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alexgriva [62]
4 years ago
15

What are three examples of objects that are too big to be studied without models

Physics
1 answer:
zysi [14]4 years ago
4 0
1.The entire Universe
2. The milky way
3.Time and Space 
hoped this helped

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What is the linear speed of a point on the equator, due to the earth's rotation?
kvv77 [185]
The equatorial radius of the earth is
r = 6378 km = 6378 x 10³ m

The earth makes 1 revolution in 24 hours.
The angular velocity is
ω = (2π rad)/(24*3600 s) = 7.2722 x 10⁻⁵ rad/s

The tangential velocity (linear velocity) at a point on the equator is
v = rω
   = (6378 x 10³ m)*(7.2722 x 10⁻⁵ rad/s)
   = 463.8 m/s

Answer: 463.8 m/s

8 0
4 years ago
Release an electron initially at rest in the presence of an electric field. The electron tends to go to the region of 1. same el
olga nikolaevna [1]

Answer:

The electron tends to go to the region of 4. higher electric potential.

Explanation:

When a charged particle is immersed in an electric field, it experiences a force given by

F=qE

where

q is the charge of the particle

E is the electric field

The direction of the force depends on the sign of the charge. In particular:

- The force and the electric field have the same direction if the charge is positive

- The force and the electric field have opposite directions if the charge is negative

Therefore, an electron (negative charge) moves in the direction opposite to the electric field lines.

However, electric field lines go from points at higher potential to points at lower potential: so, electrons move from regions at lower potential to regions of higher potential.

Therefore, the correct answer is

The electron tends to go to the region of 4. higher electric potential.

4 0
4 years ago
2. A 2,000 kg rocket is launched 12 km straight up at a constant acceleration into the sky at which point the rocket is travelli
Mice21 [21]

Answer:

797700000 J

Explanation:

From the question,

The work done by the rocket, is given as,

W = Ek+Ep............. Equation 1

Where Ek and Ep are the potential and the kinetic energy of the rocket respectively.

Ep = mgh............ Equation 2

Ek = 1/2mv²............. equation 3

Substitute equation 2 and equation 3 into equation 1

W = mgh+1/2mv².............. Equation 4

Where m = mass of the rocket, h = height, v = velocity of the rocket, g = acceleration due to gravity.

Given: m = 2000 kg, h = 12 km = 12000 m, v = 750 m/s, g = 9.8 m/s²

Substitute into equation 4

W = 2000(12000)(9.8)+1/2(2000)(750²)

W = 235200000+562500000

W = 797700000 J

6 0
3 years ago
Why can objects be made to float in a magnetic field?
dimulka [17.4K]
I would say that the answer is C
6 0
3 years ago
. In a physics lab, you attach a 0.200-kg air-track glider to the end of an ideal spring of negligible mass and start it oscilla
cluponka [151]

Answer:

Spring constant, k = 1.16 N/m

Explanation:

It is given that,

Mass of the air track, m = 0.2 kg

Time, t = 2.6 s

Let T is the time period of the spring. The expression for the time and spring constant is given by :

T=2\pi \sqrt{\dfrac{m}{k}}

k=\dfrac{4\pi^2m}{T^2}

k=\dfrac{4\pi^2\times 0.2}{(2.6)^2}

k = 1.16 N/m

So, the spring’s force constant is 1.16 N/m. Hence, this is the required solution.

6 0
3 years ago
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