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Shkiper50 [21]
3 years ago
15

Choose all the answers that apply. The continents _____. were once connected have always been in the same location are slowly dr

ifting have stopped moving are made mostly of basalt
Physics
1 answer:
svetoff [14.1K]3 years ago
7 0

Answer: the correct options are

- were once connected

- are slowly drifting

Explanation: The continent is the main land mass of the earth which includes the following:

- Asia, Africa, Europe, Australia, North America, South America, and Antarctica.

These continents were once connected forming a supercontinent called PANGAEA

during the late palaeozoic and early mesozoic eras, millions of years ago.

Alfred Wegener, who is a German meteorologist, in 1912 came up with a hypothesis that the continents are slowly drifting the earth. He called the movement continental drift.

The continents rest on massive slabs of rock called tectonic plates. The plates are always moving and interacting in a process called plate tectonics. In recent studies, through the science of plate tectonics, we where able to understand the mechanism through which continental drift occurs. The mechanism includes:

- processes of seafloor spreading,

- rift valley formation, and

- subduction (where heavier tectonic plates sink beneath lighter ones).

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In order to place a satellite into orbit, it requires enough fuel to supply the necessary mechanical energy. Into what types of
nexus9112 [7]

When a satellite is revolving into the orbit around a planet then we can say

net centripetal force on the satellite is due to gravitational attraction force of the planet, so we will have

F_g = F_c

\frac{GM_pM_s}{r^2} = \frac{M_s v^2}{r}

now we can say that kinetic energy of satellite is  given as

KE = \frac{1}{2}M_s v^2

KE = \frac{GM_sM_p}{2r}

also we know that since satellite is in gravitational field of the planet so here it must have some gravitational potential energy in it

so we will have

U = -\frac{GM_sM_p}{r}

so we can say that energy from the fuel is converted into kinetic energy and gravitational potential energy of the satellite

6 0
3 years ago
A box is sliding up an incline that makes an angle of 14.0° with respect to the horizontal. the coefficient of kinetic friction
mezya [45]
A box is sliding up an incline that makes an angle of 14.0° with respect to the horizontal. the coefficient of kinetic friction between the box and the surface of the incline is 0.180. the initial speed of the box at the bottom of the incline is 2.20 m/s. how far does the box travel along the incline before coming to rest?
6 0
3 years ago
A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpend
Sonja [21]

The time interval for which the proton remains in the field is -

Δt = $\frac{\pi R}{40}.

We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.

We have to determine time interval during which the proton is in the field.

<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>

The magnitude of force on the charged particle moving in a uniform magnetic field is given by -

F = qvB sinθ



According to the question, we have -

Entering Velocity (v) = 20 i  m/s

Magnetic field intensity (B) = 0.3 T

Leaving velocity (u) = - 20 j  m/s

Now -

The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -

d = $\frac{2\pi r}{4} = $\frac{\pi R}{2}

Since, the radius of circular path is not given, we will assume it R.

Therefore, time for which proton remained in the field is -

t = $\frac{\pi R}{2v} = \frac{\pi R}{40}

Hence, the time interval for which the proton remains in the field is -

Δt = $\frac{\pi R}{40}

To solve more questions on Force on charged particle, visit the link below-

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6 0
1 year ago
A 1430 kg car speeds up from 7.50 m/s to 11.0 m/s in 9.30 s. Ignoring friction, how much power did that require?(unit=W)PLEASE H
Aleks04 [339]

Answer:

Kinetic energy = (1/2) (mass) (speed²)

Original KE = (1/2) (1430 kg) (7.5 m/s)²  =  40,218.75 joules

Final KE  =  (1/2) (1430 kg) (11.0 m/s)²  =   86,515 joules

Work done during the acceleration = (40218.75 - 86515) = 46,296.25 joules

Power = work/time = 46,296.25 joules / 9.3 sec  =  4,978.1 watts .

Explanation:

Dont report my answer please

3 0
2 years ago
Determine the vector perpendicular to the plane of A= 31+ 6j - 2k and B=4i-j +3k
Sliva [168]

The vector perpendicular to the plane of A = 3i+ 6j - 2k and B = 4i-j +3k is 16 i - 17 j - 27 k

Let r be the vector perpendicular to A and B,

r = A * B

A = 3i + 6j - 2k

B = 4i - j + 3k

a1 = 3

a2 = 6

a3 = - 2

b1 = 4

b2 = - 1

b3 = 3

a * b = ( a2 b3 - b2 a3 ) i + ( a3 b1 - b3 a1 ) j + ( a1 b2 - b1 a2 ) k

a * b = [ ( 6 * 3 ) - ( - 1 * - 2 ) ] i + [ ( - 2 * 4 ) - ( 3 * 3 ) ] j + [ ( 3 * - 1 ) - ( 4 * 6 ) ] k

a * b = 16 i - 17 j - 27 k

The perpendicular vector, r = 16 i - 17 j - 27 k

Therefore, the vector perpendicular to the plane of A = 3i + 6j - 2k and B = 4i - j + 3k is 16 i - 17 j - 27 k

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5 0
1 year ago
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