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2 years ago

Name all the harry potter books 1-7

Physics

2 answers:

Savatey [412]2 years ago

4 0

What do you mean to name them do you need them in a certain order ?

miss Akunina [59]2 years ago

3 0

Answer:

1) Harry Potter and the Philosophers Stone (Sorcerers Stone)

2) Harry Potter and the Chamber of Scerets

3)Harry Potter and the Prizoner of Azkaban

4) Harry Potter and the Goblet of Fire

5) Harry Potter and the Order of the Phonix

6) Harry Potter and the Half-Blood Prince

7) Harry Potter and the Deathy Hallows

Explanation:

You might be interested in

A block with mass M is placed on an inclined plane with slope angle q and is connected to a second hanging block with mass m by

tensa zangetsu [6.8K]

Answer:

The mass of the block m is:

$m=M(sin(\theta)+\mu_{s}cos(\theta))$

Explanation:

Let's analyze the block by parts

For the block M

$T-W_{x}-f_{f}=0$ (1)

Where:

T is the tension
W(x) is the component of the weight in the x-direction
F(f) is the friction force

$T-Mgsin(\theta)-\mu_{s}N=0$

$T-Mgsin(\theta)-\mu_{s}Mgcos(\theta)=0$

For the block m

$T-W=0$

$T=mg$ (2)

Now, let's combines equation (1) and (2):

$mg-Mgsin(\theta)-\mu_{s}Mgcos(\theta)=0$

Finally, let's solve it for block m.

$mg-Mg(sin(\theta)+\mu_{s}cos(\theta))=0$

$m=M(sin(\theta)+\mu_{s}cos(\theta))$

I hope it helps you!

7 0

3 years ago

Which of the following accurately describe some aspect of gravitational waves? Select all that apply.

steposvetlana [31]

<h2>Answers:</h2>

-The first direct detection of gravitational waves came in 2015

-The existence of gravitational waves is predicted by Einstein's general theory of relativity

-Gravitational waves carry energy away from their sources of emission

<h2>Explanation:</h2>

Gravitational waves were discovered (theoretically) by Albert Einstein in 1916 and "observed" for the first time in direct form in 2015 (although the results were published in 2016).

These gravitational waves are fluctuations or disturbances of space-time produced by a massive accelerated body, modifying the distances and the dimensions of objects in an imperceptible way.

In this context, an excellent example is the system of two neutron stars that orbit high speeds, producing a deformation that propagates like a wave,<u> in the same way as when a stone is thrown into the water</u>. So, in this sense, gravitational waves carry energy away from their sources .

Therefore, the correct options are D, E and F.

5 0

4 years ago

Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of

Pachacha [2.7K]

Answer:

<em>The frequency changes by a factor of 0.27.</em>

Explanation:

The frequency of an object with mass m attached to a spring is given as

$f$ = $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

where $f$ is the frequency

k is the spring constant of the spring

m is the mass of the substance on the spring.

If the mass of the system is increased by 14 means the new frequency becomes

$f_{n}$ = $\frac{1}{2\pi } \sqrt{\frac{k}{14m} }$

simplifying, we have

$f_{n}$ = $\frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }$

$f_{n}$ = $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$

if we divide this final frequency by the original frequency, we'll have

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ ÷ $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ x $2\pi \sqrt{\frac{m}{k} }$

==> 1/3.742 = <em>0.27</em>

7 0

3 years ago

A 2750 kg helicopter flies horizontally at constant speed. Air resistance creates a 7510 n backward force. What is the magnitude

svp [43]

Answer:

Explanation:

Let lift force F be created by propellers at angle of Ф with the horizontal .

The vertical component of this force will balance the weight of helicopter and horizontal component will balance the air resistance .

F sinФ = mg = 2750 x 9.8 = 26950 N

F cosФ = 7510 N

Squaring and adding ,

F² = 26950² + 7510²

= 726302500 + 56400100

= 782702600

F = 27976.82 N .

3 0

3 years ago

On an asteroid, the density of dust particles at a height of 3 cm is ~30% of its value just above the surface of the asteroid. A

Anvisha [2.4K]

From the law of atmosphere

$N_v(y) = n_0*e^{-\frac{mgy}{Kb*T}}$

Where

$n_0$ = constant and is number density where the height y = 0cm

$n_V$ = Number density at height y=3cm

Kb = Boltzmann constant $= 1.38*10^{-23}J/K$

$T=20K$

$m = 10^{-19}kg$

Re-arranging the equation to have the value of the gravity,

$\frac{N_v(y)}{n_0} = e^{-\frac{mghy}{KbT}}$

$ln(\frac{N_v(y)}{n_0}) = -\frac{mgy}{KbT}$

Since it is 30% of value above surface, therefore $N_v = 0.3n_0$

$ln(\frac{0.3n_0}{n_0}) = -\frac{mgy}{KbT}$

$g = -\frac{KbT ln(0.3)}{my}$

$g = -\frac{(1.38*10^{-23}J/K)(20K)(Ln(0.3))}{10^{-19}(3*10^{-2})}$

$g = \frac{1.38*2*ln(0.3)*10^{-22}}{3*10^{-4}}$

$g = 1.104*10^{-1}m/s^2$

$g = 0.1m/s^2$

Therefore the correct answer is C.

4 0

4 years ago