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IgorC [24]
3 years ago
5

Should the pilot drop the bombs before the plane is over the airfield when the plane is over the airfield, or after the plane ha

s passed the airfield? A pilot is flying a mission to drop bombs on an enemy airfield. The plane is flying high and fast to the north, and the city is due north.
A. Before the plane reaches the airfield
B. When the plane is directly over the airfield
C. After the plane has passed the airfield
Physics
1 answer:
gizmo_the_mogwai [7]3 years ago
3 0

Answer:

A. Before the plane reaches the airfield.

Explanation:

The pilot should drop the bomb before the plane reaches the airfield. This is because the bomb will travel at the same horizontal velocity (due north) as the plane when it is falling towards the airfield. So if the pilot releases the bomb over or after the airfield, the bomb will travel north and miss the airfield.

But if the pilot drops the bomb before the airfield, it will travel north and hit its target when it falls.

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The gravitational force between two asteroids is 6.2 × 108 N. Asteroid Y has three times the mass of asteroid Z.
NemiM [27]

Answer:

1.1x10¹⁶ kg

Explanation:

Let m =  mass of asteroid y.

Because asteroid y has three times the mass of asteroid z, the mass of asteroid z is m/3.

Given:

F = 6.2x10⁸ N

d = 2100 km = 2.1x10⁶ m

Note that

G = 6.67408x10⁻¹¹ m³/(kg-s²)

The gravitational force between the asteroids is

F = (G*m*(m/3))/d² = (Gm²)/(3d²)

or

m² = (3Fd²)/G

    = [(3*(6.2x10⁸ N)*(2.1x10⁶ m)²]/(6.67408x10⁻¹¹ m³/(kg-s²))

   = 1.229x10³² kg²

m = 1.1086x10¹⁶ kg = 1.1x10¹⁶ kg (approx)

5 0
3 years ago
A 70.0 kg astronaut is training for accelerations that he will experience upon reentry. He is placed in a centrifuge (r = 15.0 m
Levart [38]

Answer:

1.3823 rad/s

20.7345 m/s

28.66129935 m/s²

a=2.92164g

2006.29095 N radially outward

Explanation:

r = Radius = 15 m

m = Mass of person = 70 kg

g = Acceleration due to gravity = 9.81 m/s²

Angular velocity is given by

\omega=13.2\times \dfrac{2\pi}{60}\\\Rightarrow \omega=1.3823\ rad/s

Angular velocity is 1.3823 rad/s

Linear velocity is given by

v=r\omega\\\Rightarrow v=15\times 1.3823\\\Rightarrow v=20.7345\ m/s

The linear velocity is 20.7345 m/s

Centripetal acceleration is given by

a_c=r\omega^2\\\Rightarrow a_c=15\times 1.3823^2\\\Rightarrow a_c=28.66129935\ m/s^2

The centripetal acceleration is 28.66129935 m/s²

Acceleration in terms of g

\dfrac{a}{g}=\dfrac{28.66129935}{9.81}\\\Rightarrow a=2.92164g

a=2.92164g

Centripetal force is given by

F_c=ma_c\\\Rightarrow F_c=70\times 28.66129935\\\Rightarrow F_c=2006.29095\ N

The centripetal force is 2006.29095 N radially outward

The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.

3 0
4 years ago
What's impulse of<br> force <br><br>​
vekshin1

Answer:

The impulse experienced by the object equals the change in momentum of the object. In equation form, F.t = m. Δv. In a collision, objects experience an impulse; the impulse causes and is equal to the change in momentum.

6 0
3 years ago
A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co
algol [13]

Answer:

\displaystyle X_T=66.6\ km

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

\displaystyle V_f=V_o+a\ t

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of 0.987 m/s^2 for 182 seconds. Then it brakes with a constant acceleration of -0.321 m/s^2 until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

\displaystyle X=V_o\ t+\frac{a\ t^2}{2}

Our data is

\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec

Let's compute the first distance X1

\displaystyle X_1=0+\frac{0.987\times 182^2}{2}

\displaystyle X_1=16,346.7\ m

Now, we find the speed at the end of the first period of time

\displaystyle V_{f1}=0+0.987\times 182

\displaystyle V_{f1}=179.6\ m/s

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0

\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}

\displaystyle t=559.5\ sec

Computing the second distance

\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}

\displaystyle X_2=50,243.2\ m

The total distance is

\displaystyle X_t=x_1+x_2=16,346.7+50,243.2

\displaystyle X_t=66,589.9\ m

\displaystyle \boxed{X_T=66.6\ km}

8 0
4 years ago
Which of the following is true? A. Drivers entering a driveway or alley should yield to bicyclists and pedestrians on the sidewa
slega [8]

Explanation:

The answer is D. Drivers must always yield to pedestrians and bicylists coming off a main road and always yield to emergency vehicles with lights on. Vehicles in a funeral must have lights on to alert other drivers of their pace and procession.

8 0
3 years ago
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