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3 years ago

11

Which one of the following does not accurately describe the universal gravitational law? Question 10 options: A) Gravitational f

orce is proportional to the square of the distance between the centers of both bodies. B) Every object in the universe that has mass attracts every other object in the universe that has mass. C) Gravitational force is proportional to the product of the two masses. D) The object with the smaller mass does most of the moving because the object with the larger mass has too much inertia to move any noticeable amount.

2 answers:

just olya [345]3 years ago

5 0

Answer:

Choice C seems to be the right answer.

Explanation:

Kaylis [27]3 years ago

4 0

Answer:

D is right answer

Explanation:

$f = \frac{Gm1m2}{ {r}^{2} }$

•Gravitational force directly proportional to product of their masses

•And is inversely proportional to square of distance between them

•Every object in the universe that has mass attracts every other object in the universe that has mass.

THEREFORE OPTION 4 WILL BE THE ANSWER

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a nitrogen laser generates a pulse containing 10.0 mj of energy at a wavelength of 340.0 nm. how many photons are in the pulse?

Wewaii [24]

A nitrogen laser generates a pulse containing 10.0 mj of energy at a wavelength of 340.0 nm and has 1785 x 10¹⁹ photons in the pulse.

<h3>How many photons are in the pulse?</h3>

Energy of a single photon is

E=hcλ

E=6.626×10⁻³⁴ J s×3×108 m/s /340×10⁻⁹ m

E=6.31×10⁻¹⁹ J

Number of photons in the laser is

n=Total Energy/Energy per photon

n=10⁷×10⁻³J /5.90×10⁻¹⁹J/photon

n= 1785 x 10¹⁹ photons

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2 years ago

In rural areas, water is often extracted from underground by pumps. Consider an underground water source whose free surface is 6

stealth61 [152]

Answer:

W = 9533.09 Watt

Explanation:

given,

diameter of pipe inlet, d₁ = 10 cm

r₁ = 5 cm

diameter of pipe outlet, d₂ = 15 cm

r₂= 7.5 cm

head upto water level is to rise = 60 + 5

= 65 m

flow rate = 0.015 m³/s

we know

A₁ v₁ = A₂ v₂ = Q

π r₁² v₁ = π r₂² v₂ = 0.015

$v_1= \dfrac{r_2^2}{r_1^2} v_2$

$v_1= \dfrac{7.5^2}{5^2} v_2$

$v_1= 2.25 v_2$

$v_2 = \dfrac{0.015}{\pi r_2^2}$

$v_2 = \dfrac{0.015}{\pi 0.075^2}$

v₂ = 0.848 m/s

v₁ = 1.908 m/s

Applying Bernoulli's equation

$P_p = \dfrac{1}{2}\rho (v_2^2-v_1^2)+ \rho g h$

$P_p= \dfrac{1}{2}\times 1000\times (0.848^2-1.908^2)+ 1000\times 9.8\times 65$

$P_p= 635539.32 Pa$

P_p is the pump pressure

Power of the pump

W = P_p x Q

W = 635539.32 x 0.015

W = 9533.09 Watt

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3 years ago

Since this object is traveling at terminal velocity, what must be the value of Fd? (HINT: how to the lengths of both arrows comp

Juli2301 [7.4K]

The magnitude of the air drag when the object is traveling at terminal velocity is C. 850 Newtons

Explanation:

There are only two forces acting on the object here:

- The force of gravity, of magnitude $F_g = 850 N$ , acting downward

- The air drag, $F_d$ , acting upward

Therefore, the equation of motion for the object is

$F_g - F_d = ma$

where m is the mass of the object and a its acceleration.

The object in this problem is traveling at terminal velocity: this means that the acceleration is zero, so

a = 0

Therefore the equation becomes

$F_d = F_g$

which means that the magnitude of the air resistance is equal to the magnitude of the force of gravity:

$F_d = 850 N$

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4 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

<u>Distance of the center from each corners</u> $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

<u>For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:</u>

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

<u>Force by each of the charges at the remaining corners:</u>

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

<u> Now, net electric field in the vertical direction:</u>

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

<u>Now, net electric field in the horizontal direction:</u>

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

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3 years ago

A small truck has gone over an embankment. The driver has a head injury and is confused, but is able to tell you that he is hau

vladimir2022 [97]

Answer:

"Emergency resource guidebook" would provide you with the best initial information about this chemical

Explanation:

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3 years ago