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Zolol [24]
3 years ago
13

Can you help me on this fast just the second one

Physics
1 answer:
Mariulka [41]3 years ago
3 0
4. E
5. D
6. F

Hope this helps
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A proton with charge 1.601019 As is moving at 2.4105 m/s through a magnetic field of 4.5 T. You want to find the force on the pr
Whitepunk [10]

The force on the proton is 17.4 N.

<h3>What is the force on the proton?</h3>

Now we know that the proton is positively charged and that the force on the charge as it moved through the magnetic field could be given by the relation; F = qvB

Where;

F = force

q = charge

v = velocity

B = magnetic field

Having said this, we can see that;

q = 1.601019 As or C

v =  2.4105 m/s

T = 4.5 T

F =  1.601019 As * 2.4105 m/s * 4.5 T

F = 17.4 N

Learn more about magnetic force:brainly.com/question/12824331

#SPJ1

6 0
1 year ago
Can someone help? Please?
zubka84 [21]

Answer:

A. Speed

Explanation:

Speed is the magnitude of velocity, which is given in the question. Velocity is a vector quantity and therefore has both a magnitude and a direction. Only the former is implied in the question.

5 0
2 years ago
Read 2 more answers
If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.
Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

s_x=v_x_0 t

t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s

Now for the vertical distance

vy_o=0

than the equation of motion becomes

s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2

Now using this acceleration the value of electric field is calculated as

E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0
3 years ago
Determine la fuerza aplicada sobre un émbolo en una prensa hidráulica si se quiere cargar una lanza de 2970kg; toma en cuenta qu
Artist 52 [7]
I’m not sure sorry I really wish I could help
4 0
2 years ago
A T-shirt cannon can shoot a 0.085 kg T-shirt at nearly 30 m/s. The T-shirt cannon has a mass of 33 kg. If the initial net momen
IgorLugansk [536]

Answer:

Approximately 0.077\; {\rm m\cdot s^{-1}} (assuming that external forces on the cannon are negligible.)

Explanation:

If an object of mass m is moving at a velocity of v, the momentum p of that object would be p = m\, v.

Momentum of the t-shirt:

\begin{aligned} p(\text{t-shirt}) &= m(\text{t-shirt}) \, v(\text{t-shirt}) \\ &= 0.085\; {\rm kg} \times 30\; {\rm m \cdot s^{-1}} \\ &= 2.55 \; {\rm kg \cdot m \cdot s^{-1}} \end{aligned}.

If there is no external force (gravity, friction, etc.) on this cannon, the total momentum of this system should be conserved. In other words, if p(\text{cannon}) denote the momentum of this cannon:

p(\text{t-shirt}) + p(\text{cannon}) = 0.

p(\text{cannon}) = -p(\text{t-shirt}) = -2.55\; {\rm kg \cdot m \cdot s^{-1}}.

Rewrite p = m\, v to obtain v = (p / m). Since the mass of this cannon is m(\text{cannon}) = 33\; {\rm kg}, the velocity of this cannon would be:

\begin{aligned} v(\text{cannon}) &= \frac{p(\text{cannon})}{m(\text{cannon})} \\ &= \frac{-2.55\; {\rm kg \cdot m \cdot s^{-1}}}{33\; {\rm kg}} \\ &\approx 0.077\; {\rm m \cdot s^{-1}}\end{aligned}.

8 0
1 year ago
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