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WITCHER
1 month ago
9

What mass needs to be attached to a spring with a force constant of 7N/m in order to make a simple harmonic oscillator oscillate

with a time period of 3s
Physics
1 answer:
Sav [3.1K]1 month ago
6 0

Answer:

The mass will be 4.437 kg

Explanation:

The force constant k is given as 7 N/m

The time period of oscillation T is 5 sec

Thus, angular frequency \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{5}=1.256rad/sec

It is known that angular frequency is computed via

\omega =\sqrt{\frac{k}{m}}

1.256 =\sqrt{\frac{7}{m}}

Squaring both sides gives us

1.577 =\frac{7}{m}

The mass equals 4.437 kg

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An object with charge q=−6.00×10−9C is placed in a region of uniform electric field and is released from rest at point A. After
kicyunya [3294]

Answer:

it’s a

Explanation:

utilized Google

8 0
2 months ago
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
inna [3103]

Answer:

the time it takes after impact for the puck is 2.18 seconds

Explanation:

initially given information

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thickness = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

temperature of air = 15°C

to determine

time needed for the puck to reduce its speed by 10%

solution

we note that velocity changes from 0 to v

assuming initial velocity = v

therefore final velocity = 0.9v

implying a change in velocity is du = v

and clearance dy = h

shear stress acting on the surface is expressed as

= µ \frac{du}{dy}

therefore

= µ \frac{v}{h}............1

substituting the values

= 1.75 ×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

the area between the air and puck is given by

Area = \frac{\pi }{4} d^{2}

area = \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

thus, the force on the puck can be represented as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v

now applying Newton's second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

solving for t = \frac{0.1 v * 0.03}{1.37*10^{-3} v}

the time needed after impact for the puck is 2.18 seconds

3 0
2 months ago
A girl drops a pebble from a high cliff into a lake far below. She sees the splash of the pebble hitting the water 2.00s later.
Yuliya22 [3333]

Answer:

19.62 ms

Explanation:

t = Time taken = 2 s

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² (we take downward direction as positive)

v=u+at\\\Rightarrow v=0+9.81\times 2\\\Rightarrow v=19.62\ m/sUsing the equations of motion

The pebble's speed upon contact with the water is 19.62 ms

3 0
1 month ago
A particular brand of gasoline has a density of 0.737 g/mLg/mL at 25 ∘C∘C. How many grams of this gasoline would fill a 14.9 gal
Yuliya22 [3333]

Answer:

The answer to your inquiry is Mass = 41230.7 g or 41.23 kg.

Explanation:

Data

Density = 0.737 g/ml

Mass = ?

Volume = 14.9 gal

1 gal = 3.78 l

Process

1.- Convert gallons to liters

1 gal ---------------- 3.78 l

14.9 gal ------------- x

x = 56.44 l

2.- Convert liters to milliliters

1 l ------------------- 1000 ml

56.44 l --------------- x

x = (56.44 x 1000) / 1

x = 56444 ml

3.- Calculate the mass

Formula

Density = \frac{mass}{volume}

Solving for mass

Mass = density x volume

Substituting values

Mass = 0.737 x 56444

Result

Mass = 41230.7 g or 41.23 kg.

3 0
3 months ago
Two charges q1 = 5 µC, q2 = -26 µC, are L = 19 cm apart. A third charge is to be placed on the line between the two charges. How
Keith_Richards [3271]
The electric force between two objects is expressed as being proportional to the product of their charges and inversely proportional to the square of the distance separating them. In this instance, the distance between the first two charges is 19 cm. We formulate the equation k q1 q3/ (x)^2 = k q2 q3/ (19-x)^2, where x denotes the separation between q1 and q3. The charge q3 cancels out, and q2 is used in absolute terms. The resulting value of x is 5.79 cm.
6 0
1 month ago
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