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xeze [42]
3 years ago
9

Which of your drawings approximates the orbit of Halley's comet about the sun?

Physics
2 answers:
ryzh [129]3 years ago
7 0

Answer:

It is a elipse and it comes super close to the sun only to be launched back past pluto and then return back

Explanation:

Alla [95]3 years ago
5 0

Answer:

What???

Explanation:

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Plz help it should be super easy
fgiga [73]

Answer:

yes it is easy

Explanation:

6 0
3 years ago
Read 2 more answers
B. Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the
goblinko [34]

We have that the instantaneous velocity of the shuttlecock when it hits the ground is

V_{int}=\sqrt{U^2+19.6H}

From the question we are told

Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the

shuttlecock when it hits the ground? Show your work below.

Generally the equation for acceleration  is mathematically given as

a=v \frac{dv}{dx}\\\\\Therefore\\\\\2ah=v^2-u^2

Where

acceleration is still -9.81 m/s2,

Hence,

V^2-U^2=2(-9.81)*-H

Therefore

V_{int}=\sqrt{U^2+19.6H}

For more information on this visit

brainly.com/question/12319416?referrer=searchResults

7 0
3 years ago
Ten identical steel wires have equal lengths L and equal "spring constants" k. The Young's modulus of each wire is Y. The wires
svlad2 [7]

Answer:

option (B)

Explanation:

Young's modulus is defined as the ratio of longitudinal stress to the longitudinal strain.

Its unit is N/m².

The formula for the Young's modulus is given by

Y=\frac{F \times L}{A\times \delta L}

where, F is the force applied on a rod, L is the initial length of the rod, ΔL is the change in length of the rod as the force is applied, A is the area of crossection of the rod.

It is the property of material of solid. So, when the 10 wires are co joined together to form a new wire of length 10 L, the material remains same so the young' modulus remains same.

8 0
3 years ago
An​ airplane, flying horizontally at an altitude of 3 miles​, passes directly over an observer. If the constant speed of the air
natima [27]
I’m not sure if this helped you so here you go

6 0
3 years ago
On a horizontal frictionless surface, a small block with mass 0.200 kg has a collision with a block of mass 0.400 kg. Immediatel
Akimi4 [234]

Answer:48.2 Joules

Explanation:

Given

two masses of 0.2 kg and 0.4 kg collide with each other

after collision 0.2 kg deflect 30 north of east and 0.4 kg deflects 53.1 south of east

Velocity of 0.2 kg mass is

v_{0.2}=12\cos (30)\hat{i}+12\sin (30)\hat{j}

|v_{0.2}|=11.99 m/s\approx 12 m/s

Velocity of 0.4 kg mass

v_{0.4}=13\cos (53.1)\hat{i}-12\sin (53.1)\hat{j}

|v_{0.4}|=12.99 m/s\approx 13 m/s

Thus total Kinetic energy =\frac{0.2\times 12^2}{2}+\frac{0.4\times 13^2}{2}

Kinetic energy=48.2 J

8 0
3 years ago
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