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Whitepunk [10]
3 years ago
11

The magnetic north pole changes location but the geographic north pole does not true or false

Physics
1 answer:
a_sh-v [17]3 years ago
3 0
The North Magnetic Pole is the point on the surface of Earth's Northern Hemisphere at which the planet's magnetic field points vertically downwards (in other words, if a magnetic compass needle is allowed to rotate about a horizontal axis, it will point straight down). There is only one location where this occurs, near (but distinct from) the Geographic North Pole and the Geomagnetic North Pole. So yes true
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A scientist discovered an elementary particle paired with another particle. It has a positive charge equal to two–thirds of the
julia-pushkina [17]
W boson has +1e or - 1e charge, Z boson has 0 charge.

Leptons have +1e, -1e or 0 charge.

Photons have 0 charge.

Only quarks have a charge of +2/3e or -1/3e of an electron charge. 

To be exact, only up-type quarks (Up, Down and Top quarks) have a +2/3e or two thirds of an electron charge.

So the correct answer is D) Quark.
6 0
3 years ago
A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface.
Alexxandr [17]

Answer:

Kinetic energy of the projectile at the vertex of the trajectory: 900\; {\rm J}.

Work done when firing this projectile: 2500\; {\rm J}.

Explanation:

Since the drag on this projectile is negligible, the horizontal velocity v_{x} of this projectile would stay the same (at 30\; {\rm m\cdot s^{-1}}) throughout the flight.

The vertical velocity v_{y} of this projectile would be 0\; {\rm m\cdot s^{-1}} at the vertex (highest point) of its trajectory. (Otherwise, if v_{y} > 0, this projectile would continue moving up and reach an even higher point. If v_{y} < 0, the projectile would be moving downwards, meaning that its previous location was higher than the current one.)

Overall, the velocity of this projectile would be v = 30\; {\rm m\cdot s^{-1}}\! when it is at the top of the trajectory. The kinetic energy \text{KE} of this projectile (mass m = 2.0\; {\rm kg}) at the vertex of its trajectory would be:

\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (30\; {\rm m\cdot s^{-1}})^{2} \\ &= 900\; {\rm J} \end{aligned}.

Apply the Pythagorean Theorem to find the initial speed of this projectile:

\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \left(\sqrt{900 + 1600}\right)\; {\rm m\cdot s^{-1}} \\ &= 50\; {\rm m\cdot s^{-1}}\end{aligned}.

Hence, the initial kinetic energy \text{KE} of this projectile would be:

\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (50\; {\rm m\cdot s^{-1}})^{2} \\ &=2500\; {\rm J} \end{aligned}.

All that energy was from the work done in launching this projectile. Hence, the (useful) work done in launching this projectile would be 2500\; {\rm J}.

7 0
2 years ago
PLEASE HELP 40 POINTS
sergij07 [2.7K]

Liar, not even 40 points....

5 0
3 years ago
2.15 Un móvil en trayectoria rectilinea ocupa las posiciones indicadas en la tabla para cada minuto:
inna [77]
To make sure I don’t has 35 012345 bishops
5 0
3 years ago
Which principle allows us to conclude that gravity acts as the same today and will tomorrow?
Svet_ta [14]

Answer:

Gravity is a constant quantity

Explanation:

4 0
3 years ago
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