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4 years ago

Universal law of gravity

1 answer:

3 0

Newton's law of universal gravitation :) states that a particle attracts every other particle in the universe using a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers

And this answer isn't only for america, its for the whole universe

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A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac

Sedbober [7]

Answer:

v = 1.2 m/s

Explanation:

The wavelength of the waves is given as the horizontal distance between the crests:

λ = wavelength = 5.5 m

Now, the time period is given as the time taken by boat to move from the highest point again to the highest point. So it will be equal to twice the time taken by the boat to travel from highest to the lowest point:

T = Time Period = 2(2.3 s) = 4.6 s

Now, the speed of the wave is given as:

$v = f\lambda$

where,

v= speed of wave = ?

f = frequency of wave = $\frac{1}{T} = \frac{1}{4.6\ s} = 0.217\ Hz$

Therefore,

$v = (0.217\ Hz)(5.5\ m)\\$

v = 1.2 m/s

5 0

3 years ago

A motor does 1000J of work in 18 seconds. What is the power of the motor?

solong [7]

Answer:

55.56 watt

Explanation:

Work (W) = 1000 J

time (T) = 18 sec

power (P) = ?

We know power is the rate of doing work so

p = w / t

= 1000 / 18

= 55. 56 watt

7 0

3 years ago

A spherical balloon is made from a material whose mass is 3.30 kg. The thickness of the material is negligible compared to the 1

stellarik [79]

Answer:

563712.04903 Pa

Explanation:

m = Mass of material = 3.3 kg

r = Radius of sphere = 1.25 m

v = Volume of balloon = $\frac{4}{3}\pi r^3$

M = Molar mass of helium = $4.0026\times 10^{-3}\ kg/mol$

$\rho$ = Density of surrounding air = $1.19\ kg/m^3$

R = Gas constant = 8.314 J/mol K

T = Temperature = 345 K

Weight of balloon + Weight of helium = Weight of air displaced

$mg+m_{He}g=\rho vg\\\Rightarrow m_{He}=\rho vg-m\\\Rightarrow m_{He}=1.19\times \frac{4}{3}\pi 1.25^3-3.3\\\Rightarrow m_{He}=6.4356\ kg$

Mass of helium is 6.4356 kg

Moles of helium

$n=\frac{m}{M}\\\Rightarrow n=\frac{6.4356}{4.0026\times 10^{-3}}\\\Rightarrow n=1607.85489$

Ideal gas law

$P=\frac{nRT}{v}\\\Rightarrow P=\frac{1607.85489\times 8.314\times 345}{\frac{4}{3}\pi 1.25^3}\\\Rightarrow P=563712.04903\ Pa$

The absolute pressure of the Helium gas is 563712.04903 Pa

3 0

3 years ago

Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

s2008m [1.1K]

Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

Conservation of linear momentum:

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

after the collision we have that

$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

Conservation of kinetic energy:

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

8 0

3 years ago

Suppose Joe, who can type 40 words per minute, takes 2.5 hours to type his report. To speed things up, his friend Mark offers to

Vika [28.1K]

Explanation:

When Joe works alone, the total number of words he typed can be given by:

Total words = (40 words per minute) x (60 minutes per hour) x (2.5 hours)

Total words = 6000 words

Now, when Joe and Mark work together, let 'y' be the number of hours for which they both work simultaneously:

Total words = Words Typed by Joe + Words Typed by Mark

6000 = {(40 words per minute) x (60 minutes per hours) x (y hours)} + {(20 words per minute) x (60 minutes per hours) x (y hours)}

6000 = 2400y + 1200y = 3600y

y = 1.67 hours = 1 hour and 40 minutes

Thus, working together simultaneously, Joe and Mark will take 1 hour and 40 minutes to complete the report.

5 0

3 years ago