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3 years ago

PLEASE HELP, I am desperate. This part of the presentation I am completely lost with. It is STEP 6, I got the rest finished.

Physics

1 answer:

yKpoI14uk [10]3 years ago

5 0

Answer:

Explanation:

Ok, pick some idea from here

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=10&ved=2ahUKEwjWoLaFypvnAhUowlkKHXZrCMkQFjAJegQIBBAB&url=https%3A%2F%2Fwww.wappingersschools.org%2Fcms%2Flib%2FNY01001463%2FCentricity%2FDomain%2F1552%2FGravity.ppt&usg=AOvVaw1N8kqPWy6ERpA3LNbbepit

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A 0.600 m long pendulum is used to determine the acceleration due to gravity on a distant plane. If 20 oscillations are complete

katrin2010 [14]

Answer:

7.50 m/s^2

Explanation:

The period of a pendulum is given by:

$T=2\pi \sqrt{\frac{L}{g}}$ (1)

where

L = 0.600 m is the length of the pendulum

g = ? is the acceleration due to gravity

In this problem, we can find the period T. In fact, the frequency is equal to the number of oscillations per second, so:

$f=\frac{N}{t}=\frac{20}{35.5 s}=0.563 Hz$

And the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{0.563 Hz}=1.776 s$

And by using this into eq.(1), we can find the value of g:

$g=\frac{4 \pi^2 L}{T^2}=\frac{4 \pi^2 (0.600 m)}{(1.776 s)^2}=7.50 m/s^2$

6 0

3 years ago

a man is 9 m behind a dorr of train when it starts moving with a=2ms^-1. how far the man have to run and after what time will he

Naya [18.7K]

Is their a multiple choice to choose from I'm not sure the answer I got is even right.
That would be very helpful.

8 0

3 years ago

What if m1 is initially moving at 3.4 m/s while m2 is initially at rest? (a) find the maximum spring compression in this case?

Lisa [10]

<span>Ans : Initial E = KE = Â˝mvÂ˛ = Â˝ * 1.2kg * (2.2m/s)Â˛ = 2.9 J max spring compression where both velocities are the same: conserve momentum: 1.2kg * 2.2m/s = (1.2 + 3.2)kg * v â†’ v = 0.6 m/s which means the combined KE = Â˝ * (1.2 + 3.2)kg * (0.6m/s)Â˛ = 0.79 J The remaining energy went into the spring: U = (2.9 - 0.79) J = 2.1 J = Â˝kxÂ˛ = Â˝ * 554N/m * xÂ˛ x = 0.0076 m â† (a)</span>

7 0

3 years ago

Each ball has a negligible size and a mass of 11.5 kg and is attached to the end of a rod whose mass may be neglected. The rod i

Digiron [165]

Answer:

3.31m/s

Explanation:

Angular momentum for 3s is

$L = L_i_n_i + L_3_s$

$L = 2(11.5kg) + \int\limits^ {3s}_ {0s} {(t^2 + 2)} \, dt$

$L = 23kg+(\frac{t^3}{3} +2t)^ {3s}_ {0s}\\\\L=38kgm/s$

Moment if inertia is

$I = 2ml^2$

$I = 2(11.5)(0.5)^2\\\\I=5.75kgm^2$

Angular speed

ω = L/I

$= 38 / 5.75\\\\=6.61$

The speed of each ball is

V = ωL

$= 6.61\times0.5\\\\= 3.31m/s$

7 0

3 years ago

A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m

Cloud [144]

Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, $I_1=900\ kg.m^2$

Initial angular velocity of the platform, $\omega=0.95\ rad/s$

Part A,

Let $\omega_2$ is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :