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3 years ago

Determine the value of the resultant and its location from O.see attach image.

Physics

1 answer:

xxTIMURxx [149]3 years ago

8 0

Answer:

Explanation:

In the x direction the force will be

½(-w₀)L/2 = -¼w₀L

acting ⅔(L/2) = L/3 below the x axis.

In the y direction the force will be

½(-w₀)L + ½w₀L/2 = -¼w₀L

the magnitude of the resultant will be

F = w₀L √((-¼)² + (-¼)²) = w₀L√⅛

in the direction

θ = arctan(-¼w₀L / -¼w₀L) = 225°

to find the distance, we balance moments

(w₀L√⅛)[d] = ½(w₀)L[⅔L] + ¼w₀L[⅔L/2] - ¼w₀L[L - ⅓L/2]

(√⅛)[d] = ½ [⅔L] + ¼ [⅔L/2] - ¼ [L - ⅓L/2]

(√⅛)[d] = ½[⅔L] + ¼[⅔L/2] - ¼[L - ⅓L/2]

(√⅛)[d] = ⅓L + ⅟₁₂L - ¼L + ⅟₂₄L

(√⅛)[d] = 5L/24

d = 5L/24 / (√⅛)

d = 5√⅛L/3

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2 years ago

A dog walking to the right at 1.5\,\dfrac{\text m}{\text s}1.5sm1, point, 5, space, start fraction, m, divided by, s, end fract

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Answer:

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3 years ago

A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are

slava [35]

Answer:

The magnitude and direction of the acceleration of the particle is $a= 0.3296\ \hat{k}\ m/s^2$

Explanation:

Given that,

Mass $m = 1.81\times10^{-3}\ kg$

Velocity $v = (3.00\times10^{4}\ m/s)j$

Charge $q = 1.22\times10^{-8}\ C$

Magnetic field $B= (1.63\hat{i}+0.980\hat{j})\ T$

We need to calculate the acceleration of the particle

Formula of the acceleration is defined as

$F = ma=q(v\times B)$

$a =\dfrac{q(v\times B)}{m}$

We need to calculate the value of $v\times B$

$v\times B=(3.00\times10^{4}\ m/s)j\times(1.63\hat{i}+0.980\hat{j})$

$v\times B=4.89\times10^{4}$

Now, put the all values into the acceleration 's formula

$a =\dfrac{1.22\times10^{-8}\times(-4.89\times10^{4}\hat{k})}{1.81\times10^{-3}}$

$a= -0.3296\ \hat{k}\ m/s^2$

Negative sign shows the opposite direction.

Hence, The magnitude and direction of the acceleration of the particle is $a= 0.3296\ \hat{k}\ m/s^2$

7 0

3 years ago

Read 2 more answers

Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 3 m3, is stirred

madam [21]

Answer:

a)P₂ =4 bar

b)W= - 1482.48 KJ

It means that work done on the system.

c)S₂ - S₁ = 3.42 KJ/K

Explanation:

Given that

T₁ = 300 K ,V₁ = 3 m³ ,P₁=2 bar

T₂ = 600 K ,V₂=V₁ 3 m³

Given that tank is rigid and insulated.It means that volume of the gas will remain constant.

Lets take the final pressure = P₂

For ideal gas P V = m R T

$\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}$

$P_2=\dfrac{T_2}{T_1}\times P_1$

$P_2=\dfrac{600}{300}\times 2$

P₂ =4 bar

Internal energy

ΔU = m Cv ΔT

Cv=0.71 KJ/kg.k for air

$m=\dfrac{PV}{RT}$

$m=\dfrac{200\times 3}{0.287\times 300}\ kg$

m= 6.96 kg

ΔU= 6.96 x 0.71 x (600 - 300)

ΔU=1482.48 KJ

From first law

Q= ΔU + W

Q= 0 Insulated

W = - ΔU

W= - 1482.48 KJ

It means that work done on the system.

Change in the entropy

$S_2-S_1=mC_v\ \ln\dfrac{T_2}{T_1}$

$S_2-S_1=6.96\times 0.71\ \ln\dfrac{600}{300}$

S₂ - S₁ = 3.42 KJ/K

5 0

3 years ago

Determine the value of the resultant and its location from O.see attach image.​

Determine the value of the resultant and its location from O.see attach image.