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2 years ago

In the water cycle, which state of matter has the particles closest together?

Physics

2 answers:

salantis [7]2 years ago

8 0

Answer:

solid state

Explanation:

The states of matter that are in the water cycle are solid, liquid and gas.

The particles of solid are closest together while that of gas are farthest apart.

Dmitry [639]2 years ago

4 0

No, options are given but I believe the answer would be
In a water cycle Solid state of matter has the particles closest together.

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A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor sto

Fed [463]

Answer:

a) a = 34.375 m / s², b) v_f = 550 m / s

Explanation:

This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.

a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed

v_f = $\frac{x-x_1}{t}$

we substitute the values

v_f = $\frac{ 6600 -x_1}{4}$

The initial part of the movement is carried out with acceleration

v_f = v₀ + a t

x₁ = x₀ + v₀ t + ½ a t²

the rocket starts from rest v₀ = 0 with an initial height x₀ = 0

x₁ = ½ a t²

v_f = a t

we substitute the values

x₁ = 1/2 a 16²

x₁ = 128 a

v_f = 16 a

let's write our system of equations

v_f = $\frac{6600 - x_1}{4}$

x₁ = 128 a

v_f = 16 a

we substitute in the first equation

16 a = $\frac{6600 -128 a}{4}$

16 4 a = 6600 - 128 a

a (64 + 128) = 6600

a = 6600/192

a = 34.375 m / s²

b) let's find the time to reach this height

x = ½ to t²

t² = 2y / a

t² = 2 5100 / 34.375

t² = 296.72

t = 17.2 s

We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part

v_f = 16 a

v_f = 16 34.375

v_f = 550 m / s

8 0

2 years ago

A 58.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 140 m/s from the top of a cliff

strojnjashka [21]

(a) $6.43\cdot 10^5 J$

The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):

$E=K+U$

The initial kinetic energy is:

$K=\frac{1}{2}mv^2$

where m = 58.0 kg is the mass of the projectile and $v=140 m/s$ is the initial speed. Substituting,

$K=\frac{1}{2}(58 kg)(140 m/s)^2=5.68\cdot 10^5 J$

The initial potential energy is given by

$U=mgh$

where g=9.8 m/s^2 is the gravitational acceleration and h=132 m is the height of the cliff. Substituting,

$U=(58.0 kg)(9.8 m/s^2)(132 m)=7.5\cdot 10^4 J$

So, the initial mechanical energy is

$E=K+U=5.68\cdot 10^5 J+7.5\cdot 10^4 J=6.43\cdot 10^5 J$

(b) $-1.67 \cdot 10^5 J$

We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.

The kinetic energy is

$K=\frac{1}{2}(58 kg)(99.2 m/s)^2=2.85\cdot 10^5 J$

while the potential energy is

$U=(58.0 kg)(9.8 m/s^2)(336 m)=1.91\cdot 10^5 J$

So, the mechanical energy is

$E=K+U=2.85\cdot 10^5 J+1.91 \cdot 10^5 J=4.76\cdot 10^5 J$

And the work done by friction is equal to the difference between the initial mechanical energy of the projectile, and the new mechanical energy:

$W=E_f-E_i=4.76\cdot 10^5 J-6.43\cdot 10^5 J=-1.67 \cdot 10^5 J$

And the work is negative because air friction is opposite to the direction of motion of the projectile.

(c) 88.1 m/s

The work done by air friction when the projectile goes down is one and a half times (which means 1.5 times) the work done when it is going up, so:

$W=(1.5)(-1.67\cdot 10^5 J)=-2.51\cdot 10^5 J$

When the projectile hits the ground, its potential energy is zero, because the heigth is zero: h=0, U=0. So, the projectile has only kinetic energy:

E = K

The final mechanical energy of the projectile will be the mechanical energy at the point of maximum height plus the work done by friction:

$E_f = E_h + W=4.76\cdot 10^5 J +(-2.51\cdot 10^5 J)=2.25\cdot 10^5 J$

And this is only kinetic energy:

$E=K=\frac{1}{2}mv^2$

So, we can solve to find the final speed:

$v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(2.25\cdot 10^5 J)}{58 kg}}=88.1 m/s$

4 0

2 years ago

A guitar string with a length of 80.0 cm is plucked. The speed of a wave in the string is 400 m/sec. Calculate the frequency of

Sholpan [36]

Answer:

The first harmonic is: 250Hz, second harmonic 500Hz, third harmonic 750Hz.

Explanation:

Use the frequency f, speed v, and wavelentgh L relationship:

$v = f\cdot L\implies f = \frac{v}{L}$

We are given the speed v=400 m/s. The base wavelength on a string of length 80cm is twice the length of the string (a "half wave" along the full length of the string), so:

$f = \frac{400\frac{m}{s}}{2\cdot0.8 m}= 250\frac{1}{s}=250 Hz$

The fundamental frequency (first harmonic) is 250 Hz

The second harmonic is produced by one full wave across the string (adding one node in the middle), so L=80cm in this case, therefore the second harmonic frequency is: f2 = 2*250=500Hz

the third harmonic add another node (and a half wave) to the pattern and the wavelength will be 2/3 of 80cm, so f3=3*250Hz = 750Hz

8 0

2 years ago

Adita lifts a book from the floor, carries it across the room, and places it on a high shelf. When is Adita doing work on the bo

noname [10]

She does work from the moment she touches the book until she lets it go. Work is anything that requires energy. Therefore, she is working as she picks up the book, carries it, and when she is lifting it onto the shelf.

5 0

2 years ago

Read 2 more answers

Where does fission regularly occur and what kind of energy is produced?

Lapatulllka [165]

Your answer is C hope i helped
have a good day :)

3 0

2 years ago

Read 2 more answers