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3 years ago

How are scientists able to predict when and where the next eclipse will occur?

Physics

1 answer:

ahrayia [7]3 years ago

5 0

Answer:

When the Earth and sun are perfectly lined up, then it will happen. They can tell when it's going to happen.

Explanation:

This is why it only happens in some places. Some days it's not sunny out, so it's not going to happen.

You might be interested in

If position vector r = bt^2i + ct^3j, where b and c are positive constants, when does the velocity vector make an angle of 450 w

vazorg [7]

$\vec r(t)=bt^2\,\vec\imath+ct^3\,\vec\jmath$

The velocity at time $t$ is

$\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=2bt\,\vec\imath+3ct^2\,\vec\jmath$

Take two vectors that point in the positive $x$ and positive $y$ directions, such as $\vec\imath$ and $\vec\jmath$ . The dot products of the velocity vector with $\vec\imath$ and $\vec\jmath$ are

$\dfrac{\mathrm d\vec r(t)}{\mathrm dt}\cdot\vec\imath=2bt=\sqrt{4b^2t^2+9c^2t^4}\cos\theta$

and

$\dfrac{\mathrm d\vec r(t)}{\mathrm dt}\cdot\vec\jmath=3ct^2=\sqrt{4b^2t^2+9c^2t^4}\cos\theta$

We want the angles between these vectors to be 45º, for which we have $\cos45^\circ=\frac1{\sqrt2}$ . So

$\begin{cases}2\sqrt2\,bt=\sqrt{4b^2t^2+9c^2t^4}\\3\sqrt2\,ct^2=\sqrt{4b^2t^2+9c^2t^4}\end{cases}\implies3\sqrt2\,ct^2-2\sqrt2\,bt=0$

$\implies t(3ct-2b)=0$

$\implies t=0\text{ or }t=\dfrac{2b}{3c}$

When $t=0$ , the velocity vector is equal to the zero vector, which technically has no direction/doesn't make an angle with any other vector. So the only time this happens is for

$\boxed{t=\dfrac{2b}{3c}}$

7 0

3 years ago

A man tries to push a 200 kg Car that moves at a acceleration 0.50 m/s2. The man is able to displace the car 10 m. How much work

yawa3891 [41]

The work done by the man pushing the car over the given distance is 1000J.

Given the data in the question;

Mass of car; $m = 200kg$

Acceleration of the car; $a = 0.5m/s^2$

Distance covered by the car; $d = 10m$

Work done; $W = \ ?$

Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;

$Work\ done = f * d$

Where f is force applied and d is distance travelled.

To determine the work done by the man, we first solve for the force applied F.

From Newton's Second Law; $Force \ F = m * a$

We substitute our given values into the expression

$F = m * a \\\\F = 200kg * 0.5m/s^2\\\\F = 100kg.m/s^2$

Next we substitute our values into the expression of work done above.

$Work \ done = f * d\\\\Work \ done = 100kg.m/s^2 * 10m\\\\Work \ done = 1000kgm^2/s^2\\\\Work \ done = 1000J$

Therefore, the work done by the man pushing the car over the given distance is 1000J.

Learn more about work done: brainly.com/question/26115962

7 0

2 years ago

Hiii please help i’ll give brainliest if you give a correct answer please thanks!

MaRussiya [10]

Answer:

i think number 4 :/ i hope its right...

6 0

3 years ago

How many seconds will elapse between seeing lightning and hearing the thunder if the lightning strikes 1mi (5280 ft) away and th

alexandr1967 [171]

Answer:

t = 4.58 s

Explanation:

In this problem, we need to find the time elapse between seeing lightning and hearing the thunder if the lightning strikes 1mi (5280 ft) away and the air temperature is 90.0°F.

T = 90.0°F = 32.2 °C

The speed of sound at temperature T is given by :

v = (331.3 +0.6T)

Put T = 32.2°C

So,

v = (331.3 +0.6(32.2))

= 350.62 m/s

We have, distance, d = 1 mile = 1609.34

So,

$t=\dfrac{d}{v}\\\\t=4.58\ s$

So, the required time is equal to 4.58 seconds.

8 0

3 years ago

A wire loop of radius 0.50 m lies so that an external magnetic field of magnitude 0.40 T is perpendicular to the loop. The field

vazorg [7]

The magnitude of the induced emf is given by:

ℰ = |Δφ/Δt|

ℰ = emf, Δφ = change in magnetic flux, Δt = elapsed time

The magnetic field is perpendicular to the loop, so the magnetic flux φ is given by:

φ = BA

B = magnetic field strength, A = loop area

The area of the loop A is given by:

A = πr²

r = loop radius

Make a substitution:

φ = B2πr²

Since the strength of the magnetic field is changing while the radius of the loop isn't changing, the change in magnetic flux Δφ is given by:

Δφ = ΔB2πr²

ΔB = change in magnetic field strength

Make another substitution:

ℰ = |ΔB2πr²/Δt|

Given values:

ΔB = 0.20T - 0.40T = -0.20T, r = 0.50m, Δt = 2.5s

Plug in and solve for ℰ:

ℰ = |(-0.20)(2π)(0.50)²/2.5|

ℰ = 0.13V

3 0

3 years ago