Students set up a controlled experiment (please help I am so tired and unable to function lol)

A 297 g block is connected to a light spring with spring constant 4.34 N/m, and displaced 7.45 cm from equilibrium. It is then r

vagabundo [1.1K]

Answer:

x = A sin ω t describes the displacement of the particle

v = A ω cos ω t

a = -A ω^2 sin ω t

a (max) = -A ω^2 is the max acceleration (- can be ignored here)

ω = (K/ m)^1/2 for SHM

F = - K x^2 restoring force of spring

K = 4.34 / .0745^2 = 782 N / m

ω = (782 / .297)^1/2 = 51.3 / sec

a (max) = .0745 * 782 / .297 = 196 m / s^2

4 0

1 year ago

Suppose that the sound level of a conversation is initially at an angry 71 dB and then drops to a soothing 54 dB. Assuming that

Goryan [66]

Answer:

Explanation:

In the decibel scale , intensity of sound changes logarithmically as follows

$10log\frac{I}{I_0} =$ Value in decibel scale , the value of I₀ = 10⁻¹² W /m².

Putting the values

$10log\frac{I}{10^{-12}} = 71$

$log\frac{I}{10^{-12}} = 7.1$

$\frac{I}{10^{-12}} = 10^{7.1}$

$I= 10^{-4.9}$ W/m²

Similarly for 54 dB sound intensity can be given as follows

I = 10⁻¹² x $10^{5.4}$

$I= 10^{-6.6 }$ W / m²

For intensity of sound the relation is as follows

I = 2π²υ²A²ρc where υ is frequency , A is amplitude , ρ is density of air and c is velocity of sound .

Putting the given values for 71 dB

$I= 10^{-4.9}$ = 2π² x 504²xA²x 1.21 x 346

A² = 60.03 x 10⁻¹⁶

A = 7.74 x 10⁻⁸ m

For 54 dB sound

$10^{-6.6}$ = 2π² x 504²xA²x 1.21 x 346

A² = 1.1978 x 10⁻¹⁶

A = 1.1 x 10⁻⁸ m

6 0

3 years ago

A 5.0-kilogram sphere, starting from rest, falls freely 22 meters in 3.0 seconds near the surface of a planet. Compared to the a

Whitepunk [10]

Answer:

C) one-half as great

Explanation:

We can calculate the acceleration of gravity in that planet, using the following kinematic equation:

$\Delta x=v_0t+\frac{gt^2}{2}$

In this case, the sphere starts from rest, so $v_0=0$ . Replacing the given values and solving for g':

$g'=\frac{2\Delta x}{t^2}\\g'=\frac{2(22m)}{(3s)^2}\\g'=4.89\frac{m}{s^2}$

The acceleration due to gravity near Earth's surface is $g=9.8\frac{m}{s^2}$ . So, the acceleration due to gravity near the surface of the planet is approximately one-half of the acceleration due to gravity near Earth's surface.

5 0

2 years ago

Can someone please help

ki77a [65]

Answer:

Acceleration of that planet is 30 $\frac{m}{s^{2} }$ .

Given:

initial speed of hammer = 0 $\frac{m}{s}$

time = 1 s

distance = 15 m

To find:

Acceleration due to gravity = ?

Formula used:

Distance covered by hammer is given by,

s = ut + $\frac{1}{2} a t^{2}$

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

Solution:

Distance covered by hammer is given by,

s = ut + $\frac{1}{2} a t^{2}$

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

u = 0

t = 1 s

s = 15 m

a = g

Thus substituting these value in above equation.

15 = 0 + $\frac{1}{2} g 1^{2}$

g = 15 × 2

g = 30 $\frac{m}{s^{2} }$

Thus, acceleration of that planet is 30 $\frac{m}{s^{2} }$ .

8 0

3 years ago

What is the relationship between force and motion described by Newton's first law

prisoha [69]

Newtons First Law of Motion:
An object at rest stays at rest and an object in motion<span> stays in </span>motion <span>with the same speed and in the same direction unless acted upon by an unbalanced force.</span>

Therefore, the relationship between force and motion is that it takes force to change the speed or direction of any object in motion.

4 0

3 years ago