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kotykmax [81]
2 years ago
9

A helicopter is accelerating up into the air while it is also accelerating due north. Which statement about the forces acting on

the helicopter is true?
Physics
1 answer:
stich3 [128]2 years ago
4 0

Answer:

The drag on the helicopter is directed downward and to the south.

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A. The potential energy stored in the compressed spring of a dart gun, with a spring constant of 62.00 N/m, is 0.940 J. Find by
valina [46]

a. The spring is compressed  by  0.174 m.

b.  The vertical distance the dart travels from its position when the spring is compressed to its highest position is 9.6 m.

c. The horizontal velocity of the dart at that time is  13.74 m/s.

d. The horizontal distance from the equilibrium position at which the dart hits the ground is 19.236 m.

<h3>What is potential energy?</h3>

The energy by virtue of its position is called the potential energy.

a. Given is the potential energy stored in the compressed spring of a dart gun, with a spring constant of 62.00 N/m, is 0.940 J.

PE  of spring = 1/2 kx²

Put the values, we get

P.E = 0.940 = 1/2 x 62 x²

x = 0.174 m

Thus, the spring is compressed by 0.174 m.

b. Given is a 0.010 kg dart is fired straight up.

The vertical height is find out by

0.940 J = (0.010 kg) (9.8 m/s²) h

h = 9.6 m

Thus, the vertical distance the dart travels from its position is 9.6 m

c. From the conservation of energy principle, total mechanical energy is conserved.

1/2 mv² =mgh

v = √2gh

Plug the values, we get

v = √2 x 9.8x 9.6

v = 13.74 m/s

Thus, the horizontal velocity is  13.74 m/s.

d.  Time that dart spends in air, t = √2h/g

t = √(2x9.6)/9.81

t = 1.4 s

The horizontal distance from the equilibrium position at which the dart hits the ground.

Horizontal distance = (Velocity on x direction) x time

Horizontal distance = 13.74 m/s x 1.4s

Horizontal distance = 19.236 m

Thus, the horizontal distance is 19.236 m.

Learn more about potential energy.

brainly.com/question/24284560

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7 0
2 years ago
If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h
raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory (u_x) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

u^2=u_x^2+u_y^2

Where, u_x\ and\ u_y are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1

Now, we know that, for a projectile motion, the maximum height is given as:

H=\frac{u_y^2}{2g}

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m

Therefore, the maximum height reached is 4.63 m.

3 0
4 years ago
Hi Hi. How are you all?
kap26 [50]

Answer:

Amazing Thanks. What about u?

Explanation:

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3 years ago
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. Take a stick of length one and denote it by the unit interval [0, 1]. Pick two i.i.d. Uniform(0, 1) random variables U and V a
BartSMP [9]

Answer:

What is the expected value of L = 1/3

Explanation:

kindly check the attached file below.

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4 years ago
What is a small body that follows a highly elliptical orbit around the Sun
Ipatiy [6.2K]

That could be a comet, or any one of the billions of meteoroids
moving in a cloud that's actually the remains of a shattered comet.

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3 years ago
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