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REY [17]
3 years ago
10

A plane is flying east at 115 m/s. The wind accelerates it at 2.88 m/s^2 directly northwest. After 25.0 seconds what is the dire

ction of the velocity of the plane?
Physics
1 answer:
Furkat [3]3 years ago
7 0

Answer:

38.5° north of east

Explanation:

Directly northwest is 135° from east.  If we say east is +x and north is +y, then the components of the acceleration are:

ax = 2.88 cos 135° ≈ -2.036 m/s²

ay = 2.88 sin 135° ≈ 2.036 m/s²

And the initial velocity is:

v₀x = 115 m/s

v₀y = 0 m/s

After, 25.0 seconds, the final velocity components are:

vx = at + v₀x

vx = (-2.036)(25.0) + 115

vx ≈ 64.1 m/s

vy = at + v₀y

vy = (2.036)(25.0) + 0

vy ≈ 50.9 m/s

The plane's direction is:

θ = atan(vy / vx)

θ = atan(50.9 / 64.1)

θ ≈ 38.5°

The direction of the plane's velocity is approximately 38.5° north of east.

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Calculate the Force required to give a bullet of mass 50 g an acceleration of 300 m/s2
NNADVOKAT [17]

Answer:

<h3>The answer is 15 N</h3>

Explanation:

The force acting on an object can be found by using the formula

<h3>Force = mass × acceleration</h3>

From the question

mass = 50 g = 0.05 kg

acceleration = 300 m/s²

We have

force = 0.05 × 300

We have the final answer as

<h3>15 N</h3>

Hope this helps you

7 0
3 years ago
Give examples of not useful high friction
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6 0
3 years ago
A box weighing 12,000 N is parked on a 36° slope. What will be the component of the weight parallel to the plane that balances f
sweet-ann [11.9K]

Given :

A box weighing 12,000 N is parked on a 36° slope.

To Find :

What will be the component of the weight parallel to the plane that balances friction.

Solution :

The component of that will be parallel to the plane to balance friction is :

F \ cos ( 90 - 36)^o\\\\F\ ( sin\ 36^o)

Therefore, component of force to balance friction is F sin 36° .

Hence, this is the required solution.

5 0
2 years ago
A large rocket has a mass of 2.00×10⁶ kg at takeoff, and its engines produce a thrust of 3.50×10⁷ N. Find its initial accelerati
Kazeer [188]

Answer:

17.5 m/s²

1.90476 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Force

F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{3.5\times 10^7}{2\times 10^6}\\\Rightarrow a=17.5\ m/s^2

Initial acceleration of the rocket is 17.5 m/s²

v=u+at\\\Rightarrow \frac{120}{3.6}=0+17.5t\\\Rightarrow t=\frac{\frac{120}{3.6}}{17.5}=1.90476\ s

Time taken by the rocket to reach 120 km/h is 1.90476 seconds

Change in the velocity of a rocket is given by the Tsiolkovsky rocket equation

\Delta v=v_{e}\ln \frac{m_0}{m_f}

where,

m_0 = Initial mass of rocket with fuel

m_f = Final mass of rocket without fuel

v_e = Exhaust gas velocity

Hence, the change in velocity increases as the mass decreases which changes the acceleration

4 0
3 years ago
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