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vodka [1.7K]
3 years ago
15

Need a lot of help plz

Physics
1 answer:
Over [174]3 years ago
3 0
Im sorry i do not understand.. But if im reading your paper close enough, im pretty sure you're supposed to put and item and describe it..Hope this helps!
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Which of the following are features of the nucleus of an atom
vichka [17]
We r made of atom so v can’t touch anything hehe I just joking
4 0
2 years ago
Applying newton's version of kepler's third law (or the orbital velocity law) to the a star orbiting 40,000 light-years from the
Ugo [173]

Applying Newtons version of Kepler's third law or the orbital velocity law to the star orbiting 40000 light years from the center of the Milky Way Galaxy allows us to determine the mass of the Milky Way Galaxy that lies within 40000 light years in the galactic center.

<h3></h3><h3>What is orbital velocity law?</h3>

The orbital velocity law states that, the orbital velocity is directly proportional to the mass of the body for which it is being calculated and inversely proportional to the radius of the body. Earths orbital velocity near its surface is around 8km/sec if the air resistance is disregarded.

In space exploration, orbital velocity is a crucial topic. Space authorities heavily rely on it to comprehend how to launch satellites. It aids scientists in figuring out the velocities at which satellites must orbit a planet or other celestial body to prevent collapsing into it. The speed at which one body orbits the other body is known as the orbital velocity. The term "orbit" refers to an object's consistent circular motion around the Earth. The distance between the object and the earth's centre determines the orbit's velocity.

To know more about orbital velocity law, refer brainly.com/question/11353717

#SPJ4

5 0
2 years ago
What is the freezing point of an aqueous solution that boils at 101 degrees C? Kfp = 1.86 K/m and Kbp = 0.512 K/m. Enter your an
astra-53 [7]

Answer:

-3.63 degree Celsius

Explanation:

We are given that

Boiling point of solution=T_b=101^{\circ} C

Boiling point water=100 degree Celsius

K_f=1.86K/m

K_b=0.512 K/m

\Delta T_b=T-T_0

Where T=Boiling point of solution

T_0=Boiling point of pure solvent

\Delta T_b=101-100=1^{\circ}C

\Delta T_b=k_bm

Using the formula

1=0.512\times m

Molality,m=\frac{1}{0.512} m

\Delta T_f=k_fm

Using the formula

\Delta T_f=\frac{1}{0.512}\times 1.86

\Delta T_f=3.63 C

We know that

\Delta T_f=T_0-T_1

Where T_0 =Freezing point of solvent

T_1= Freezing point of solution

Using the formula

3.63=0-T_1

Freezing point of water=0 degree Celsius

T_1=0-3.63=-3.63 C

Hence, the freezing point of solution=-3.63 degree Celsius

7 0
3 years ago
A particle’s position is ~r = (ct2 − 2dt3 )iˆ+ (2ct2 − dt2 )jˆ, where c and d are positive constants. Find the expressions for t
Mars2501 [29]

The velocity of the particle is given by the derivative of the position vector:

\vec v = \dfrac{\mathrm d\vec r}{\mathrm dt} = (2ct-6dt^2)\,\vec\imath + (4ct-2dt)\,\vec\jmath

(a) The particle is moving in the <em>x</em>-direction when the <em>y</em>-component of velocity is zero:

4ct-2dt = 2t (2c - d) = 0 \implies t=0

But we want <em>t</em> > 0, so this never happens, unless 2<em>c</em> = <em>d</em> is given, in which case the <em>y</em>-component is always zero.

(b) Similarly, the particle moves in the <em>y</em>-direction when the <em>x</em>-component vanishes:

2ct-6dt^2 = 2t (c - 3dt) = 0 \implies t=0 \text{ or }c-3dt = 0

We drop the zero solution, and we're left with

c-3dt = 0 \implies c=3dt \implies \boxed{t = \dfrac c{3d}}

In the case of 2<em>c</em> = d, this times reduces to <em>t</em> = <em>c</em>/(6<em>c</em>) = 1/6.

7 0
3 years ago
The center of a moon of mass m is a distance D from the center of a planet of mass M. At some distance x from the center of the
Eddi Din [679]

Answer:

x = D (M/M-m) 2.41

Explanation:

a) Let's apply Newton's second law to find the summation of force, where each force is given by the law of universal gravitation

        F = g m₁m₂ / r²

        Σ F = 0

       F1- F2 = 0

       F1 = F2

We set the reference system in the body of greatest mass (M) the planet

       F1 = g m₁ M / x²

       F2 = G m1 m / (D-x)²

      G m₁ M / x² = G m₁ m / (D-x)²

      M (D-x)² = m x²

      MD² -2MD x + M x² = m x²

     x² (M-m) -2MD x + MD² = 0

We solve the second degree equation

     x = [2MD  ±√ (4M²D² - 4 (M-m) MD²)] / 2 (M-m)

     x = {2MD ± 2D √ (M² + (M-m) M)} / 2 (M-m)

     x = D {M  ±  Ra (2M²-mM)} / (M-m)

    x = D (M ± M √ (2-m/M)) / (M-m)

    x = D (M / (M-m)) (1 ±√ (2-m/M)

Let's analyze this result, the value of M-m >> 1, so if we take the negative root, the value of x would be negative, it is out of the point between the two bodies, so the correct result must be taken with the positive root

 

    x = D (M / (M-m)) (1 + √2)

     x = D (M/M-m) 2.41

b) X = 2/3 D

     x = D (M/M-m) 2.41

     2/3 D = D (M/(M-m)) 2.41

     2/3 (M-m) = M 2.41

     2/3 M - 2/3 m = 2.41 M

     1.743 M = 0.667 m

     M/m = 0.667/1.743

     M/m =  0.38

3 0
3 years ago
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