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2 years ago

Which statement best explains how a planet affects the orbit of a comet as the comet passes by the planet?

1 answer:

8 0

Comets orbit the sun just like planets do. Except a comet usually has a very elongated orbit. Thanks to the laws of gravity comets obey the same laws. A comets orbit takes it very close to the sun and then far away again.

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An object is falling from a height of 7.5 meters. At what height will it’s velocity be 7 meters/second?

Nady [450]

One of the equations of gravity is this:
${v}^{2} = {u}^{2} + 2gh$
Where v = final velocity which is 7m/s
u = initial velocity which is 0 for objects falling from a height
g = acceleration due to gravity and it is approximately 10m/s^2. It's a constant so pretty much remember this number. It's positive since the work being done is caused by gravity (in other words, it's falling down). It can also be negative if the work being down is against gravity (in other words, it's going up)
h = height of object

Substitute for the values and you should have something like this
${7}^{2} = {0}^{2} + 2 \times 10 \times h$
$49 = 0 + 20h$
$h = \frac{49}{20}$
$h = 2.5m$

6 0

3 years ago

Starting from rest, a truck travels in a straight line for 8.0 s with a uniform acceleration of +1.6 m/s2. The driver then appli

lana [24]

Because it should be the total number

4 0

3 years ago

I.Solve the following problems and answer the following questions. Show all your work and provide answers rounded off to the app

Ilia_Sergeevich [38]

The sprinter’s average acceleration is 1.98 m/s²

The given parameters;

initial velocity of the sprinter, u = 18 km/h

final velocity of the sprinter, v = 27 km/h

time of motion of the sprinter, t = 3.5 x 10⁻⁴ h

Convert the velocity of the sprinter to m/s;

$initial \ velocity, u = 18 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 5 \ m/s\\\\final \ velocity, v =27 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 7.5 \ m/s\\\\$

The time of motion is seconds;

$t = 3.5 \times 10^{-4} \ h \times \frac{3600 \ s}{1 \ h} = 1.26 \ s$

The sprinter’s average acceleration is calculated as follows;

$a = \frac{v- u}{t} \\\\a = \frac{7.5 \ m/s \ - \ 5 \ m/s}{1.26 \ s} \\\\a = 1.98 \ m/s^2$

Thus, the sprinter’s average acceleration is 1.98 m/s²

Learn more here:brainly.com/question/17280180

6 0

3 years ago

Help for physical science u4 limiting reactants

Yuliya22 [10]

The reactants are on the left and the products are on the right of the equation

3 0

3 years ago

A 0.500-nm x-ray photon is deected through 134 in a Compton scattering event. At what angle (with respect to the incident beam)

natita [175]

Answer:

The angle of recoil electron with respect to incident beam of photon is 22.90°.

Explanation:

Compton Scattering is the process of scattering of X-rays by a charge particle like electron.

The angle of the recoiling electron with respect to the incident beam is determine by the relation :

$\cot\phi = (1+\frac{hf}{m_{e}c^{2} })\tan\frac{\theta }{2}$ ....(1)

Here ∅ is angle of recoil electron, θ is the scattered angle, h is Planck's constant, $m_{e}$ is mass of electron, c is speed of light and f is the frequency of the x-ray photon.