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3 years ago

Does the orbital period of a planet depend on the mass of the planet or on the mass of the star that it orbits?

Physics

1 answer:

jasenka [17]3 years ago

8 0

Answer:

The orbital period of a planet depends on the mass of the planet.

Explanation:

A less massive planet will take longer to complete one period than a more massive planet.

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A leaky faucet drips once each second. at this rate, it takes 1 hour to fill a 1-liter graduated cylinder to the 200-cm3 mark. w

UNO [17]

Define
v = volume of a drop per second, cm³/s

The time taken to fill 200 cm³ is 1 hour.
Let V = 200 cm³, the filled volume.
Let t = 1 h = 3600 s, the time required to fill the volume.

Therefore,
$\frac{200 \, cm^{3}}{v \, cm^{3}/s} = 3600 \, s \\ v = \frac{200}{3600} =0.0556 \, cm^{3}$

The average volume of a single drop is approximately 0.0556 cm³.

Answer: 0.0556 cm³

8 0

3 years ago

Compute the kinetic energy of a proton (mass 1.67×10−27kg ) using both the nonrelativistic and relativistic expressions for spee

JulsSmile [24]

Answer:

The non-relativistic kinetic energy of a proton is $6.76\times10^{-12}\ J$

The relativistic kinetic energy of a proton is $7.25\times10^{-12}\ m/s$

Explanation:

Given that,

Mass of proton $m=1.67\times10^{-27}\ kg$

Speed $v= 9.00\times10^{7}\ m/s$

We need to calculate the kinetic energy for non relativistic

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times1.67\times10^{-27}\times(9.00\times10^{7})^2$

$K.E=6.76\times10^{-12}\ J$

We need to calculate the kinetic energy for relativistic

Using formula of kinetic energy

$K.E=mc^2(\sqrt{(\dfrac{1}{1-\dfrac{v^2}{c^2}})}-1)$

$K.E=1.67\times10^{-27}\times(3\times10^{8})^{2}\cdot\left(\sqrt{\frac{1}{1-\frac{\left(9.00\times10^{7}\right)^{2}}{(3\times10^{8})^{2}}}}-1\right)$

$K.E=7.25\times10^{-12}\ m/s$

Hence, The non-relativistic kinetic energy of a proton is $6.76\times10^{-12}\ J$

The relativistic kinetic energy of a proton is $7.25\times10^{-12}\ m/s$

7 0

3 years ago

A wave front has the form of a

podryga [215]

Your answer is "<span>surface of a sphere"

Hope this helps.</span>

3 0

3 years ago

In 4 -5 sentences, summarize the science of the sport of curling and how it relates to friction

zalisa [80]

Well, a curler uses heat to calm down friction so I am not sure

5 0

3 years ago

Can you find the magnetic field based on force? a straight segment of wire 35.0 cm long carrying a current of 1.40 a is in a uni

Lostsunrise [7]

The force exerted by a magnetic field on a wire carrying current is:
$F=ILB \sin \theta$
where I is the current, L the length of the wire, B the magnetic field intensity, and $\theta$ the angle between the wire and the direction of B.

In our problem, the force is F=0.20 N. The current is I=1.40 A, while the length of the wire is L=35.0 cm=0.35 m. The angle between the wire and the magnetic field is $53 ^{\circ}$ , so we can re-arrange the formula and substitute the numbers to find B:
$B= \frac{F}{IL \sin \theta}= \frac{0.20 N}{(1.40 A)(0.35 m)(\sin 53^{\circ})}=0.51 T$

3 0

3 years ago