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Annette [7]
2 years ago
9

What does the y-intercept mean in a Period vs. Length graph?

Physics
1 answer:
nadya68 [22]2 years ago
4 0
You should search it up in the internet
You might be interested in
Need help ???? Only answer if you think you know it
charle [14.2K]

Answer:

I believe it would be Direction B

Explanation:

After passing A, gravity would pull the ball downwards

5 0
3 years ago
A spring stretches 2.6 cm when a 7 g object is hung from it. The object is replaced with a block of mass 28 g which oscillates i
PSYCHO15rus [73]

Answer:

period = 0.65 sec

Explanation:

from the question we are given the following

extension (x) = 2.6 cm = 0.026 m

mass of object (Mo) = 7 g = 0.007 kg

mass of block (Mb) = 28 g = 0.028 g

acceleration due to gravity (g) = 9.8 m/s^{2}

period = 2π x \sqrt{\frac{Mb}{k}}

where k is the spring constant of the spring

and k = \frac{Mo x g}{x}

k =  \frac{0.007 x 9.8}{0.026}

k = 2.64 N/m

now period = 2π x \sqrt{\frac{0.028}{2.64}}

period = 0.65 sec

4 0
3 years ago
A wave of wavelength 0.3 m travels 900 m in 3.0 s. Calculate its frequency.
Crank

Answer: 1000 Hz

Explanation:

You can calculate frequency by dividing velocity by wavelength

Frequency = velocity/wavelength

Find velocity first.

900 m/3 s = 300 m/s

Plug values in to find frequency.

F = (300 m/s)/0.3 m

F = 1000 Hz

8 0
2 years ago
A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the
irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

x=V_{o} cos \theta t (1)

y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2} (2)

Where:

x is the horizontal distance between the cannon and the ball

V_{o}=23 m/s is the cannonball initial velocity

\theta=0\° since the cannonball was shoot horizontally

t is the time

y=0 is the final height of the cannonball

y_{o}=1.96 m is the initial height of the cannonball

g=9.8 m/s^{2} is the acceleration due gravity

Isolating t from (2):

t=\sqrt{-\frac{2(y-y_{o})}{g}} (3)

t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}} (4)

t=0.63 s (5)

Substituting (5) in (1):

x=(23 m/s) cos(0\°) 0.63 s (6)

Finally:

x=14.49 m

5 0
2 years ago
La tensión en newtons necesaria para que una onda transversal cuya longitud de onda es 3.33 cm vibre a razón de 625 ciclos por s
NemiM [27]

Answer:

9.34 N

Explanation:

First of all, we can calculate the speed of the wave in the string. This is given by the wave equation:

v=f \lambda

where

f is the frequency of the wave

\lambda is the wavelength

For the waves in this string we have:

f=625 Hz, since it completes 625 cycles per second

\lambda=3.33 cm = 0.033 m is the wavelength

So the speed of the wave is

v=(625)(0.0333)=20.6 m/s

The speed of the waves in a string is related to the tension in the string by

v=\sqrt{\frac{T}{\mu}} (1)

where

T is the tension in the string

\mu=\frac{m}{L} is the linear density

In this problem:

m=16.5 g = 16.5\cdot 10^{-3} kg is the mass of the string

L = 0.75 m is the its length

Solving the equation (1) for T, we find the tension:

T=\mu v^2 = \frac{m}{L} v^2 = \frac{16.5\cdot 10^{-3}}{0.75}(20.6)^2=9.34 N

8 0
3 years ago
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