Im B O R E D someone please entertain me :(

An object with mass 3.5 kg is attached to a spring with spring stiffness constant k = 270 N/m and is executing simple harmonic m

Elanso [62]

Answer:

Part a)

A = 0.066 m

Part b)

maximum speed = 0.58 m/s

Explanation:

As we know that angular frequency of spring block system is given as

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

here we know

m = 3.5 kg

k = 270 N/m

now we have

$\omega = \sqrt{\frac{270}{3.5}}$

$\omega = 8.78 rad/s$

Part a)

Speed of SHM at distance x = 0.020 m from its equilibrium position is given as

$v = \omega \sqrt{A^2 - x^2}$

$0.55 = 8.78 \sqrt{A^2 - 0.020^2}$

$A = 0.066 m$

Part b)

Maximum speed of SHM at its mean position is given as

$v_{max} = A\omega$

$v_{max} = 0.066(8.78) = 0.58 m/s$

4 0

3 years ago

Ksp for agbr is 5x10-13. what is the maximum concentration of silver ion that you can have in a 0.1 m solution of nabr?

liberstina [14]

Answer : The maximum concentration of silver ion is $5\times 10^{-12}m$

Solution : Given,

$K_{sp}$ for AgBr = $5\times 10^{-13}$

Concentration of NaBr solution = 0.1 m

The equilibrium reaction for NaBr solution is,

$NaBr(aq)\rightleftharpoons Na^++Br^-$

The concentration of NaBr solution is 0.1 m that means,

$[Na^+]=[Br^-]=0.1m$

The equilibrium reaction for AgBr is,

$AgBr\rightleftharpoons Ag^++Br^-$

At equilibrium s s

The expression for solubility product constant for AgBr is,

$K_{sp}=[Ag^+][Br^-]$

The concentration of $Ag^+$ = s

The concentration of $Br^-$ = 0.1 + s

Now put all the given values in $K_{sp}$ expression, we get

$5\times 10^{-13}=(s)(0.1+s)$

By rearranging the terms, we get the value of 's'

$s=5\times 10^{-12}m$

Therefore, the maximum concentration of silver ion is $5\times 10^{-12}m$ .

4 0

3 years ago

Read 2 more answers

The chances of being involved in a fatal crash is higher at night then during the day

Ierofanga [76]

Indeed because some leave headlights off and ignore that fact since there are street lights around.

3 0

3 years ago

Please help me with this question

snow_lady [41]

politics. Famous possible eg is of Werner Heisenberg in WW2. He delayed German attempt to build a nuclear bomb. US did build one ... hiroshima and nagasaki.

debbie may have got skilfully lucky by trial and error

6 0

3 years ago

Read 2 more answers

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti

Colt1911 [192]

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

5 0

2 years ago