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frez [133]
3 years ago
11

Two forces, F1 and F2, act at a point. F1 has a magnitude of 8.60 N and is directed at an angle of 55.0 ∘ above the negative x a

xis in the second quadrant. F2 has a magnitude of 7.00 N and is directed at an angle of 53.2 ∘ below the negative x axis in the third quadrant. What is the x component Fx of the resultant force?
Physics
1 answer:
9966 [12]3 years ago
5 0

Answer:9.125 N

Explanation:

Given

Magnitude of Force F_1=8.60 N at  angle of 55^{\circ} above negative x axis

Magnitude of  Force F_2=7 N at  angle of 53.2^{\circ} below negative x axis

\theta _1=180^{\circ}-55^{\circ}=125^{\circ}

\theta _2=180^{\circ}+53.2^{\circ}=233.2^{\circ}

Thus net force in x direction

F_x=F_1\cos \theta _1+F_2\cos \theta _2

F_x=8.6\times \cos (125)+7\times \cos (233.2)

F_x=-9.1251 N towards negative x axis

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Explanation:

Given that,

Distance 1, r = 100 m

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Let I' is the intensity at r'. So,

I'=\dfrac{P}{4\pi r'^2}............(2)

From equation (1) and (2) :

I'=\dfrac{Ir}{r'^2}

I'=\dfrac{1.24\times 10^{-8}\times 100}{25^2}

I'=1.98\times 10^{-9}\ W/m^2

Intensity level is given by :

dB=10\ log(\dfrac{I'}{I_o}), I_o=10^{-12}\ W/m^2

dB=10\ log(\dfrac{1.98\times 10^{-9}}{10^{-12}})

dB = 32.96 dB

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Correct answer choice is :


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Explanation:


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4 0
3 years ago
Read 2 more answers
2) Two ice skaters have masses m1 and m2 and are initially stationary. Their skates are identical. They push against one another
worty [1.4K]

Answer:

m_1 / m_2 = sqrt (1 / 2)

Explanation:

Given:

- Initial velocity of both skaters V_i = 0

- Velocity of skater 1 after push = V_1

- Velocity of skater  after push = V_2

- Distance traveled by skater 1 = s_1

- Distance traveled by skater 2 = s_2

- s_1 = 2*s_2

- Accelerations of both skaters to halt is equal

Find:

What is the ratio m1/m2 of their masses

Solution:

- Apply conservation of momentum for two skaters just before and after the push as follows:

                                              P_i = P_f

                                  0 = m_1*V_1 - m_2*V_2

- Evaluate:                 m_1 / m_2 = ( V_2 / V_1 )

- Apply Conservation of Energy on both skaters as follows:

- Skater 1:

                               0.5*m_1*V_1^2 = u_k*m_1*g*s_1

-Simplify:                      0.5*V_1^2 = u_k*g*(2*s_2)

- Skater 2:

                               0.5*m_2*V_2^2 = u_k*m_2*g*s_2

-Simplify:                      0.5*V_2^2 = u_k*g*s_2

- Divide the two energy equations for skaters:

                                    (V_1 / V_2)^2 = 2

                                    (V_2 / V_1)^2 = 1 / 2

- simplify:                     (V_2 / V_1) = sqrt (1 / 2)

-Hence from earlier momentum conservation results:

                                  m_1 / m_2 = ( V_2 / V_1 ) = sqrt (1 / 2)

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