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irakobra [83]
3 years ago
7

Can someone help me?!!!!!

Physics
1 answer:
saveliy_v [14]3 years ago
3 0
<h2>Hello!</h2>

The answer is:

The third option, the approximate magnitude of the given vector is 9.9 units.

|A|=9.9units

<h2>Why?</h2>

To calculate the magnitude (length) of a vector, we need to apply the following formula:

|A|=\sqrt{(A_x)^{2} +(A_y)^{2} }

So, we are given the vector:

A=(7.6,-6.4)

Then,  substituting and calculating the magnitude of the vector, we have:

|A|=\sqrt{(7.6)^{2} +(-6.4)^{2}}=\sqrt{57.76+40.96}=\sqrt{98.72}\\\\|A|=\sqrt{98.72}=9.9units

Hence, we have that the correct option is the third option, the approximate magnitude of the given vector is 9.9 units.

|A|=9.9units

Have a nice day!

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The four-wheel-drive all-terrain vehicle has a mass of 385 kg with center of mass G2. The driver has a mass of 75 kg with center
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The coefficient of friction is missing and it has a value of μ = 0.4

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a = 3.924 m/s²

Explanation:

I've attached the kinematic free body diagram.

Taking the sum of all upward and downward forces,

EFy = 0;

N1 + N2 - m_p•g - m_v•g = 0

N1 + N2 = m_p•g + m_v•g

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N1 and N2 are the normal reactions at the wheels

m_p is the mass of the driver

m_v is the mass of the vehicle

g is the acceleration due to gravity.

Plugging in the relevant values in the question,we obtain;

N1 + N2 = (385 + 75) x 9.81

N1 + N2 = 4512.6N - - - (eq1)

Now, taking sum of all horizontal forces;

EFx = (m_p + m_v) x a

So,

μ(N1 + N2) = (mp + mv) x a

Thus,

0.4(N1 + N2) = (385 + 75)a

0.4(N1 + N2) = 460a

N1 + N2 = 1150a

From eq(1),N1 + N2 = 4512.6N

Thus,

1150a = 4512.6N

a = 4512.6/1150

a = 3.924 m/s²

Therefore, the acceleration, a = 3.924 m/s²

7 0
3 years ago
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