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Since this object is traveling at terminal velocity, what must be the value of Fd? (HINT: how to the lengths of both arrows comp

Juli2301 [7.4K]

The magnitude of the air drag when the object is traveling at terminal velocity is C. 850 Newtons

Explanation:

There are only two forces acting on the object here:

- The force of gravity, of magnitude $F_g = 850 N$ , acting downward

- The air drag, $F_d$ , acting upward

Therefore, the equation of motion for the object is

$F_g - F_d = ma$

where m is the mass of the object and a its acceleration.

The object in this problem is traveling at terminal velocity: this means that the acceleration is zero, so

a = 0

Therefore the equation becomes

$F_d = F_g$

which means that the magnitude of the air resistance is equal to the magnitude of the force of gravity:

$F_d = 850 N$

Learn more about acceleration and Newton laws of motion here:

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8 0

4 years ago

A yellow train of mass 100 kg is moving at 8 m/s towards an orange train of mass 200 kg traveling on the opposite direction on t

vladimir1956 [14]

Mass of yellow train, my = 100 kg

Initial Velocity of yellow train, = 8 m/s

mass of orange train = 200 kg

Initial Velocity of orange train = -1 m/s (since it moves opposite direction to the yellow train, we will put negative to show the opposite direction)

To calculate the initial momentum of both trains, we will use the principle of conservation of momentum which

The sum of initial momentum = the sum of final momentum

Since the question only wants the sum of initial momentum,

(100)(8) + (200)(-1) = 600 m/s

8 0

2 years ago

Please help me very urgent :((

Stells [14]

Answer:

-150 N

Explanation:

(Newton's second law) F=ma

Sum of forces in Y direction= (+200 N)+(-200 N)= 0...

forces cancel, object does not accelerate up/down

Sum of forces in X direction= (+65 N)+(-65 N)+(-150 N)

= -150 N

notice that the +/- 65 components cancel, leaving a net force of 150 N in the LEFTwards direction (which is typically defined as negative)

Overall, the net force is -150 N

7 0

3 years ago

A force AB of length 100m and weight 600N has its centre of gravity 4.0 m from the end, and lies on horizontal ground calculate

Stells [14]

Answer:

F = 24 N

Explanation:

In this exercise we have a bar l = 100 m with a center of gravity x = 4 m, which force is needed to lift it from the other end

Let's use the rotational equilibrium relationship, where we consider the counterclockwise rotations as positive and fix the reference system at the point closest to the center of gravity

∑ τ = 0

F l -x W = 0

F = $\frac{x}{l} \ W$

let's calculate

F = $\frac{4}{100}$ 4/100 600

F = 24 N

8 0

3 years ago

A soccer ball is kicked with a speed of 22m/s at an angle of 35.0º above the horizontal. If the ball lands at the same level fro

Lisa [10]

Answer:

2.57 seconds

Explanation:

The motion of the ball on the two axis is;

x(t) = Vo Cos θt

y(t) = h + Vo sin θt - 1/2gt²

Where; h is the initial height from which the ball was thrown.

Vo is the initial speed of the ball, 22 m/s , θ is the angle, 35° and g is the gravitational acceleration, 9.81 m/s²

We want to find the time t at which y(t) = h

Therefore;

y(t) = h + Vo sin θt - 1/2gt²

Whose solutions are, t = 0, at the beginning of the motion, and

t = 2 Vo sinθ/g

= (2 × 22 × sin 35°)/9.81

= 2.57 seconds

6 0

3 years ago