How many universes out there?

Physics

2 answers:

bonufazy [111]3 years ago

8 0

Only one but there are theories that there are more

Montano1993 [528]3 years ago

7 0

Answer:

onlu 1 universe

Explanation:

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In a bicycle dynamo,does 1. The permanent magnet surrounds a conducting coil 2. The conducting coil rotates when the rear wheel

atroni [7]

Explanation:

3: the electricity is generated in the permanent magnet

7 0

3 years ago

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An airplane is heading south at a speed of 600km/h. If a wind begins blowing from the southwest at a speed of 100km/h, calculate

Karo-lina-s [1.5K]

Answer:

Here's what I got:

Let's assume that N and E are + directions while S and W are - directions.

Wind is blowing from SW; thus, it is blowing towards NE (or at 45 deg N of E).

Dividing the wind's speed into components:y-component: +70.71 km/h; x-component: +70.71 km/h

Dividing the airplane's speed into components:y-component: -600 km/h; x-component: 0 km/h

Adding the components to get the resulting components:y-component: -529.29 km/h; x-component: +70.71

Using the Pythagorean Theorem to find the resulting speed:v^2 = y^2 + x^2 so v = 533.99 km/h

To find the angle of direction, use arctan (y/x):arctan (529.29/70.71) = 82.39 deg

ANSWER: velocity = 533.99 km/h at 82.39 deg S of E

Explanation:

8 0

4 years ago

The magnitude of the weight of a 3.0 kg object on the surface of the earth is 29 N. True False

madreJ [45]

True

In fact, the weight of an object on the surface of the Earth is given by:
$F=mg$
where m is the mass of the object and $g=9.81 m/s^2$ is the gravitational acceleration on Earth's surface. If we use the mass of the object, m=3.0 kg, we find
$F=mg=(3.0 kg)(9.81 m/s^2)=29 N$

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3 years ago

2) The horizontal and vertical components of the initial velocity of a football are 16 m/s and 20 m/s respectively. How long doe

ValentinkaMS [17]

Answer: 2.04 s

Explanation:

Let the initial velocity be v, Angle of projectile be

Then the horizontal component = v cos θ = 16 m/s

Vertical component of velocity = v sin θ = 20 m/s

Time taken to reach the highest point is half the time taken for total flight.

Time for total flight,

$t = \frac{2vsin \theta}{g}$