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GREYUIT [131]
3 years ago
14

SHOW WORK

Physics
1 answer:
Helga [31]3 years ago
5 0

Answer:

Follows are the solution to the given question:

Explanation:

For point a:

T= 2\pi \sqrt{\frac{m}{k}}\\\\k = \frac{4 \pi^2 m}{T^2}\\\\= \frac{4 \times (3.14)^2 \times 3}{2^2}\\\\=29.578 \ \frac{N}{m}\\\\

For Point b:

E=\frac{1}{2} m a^2 w^2\\\\

   =\frac{1}{2} \times m \times a^2 \times \frac{4\pi^2}{T^2}\\\\=\frac{1}{2} \times 3 \times (0.15)^2 \times \frac{4\times 3.14^2}{2^2}\\\\=0.332 \ J

For Point C:

V_{max}= a w

        = (0.15) \times \frac{2\pi}{T}\\\\= (0.15) \times \frac{2\times 3.14}{2}\\\\=0.471 \frac{m}{s}

For point D:

X= a \sin (wt+ \phi)\\\\0.91=0.15 \sin(\frac{2\pi}{T} \times t+\phi)\\\\0.91=0.15 \sin(\frac{2\times 3.14}{2} \times 0.5+\phi)\\\\0.60 = \sin(3.14 \times 0.5+\phi)\\\\0.60 = \sin(1.57+\phi)\\\\1.57 +\phi =\sin^{-1} 60^{\circ}\\\\1.57 +\phi = 36.86^{\circ}\\\\=35.29^{\circ}\\\\So, X=15 \sin(3.14t+35.29^{\circ}) \ cm

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I think these two are simple questions.. but I need help asap..... TT
frosja888 [35]

1) The average velocity is 56 m/min

2) The average velocity is -83 m/min

Explanation:

1)

The average velocity of an object is given by

v=\frac{d}{t}

where

d is the displacement (change in position)

t is the time elapsed

In the graph in the problem, the displacement corresponds to the distance, therefore to the change in the y-variable (\Delta y), while the time elapsed is the change in the x-variable (\Delta x), so the average velocity can be written as

v=\frac{\Delta y}{\Delta x}

At point A, we have:

y_A = 5 m\\x_A = 0.1 min

At point B, we have:

y_E = 55 m\\x_A = 1 min

So, we have

\Delta y= 55 -5 = 50 m\\\Delta x = 1.0-0.1 = 0.9 min

So the average velocity is

v=\frac{50 m}{0.9 min}=56 m/min

2)

In this part instead, we have the following:

At point F, we have:

y_F = 55 m\\x_A = 1.3 min

At point H, we have:

y_H = 30 m\\x_A = 1.6 min

So, we have

\Delta y= 30 -55 = -25 m\\\Delta x = 1.6-1.3 = 0.3 min

So the average velocity is

v=\frac{-25 m}{0.3 min}=-83 m/min

Learn more about velocity:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

4 0
2 years ago
The charge on a parallel plate capacitor is 7.68 C, and the plate is connected to a 9.0 V battery. What is the capacitance of th
Natali5045456 [20]

Charge on a capacitor = (capacitance) x (voltage)

7.68 Coulomb = (capacitance) x (9.0 V)

Capacitance = (7.68 Coul) / (9.0 V)

<em>Capacitance = 0.83 Farad</em>

<em></em>

Apparently, the " 10^something " after the 7.68 C got lost.

If the charge is actually 7.68 <u>micro</u>coulombs, then the answer is<em> (A) .</em>

8 0
3 years ago
100 POINTS 100 POINTS 100 POINTS!!!!!<br> HELP PLEASE I DON'T KNOW WHAT TO DO!!!!!
Westkost [7]

Answer:

block 2 or 4

because of the distribution of weight and force being applied to the object

3 0
3 years ago
Read 2 more answers
A horizontal clothesline is tied between 2 poles, 16 meters apart. When a mass of 3 kilograms is tied to the middle of the cloth
Alla [95]

Answer:

The magnitude of the tension on the ends of the clothesline is 41.85 N.

Explanation:

Given that,

Poles = 2

Distance = 16 m

Mass = 3 kg

Sags distance = 3 m

We need to calculate the angle made with vertical by mass

Using formula of angle

\tan\theta=\dfrac{8}{3}

\thta=\tan^{-1}\dfrac{8}{3}

\theta=69.44^{\circ}

We need to calculate the magnitude of the tension on the ends of the clothesline

Using formula of tension

mg=2T\cos\theta

Put the value into the formula

3\times9.8=2T\times\cos69.44

T=\dfrac{3\times9.8}{2\times\cos69.44}

T=41.85\ N  

Hence, The magnitude of the tension on the ends of the clothesline is 41.85 N.

4 0
3 years ago
Read 2 more answers
The acceleration due to gravity, g , is constant at sea level on the Earth's surface. However, the acceleration decreases as an
blsea [12.9K]

Answer:

  g    = g₀   [1- 2 h / Re + 3 (h / Re)²]

Explanation:

The law of universal gravitation is

        F = G m Me / Re²

Where g is the universal gravitational constant, m and Me are the mass of the body and the Earth, respectively and R is the distance between them

      F = G Me /Re²  m

We call gravity acceleration a

       g₀ = G Me / Re².

When the body is at a height h above the surface the distance is

            R = Re + h

Therefore  the attractive force is

      F = G Me m / (Re + h)²

Let's take Re's common factor

      F = G Me / Re²  m / (1+ h / Re)²

As Re has a value of 6.37 10⁶ m and the height of the body in general is less than 10⁴ m, the h / Re term is very small, so we can perform a series expansion

         (1+ h / Re)⁻² = 1 -2 h / Re + 6/2 (h / Re) 2 + ...

Let's replace

       F = G Me /Re²   m [1- 2 h / Re + 3 (h / Re)²]

       F = g₀   m  [1- 2 h / Re + 3 (h / Re)²]

If we call the force of attraction at height

     m g =g₀ m  [1- 2 h / Re + 3 (h / Re)²]

       g    = g₀   [1- 2 h / Re + 3 (h / Re)²]

3 0
3 years ago
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