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vichka [17]
3 years ago
8

Neptune (currently the most distant planet from the Sun!) lies about 4.5x109 kilometers from the Earth. If you were traveling to

ward Neptune at typical highway speeds (about 100 kilometers/hour) how many years would it take to reach Neptune?
Physics
1 answer:
OLEGan [10]3 years ago
3 0

Answer:

5,137 years.

Explanation:

we have a distance and the velocity, so we calculate the time:

time = \frac{distance}{velocity}

the distance is the distance between earth and neptune:

distance = 4.5x10^{9}km

and the velocity is the typical highway speed:

velocity = 100km/h

we divide this quantities:

time = \frac{4.5x10^9km}{100km/h} =4.5x10^7hours

it would take this many hours to get to Neptune at this speed.

Let's convert this to years:

there are 8760 hours in one year:

so the time in years is:

time =\frac{4.5x10^7hours}{8760 hours/year}=5,136.99 years

it would take about 5,137 years.

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Two instruments produce a beat frequency of 5 Hz. If one has a frequency of 264 Hz, what could be the frequency of the other ins
Lerok [7]

Answer:

259 Hz or 269 Hz

Explanation:

Beat: This is the phenomenon obtained when two notes of nearly equal frequency are sounded together. The S.I unit of beat is Hertz (Hz).

From the question,

Beat = f₂-f₁................ Equation 1

Note: The frequency of the other instrument is either f₁ or f₂.

If the unknown instrument's frequency is f₁,

Then,

f₁ = f₂-beat............ equation 2

Given: f₂ = 264 Hz, Beat = 5 Hz

Substitute into equation 2

f₁ = 264-5

f₁ = 259 Hz.

But if the unknown frequency is f₂,

Then,

f₂ = f₁+Beat................. Equation 3

f₂ = 264+5

f₂ = 269 Hz.

Hence the beat could be 259 Hz or 269 Hz

8 0
3 years ago
An object in circular motion has velocity that is constantly changing. The direction of the acceleration is
Keith_Richards [23]
<h2>Answer: Toward the center of the circle.</h2>

This situation is characteristic of the uniform circular motion , in which the movement of a body describes a circumference of a given radius with constant speed.  

However, in this movement the velocity has a constant magnitude, but its direction varies continuously.

Let's say \vec{V} is the velocity vector, whose direction is perpendicular to the radius r of the trajectory, therefore   the acceleration \vec{a} is directed toward the center of the circumference.

 

7 0
3 years ago
In case of collision of objects in two dimensions which statement is true after the collision?
Grace [21]

The correct answer to the question is : C) The horizontal momentum and the vertical momentum are both conserved. 

EXPLANATION :

Before coming into any conclusion, first we have to understand the law of conservation of momentum.

As per the law of conservation of momentum, the total linear as well as angular momentum of an isolated system is always conserved . The law of conservation of energy is a universal fact.

Hence, during any type of collision, the total momentum is always conserved.

Hence, the total horizontal momentum as well as total vertical momentum are always conserved during both elastic as well as inelastic collision.



5 0
3 years ago
Read 2 more answers
What is the hang time when the person moves 6 m horizontally during a 1.25 m high jump?
AlekseyPX

Answer:

1 sec

Explanation:

Horizontal distance (x) = 6m

Vertical distance (y) = 1.25m

Hang time is the duration the object is in the air before it reaches maximum height.

The time of free fall is given by

t = √2y/g

g = acceleration due to gravity

t = √(2*1.25)/9.8

t = √2.5/9.8

t = 0.5secs

Hang time = 2*0.5

= 1 sec

3 0
3 years ago
A marble resting near the edge of .90 m high table is given an initial horizontal speed of 1.24 m/s. What will be it’s horizonta
Maru [420]

Answer:

0.53 m

Explanation:

First of all, we have to consider the vertical motion of the ball, in order to find the time it takes for the marble to reach the ground. The initial height is h=0.90 m, the initial vertical velocity is zero, while the acceleration is g=-9.81 m/s^2, so the vertical position at time t is given by

y(t)=h-\frac{1}{2}gt^2

By demanding y(t)=0, we find the time t at which the ball reaches the ground:

0=h-\frac{1}{2}gt^2

t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(0.9 m)}{9.81 m/s^2}}=0.43 s

Now we can find the horizontal range of the marble: we know the initial horizontal speed (v=1.24 m/s), we know the total time of the motion (t=0.43 s), and since the horizontal speed is constant, the total distance traveled on the horizontal direction is

x=vt=(1.24 m/s)(0.43 s)=0.53 m

8 0
4 years ago
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