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Brut [27]
3 years ago
7

When a pendulum is pulled back from its equilibrium position by 10∘, the restoring force is 1.0 N. When it is pulled back to 30∘

, the force increases to 2.8 N. If the pendulum is then released from 30∘, will its motion be periodic? If the pendulum is then released from 30, will its motion be periodic?a) No, it will be linear motion.b) No, the motion will be chaotic.c)Yes and it will be exactly harmonic.d)Yes, but it will not be exactly harmonic.
Physics
2 answers:
jarptica [38.1K]3 years ago
5 0

Answer: B

Explanation: I said B because if you pull something back what is going to be more of a force pulling back or letting it go for a rubier band yes it will have more force if you let it go

fenix001 [56]3 years ago
3 0

Answer:

The answer is d) Yes, but it will not be exactly harmonic: the pendulum’s motion will be periodic, but will not be exactly harmonic because the condition for simple harmonic motion is not satisfied when angular displacements become larger. Typically, a pendulum’s motion will exhibit simple harmonic motion if angular displacements are smaller.

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Jules verne wrote a book called twenty thousand leagues under the sea. is the ocean really that deep?explain.
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7 0
4 years ago
A wave has a velocity of 24 m/s and a period of 3.0 s. Calculate the wavelength of the wave.
Katyanochek1 [597]

Velocity (unit:m/s) of the wave is given with the formula:

v=f∧,

where f is the frequency which tells us how many waves are passing a point per second (unit: Hz) and ∧ is the wavelength, which tells us the length of those waves in metres (unit:m)

f=1/T , where T is the period of the wave.

In our case: f=1/3

∧=v/f=24m/s/1/3=24*3=72m

5 0
3 years ago
The height of a triangle is 4 in. greater than twice its base. The area of the triangle is no more than 168 in.2. Which inequali
olga2289 [7]

<span>The height (h) of a triangle is 4 in. greater than twice its base (x): h = 2x + 4
</span>The area of the triangle is no more than 168 in.2: A ≤ 168 in²

The area of the triangle is A = x*h/2
Hence x*h/2 ≤ 168 in²

Substitute h in the formula for the area:
h = 2x + 4 = 2(x + 2)
x*h/2 ≤ 168 in²

x * 2(x + 2)/2 ≤ 168
x * (x + 2) ≤ 168
x² + 2x ≤ 168
x² + 2x - 168 ≤ 0

Using the formula for quadratic equation:
x_{1,2} = \frac{-b+/- \sqrt{ b^{2}-4ac } }{2a} = \frac{-2+/- \sqrt{ 2^{2}-4*1*(-168) } }{2*1}=  \frac{-2+/- \sqrt{ 4+672 } }{2}=  \\  \\ \frac{-2+/- \sqrt{ 676 } }{2}=\frac{-2+/- 26 }{2} \\  \\ &#10;x_1 =  \frac{-2+26}{2} = \frac{24}{2} =12 \\  \\ x_1 =  \frac{-2-26}{2} = \frac{-28}{2} =-14&#10;

6 0
3 years ago
Read 2 more answers
A car with a speed of 50ms is stopped when its brakes are pressed. The displacement passed before stopping is 150 m. How much de
monitta
I dont think it can be solved as time is not mentioned
8 0
3 years ago
A 5.0 cm object is 12.0 cm from a concave mirror that has a focal length of 24.0 cm. The distance between the image and the mirr
MA_775_DIABLO [31]

Answer:

The answer is 24cm

Explanation:

This problem bothers on the curved mirrors, a concave type

Given data

Object height h= 5cm

Object distance = 12cm

Focal length f=24cm

Let the image distance be v=?

Applying the formula we have

1/v +1/u= 1/f

Substituting our given data

1/v+1/12=1/24

1/v=1/24-1/12

1/v=1-2/24

1/v=-1/24

v= - 24cm

This implies that the image is on the same side as the object and it is real

7 0
3 years ago
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