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RUDIKE [14]
2 years ago
8

A proton moves in the negative x-direction through a uniform magnetic field in the negative y-direction what is the direction of

the magnetic force acting on the proton?
out of the page
into the page
in the positive y-direction
in the negative y-direction
Physics
1 answer:
Tom [10]2 years ago
5 0
Okay thank goodness for the reason why I am not trying it out lol I love
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You can use one or more of your five senses to make ________ during an inquiry activity.
TiliK225 [7]
A judgement or an observation?
6 0
4 years ago
A 1500 kg weather rocket accelerates upward at 10m/s2. It explodes 2.0 s after liftoff and breaks into two fragments, one twice
ladessa [460]

Answer:

Approximately \rm 19.8\; m\cdot s^{-1} (downwards.)

Assumptions:

  • the rocket started from rest;
  • the gravitational acceleration is constantly \rm -9.8\; m \cdot s^{-2};
  • there's no air resistance on the rocket and the two fragments.
  • Both fragments traveled without horizontal velocity.

Explanation:

The upward speed of the rocket increases by \rm 10\; m \cdot s^{-1}. If the rocket started from rest, the vertical speed of the rocket should be equal to \rm 20\; m \cdot s^{-1}.

The mass of the rocket (before it exploded) is 1500 kilograms. At 20 m/s, its momentum will be equal to \rm 20 \times 1500 = 30,000\; kg \cdot m\cdot s^{-1}.

What's the initial upward velocity, u, of the lighter fragment?

The upward velocity of the lighter fragment is equal to v = 0 once it reached its maximum height of x = \rm 530\; m.

v^2 - u^2 = 2g \cdot x.

\begin{aligned}u &= \sqrt{v^2 - 2g\cdot x} \\ &= \sqrt{-2 (-9.8) \times 530}\\ &\approx \rm 101.922\; m \cdot s^{-1}\end{aligned}.

Mass of the two fragments:

  • Lighter fragments: \displaystyle \frac{1}{1 + 2} \times 1500 =\rm 500\; kg.
  • Heavier fragment: \displaystyle \frac{2}{1 + 2} \times 1500 =\rm 1000\; kg.

Initial momentum of the lighter fragment:

m \cdot v = \rm 10192.2\; kg \cdot m \cdot s^{-1}.

If there's no air resistance, momentum shall conserve. The momentum of the lighter fragment, plus that of the heavier fragment, should be equal to that of the rocket before it exploded.

The initial momentum of the heavier fragment should thus be equal to the momentum of the two pieces, combined, minus the initial momentum of the lighter fragment.

\rm 30000 - 10192.2 = 19807.8\;kg \cdot m \cdot s^{-1}.

Velocity of the heavier fragment:

\displaystyle \rm \frac{19807.8\;kg \cdot m \cdot s^{-1}}{1000\; kg} \approx 19.8\; m \cdot s^{-1}.

5 0
4 years ago
Read 2 more answers
5. If Joe spend 250 W power to lift a box weighing 25.5 kg in 3.0 s, how high did he lift the box?
Serjik [45]

Answer:

3m

Explanation:

p=250w

m=25.5

t=3.0s

D=x

F=mxa   F= 25.5kg x 9.8m/s^2 =250N

D= 250wx3.0s/250w

3m

5 0
3 years ago
A ball rolls from point A to point B. The total energy of the ball at point A isn’t the same as the sum of its potential energy
murzikaleks [220]
B. Some of the ball’s energy is transformed to thermal energy.

Hope this helps you!
3 0
4 years ago
Read 2 more answers
The system needs an ordinary friction-based brake to bring the train to a full stop. Explain why the magnetic brake is not very
BabaBlast [244]

Answer:

The slower the train is moving, the less are the changes of the magnetic flux, thus the eddy currents become weaker.

Explanation:

A magnetic brakes is not a very efficient way of braking when a train is moving slowly because at low speeds, the changes in the magnetic flux are very less and so it causes the eddy current to become weaker.

Let us find the drag force which is proportional to the velocity of two conducting plates.

The EMF that is induced in the eddy currents are : $E=v(B \times L)$

The force which is due to the induced magnetic field is, $F=l(L \times B)$

Therefore, $F=\frac{E}{R} \times (L \times B)$

                 $F=\frac{v(B \times L)}{R} \times (L \times B)$

Here, force is directly proportional to the velocity of the two conducting plates.

Therefore, we can say that when the speed of the train is low, the magnetic flux changes are less and thus the eddy currents are weaker.  

6 0
3 years ago
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