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2 years ago

If a helicopter's mass is 4,500kg and the net force on it is 18,000 N upward, what is its acceleration?

Physics

2 answers:

melamori03 [73]2 years ago

5 0

Answer:

Acceleration, $a=4\ m/s^2$

Explanation:

It is given that,

Mass of the helicopter, m = 4500 kg

Net force acting on the helicopter, F = 18000 N

Let a is the acceleration of the helicopter. The force acting on an object is equal to the product of mass and acceleration. It is calculated as :

$F=ma$

$a=\dfrac{F}{m}$

$a=\dfrac{18000\ N}{4500\ kg}$

$a=4\ m/s^2$

So, the acceleration of the helicopter is $4\ m/s^2$ . Hence, this is the required solution.

Vinvika [58]2 years ago

4 0

The acceleration is0.25m/s^2

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A 900-kg car traveling east at 15.0 m/s collides with a 750-kg car traveling north at 20.0 m/s. The cars stick together. Assume

dalvyx [7]

Explanation:

It is given that,

Mass of the car 1, $m_1=900\ kg$

Initial speed of car 1, $u_1=15i\ m/s$ (east)

Mass of the car 2, $m_2=750\ kg$

Initial speed of car 2, $u_1=20j\ m/s$ (north)

(b) As the cars stick together. It is a case of inelastic collision. Let V is the common speed after the collision. Using the conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)V$

$900\times 15i +750\times 20j=(900+750)V$

$13500i+15000j=1650V$

$V=(8.18i+9.09j)\ m/s$

The magnitude of speed,

$|V|=\sqrt{8.18^2+9.09^2}$

V = 12.22 m/s

(b) Let $\theta$ is the direction the wreckage move just after the collision. It is given by :

$tan\theta=\dfrac{v_y}{v_x}$

$tan\theta=\dfrac{9.09}{8.18}$

$\theta=48.01^{\circ}$

Hence, this is the required solution.

4 0

2 years ago

Determine the maximum weight of the bucket that the wire system can support so that no single wire develops a tension exceeding

stellarik [79]

Let there be N number of wires.

Maximum tension a wire can withstand = 100 lb

so, Total tension N wires can withstand = 100 N

now, total tension in N wires = Maximum weight of bucket

100 N = W

so, W = 100N

W is the weight of bucket and 100N is its maximum value.

8 0

2 years ago

Suppose you first walk 12.0 m in a direction 20 owest of north and then 20.0 m in a direction 40.0osouth of west. How far are yo

alexgriva [62]

Answer:

R=19.5m

$\theta$ = 4.65° S of W

Explanation:

Refer the attached fig.

displacement of the x and y components

x-component displacement is ( $R_{x}$ ) = $A_{x}+B_{x}$

= A $\sin$ (20°) + B $\cos$ (40°)

= -12.0 $\sin$ (20°) + 20.0 $\cos$ (40°)

= -19.425m

x-component displacement is ( $R_{y}$ ) = $A_{y}+ B_{y}$

= A $\cos$ (20°) - B $\sin$ (40°)

= 12.0 $\cos$ (20°) - 20.0 $\sin$ (40°)

= -1.579

resultant displacement

∴

R = $\sqrt{R_{x}^{2} +R_{y}^{2} } }$

= $\sqrt{(-19.425)^{2}+(-1.579)^{2} }$

=19.5m

$\theta$ = $\tan^{-1}\left | \frac{R_{x}}{R_{y}} \right |$

$\theta$ = $\tan^{-1}\left | \frac{1.579}{19.425} \right |$

$\theta$ = 4.65° S of W

6 0

2 years ago

A common carnival ride, called a gravitron, is a large r = 11 m cylinder in which people stand against the outer wall. the cylin

Leya [2.2K]

Answer:

v = 13.19 m / s

Explanation:

This problem must be solved using Newton's second law, we create a reference system where the x-axis is perpendicular to the cylinder and the Y-axis is vertical

X axis

N = m a

Centripetal acceleration is

a = v² / r

Y Axis

fr -W = 0

fr = W

The force of friction is

fr = μ N

Let's calculate

μ (m v² / r) = mg

μ v² / r = g

v² = g r / μ

v = √ (g r /μ)

v = √ (9.8 11 / 0.62)

v = 13.19 m / s

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2 years ago

A soccer ball player bounces the ball off her head, changing the velocity of the ball. She changes the x-component of the veloci

Nadya [2.5K]

The change in horizontal velocity is (4.7 - 8.1) = -3.4 m/s

The change in vertical velocity is (3.2 + 3.3) = 6.5 m/s

These are the components of velocity DELIVERED to the ball by the player's pretty head during the collision.

The magnitude of the change in velocity is √(-3.4² + 6.5²) = 7.336 m/s .

The magnitude of the ball's change in momentum is (m · v) = (0.44 · 7.336) = 3.228 kg-m/s .

==> The change in the ball's momentum is exactly the impulse during the collision. . . . . . 3.228 kg-m/s .

==> The direction of the impulse is the direction of the change in momentum: (-3.4)i + (6.5)j

The direction is arctan (6.5 / -3.4) = -62.39°

That's clockwise from the +x axis, which is roughly "southeast". The question wants it counterclockwise from the +x axis. That's (360-62.39) =

Direction of the impulse = 297.61°

We know that impulse is equivalent to the change in momentum, and that's how I approached the solution. Impulse is also (force x time) during the collision. We're given the time in contact, but I didn't need to use it. I guess I would have needed to use it if we were interested in the FORCE she exerted on the ball with her head, but we didn't need to find that.

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2 years ago