is g = 9.8 m/s².

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Related Questions

If the mass of an object is 44 kilograms and its velocity is 10 meters per second east, how much Kinetic Energy does it have?

Answers

Answer: 2200J

Explanation:

M = 44kg

V = 10m/s

K.E =?

K.E = 1/2MV2 = 1/2 x 44 x (10)^2

K.E = 22 x 100

K.E = 2200J

. A car moves forward up a hill at 12 m/s with a uni-
form backward acceleration of 1.6 m/s2.
a. What is its displacement after 6.0 s?
b. What is its displacement after 9.0 s?

Answers

A) Displacement after 6.0 s 43.2 m uphill.

B) Displacement after 9.0 s 43.2 m uphill.

Explanation:

A car moving upwards in a hill is [tex]12 ms^{-1}[/tex].

Its uniform backward acceleration is [tex]-1.6ms^{-2}[/tex]. (since backward acceleration is a negative acceleration, it is mentioned in negative)

We need to find the displacement of the car after some time.

Using the equation of the motion formula, we know can identify the displacement.

D=[tex]vt+\frac{1}{2} at^2[/tex].

a) Displacement after 6.0 seconds,

D = [tex]12(6.0)+\frac{1}{2}(-1.6)(6.0)^2[/tex].

=[tex]72+\frac{1}{2} (36)(-1.6).[/tex]

=[tex]72+\frac{1}{2}(-57.6).[/tex]

=72-28.8.

D=43.2 m.

b) Displacement after 9.0 seconds,

D= [tex]12(9.0)+\frac{1}{2}(-1.6)(9.0)^2[/tex].

=[tex]108+\frac{1}{2} (81)(-1.6).[/tex]

=[tex]108+\frac{1}{2}(-129.6).[/tex]

= 108-64.8.

D=43.2 m.

The car's displacement after 6.0 s is 14.4 m and after 9.0 s is 43.2 m.

To find the displacement of the car after a given time, we can use the equation:
Displacement (d) = Initial velocity (v) * time (t) + (1/2) * acceleration (a) * time^2

a. After 6.0 s:
Initial velocity (v) = 12 m/s
Acceleration (a) = -1.6 m/s^2 (negative because it's a backward acceleration)
Substituting the values into the equation:
d = (12 m/s) * (6.0 s) + (1/2) * (-1.6 m/s^2) * (6.0 s)^2 = 72 m - 57.6 m = 14.4 m

b. After 9.0 s:
Using the same equation and substituting the new time, we can calculate the displacement:
d = (12 m/s) * (9.0 s) + (1/2) * (-1.6 m/s^2) * (9.0 s)^2 = 108 m - 64.8 m = 43.2 m

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Which of the following is a result of gravitational forces in the Solar System?
A.
the radiation given off by Jupiter
B.
Saturn is further away from the Sun than Earth
C.
the difference in surface temperature on each of the planets
D.
the orbit of moons around their planets in the Solar system

Answers

Answer:

D

Explanation:

the answer is d because gravitational force is what allows them to rotate

hope this was helpful

a child pulls on a string that is attached to a car. if the child does 80.2 J of work while pulling the car 25.0 m, with what force is the child pulling?

Answers

Answer:

F = 3.20 N

Explanation:

Given:

Work done by child = 80.2 j

Distance that the car moves = 25.0 m

We need to find the force acting on the car.

Solution:

Using work done formula as.

[tex]W = F\times d[/tex]

Where:

W = Work done by any object.

F = Force (push or pull)

d = distance that the object moves.

Substitute [tex]W = 80.2\ J\ and\ d =25.0\ m[/tex] in work done formula.

[tex]80.2 = F\times 25[/tex]

[tex]F=\frac{80.2}{25}[/tex]

F = 3.20 N

Therefore, force acting on the car F = 3.20 N

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