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3 years ago

An aircraft on its take-off run has a steady acceleration of 3 m/s2.

Physics

1 answer:

lakkis [162]3 years ago

6 0

Answer:

Acceleration(a)=3m/s^2

Time(t)=1min[1min=60s]

=1×60

=60s

Initial velocity(u)=0m/s

Velocity(v)=?

We know that,

Acceleration=v-u/t

Or,3m/s^2=v-0/60

Or,3×60=v-0

Or,v=180m/s

So,the velocity is 180m/s.

Hope,it will helpyouu!

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Newton's second law allows us to find the results for the string tensions are:

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Newton's second law gives a relationship between force, mass and acceleration of bodies

∑ F = ma

Where the bold letters indicate vectors, F is the force, m the mass and the acceleration.

Free-body diagrams are representations of the forces applied to bodies without the details of them.

The reference system is a coordinate system with respect to which the forces decompose, in this case the x-axis is parallel to the plane and the positive direction in the direction of movement, the y-axis is perpendicular to the plane.

In the attachment we see a free-body diagram of the three-block system.

Let's apply Newton's second law to each body.

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Y-axis

$W_c -T_2 = m_c a$

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X axis

$T_2 - T_1 - W_a_x = m_a a$

Y axis

$N_a - W_a_y = 0$

Block B

X axis

$T_1 - W_b_x = m_b a$

Y axis

$N_b - W_b_y =0$

Let's use trigonometry to find the components of the weight.

Block A

cos θ = $\frac{W_a_y}{W_a}$

sin θ = $\frac{W_a_x}{W_a}$

$W_a_y = W_a cos \theta$

$W_a_x= W_a sin \theta$

Block B

cos θ = $\frac{W_b_y}{W_b}$

sin θ = $\frac{W_b_x}{W_b}$

$W_b_y = W_b cos \theta \\W_b_x = W_b sin \theta$

Let's write our system of equations.

$W_c - T_2 = m_c a \\ T_2 - T_1 - W_a_x = m_a a \\T_1 - W_b_x = m_b a$

Let's find the acceleration of the bodies, adding the equations.

$W_c - W_a_x - W_b_x = ( m_a+m_b+m_c) a\\$

The weight is

W = mg

Let's substitute

$(m_c - m_a -m_b ) g \ sin \theta = ( m_c+m_a+m_b) \ a \\a= \frac{ m_c-m_a-m_b }{ m_a+m_b+m_c} \ g sin \theta$

Indicate ma mass of the block a ma = 1.00 kg, the mass of the block b mb = 2.2 kg and the weight of the block c Wc = 16.2 N, let's find the mass of block c.

m_c = Wc / g

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m_c = 1.65 kg

we substitute the values

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a = - 0.3196 sin 45

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Taking the acceleration we are going to look for the tensions.

From the equation of block C

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$T_1 - W_b_x = m_b a\\T_1 = m_b (a + g sin \theta)\\T_1 = 1.00 (-0.226 + 9.8 \ sin 45)$

T₁ = 6.7 N

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T₁ = 6.7 N

T₂ = 16.54 N

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