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3 years ago

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A negatively charged particle (charge = -1e) has a velocity of 6.0 × 106 m/s in the positive x direction at a point where the ma

gnetic field has the components Bx = 3.0 T, By = 1.5 T, and Bz = 2.0 T. What is the magnitude of the acceleration of the particle?

1 answer:

Anni [7]3 years ago

7 0

Answer:

$a=2.6x10^{18} m/s^2$

Explanation:

$v=6.0x10^6 m/s$ , $q=1.6x10^{-19}C$ , $q=9.11x10^{-31}kg$

$\beta=\beta_x=3.0T$ , $\beta_y=1.5 T$ , $\beta_y=2.0 T$

$F_b=q*\left[\begin{array}{ccc}i&j&k\\6x10^6&0&0\\3&1.5&2\end{array}\right]$

$F_b=-1.6x10^{-19}C*[-12x10^6 j + 9x10^6 k]$

$F_b=m*a$

$19.2x10^{-13}j-14.4x10^{-13}k=m*a$

$a=\frac{19.2x10^{-13}}{9.11x10{-31}}-\frac{14.4x10^{-13}}{9.11x10{-31}}$

$a=2.107 x10^{18}j-1.58x10^{18}k$

$a=\sqrt{(2.107x10^{18})^2-(1.57x10^{18})^2}$

$a=2.6x10^{18} m/s^2$

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An airplane flies eastward and always accelerates at a constant rate. At one position along its path, it has a velocity of 34.5

Inessa05 [86]

Answer:

the acceleration of the airplane is 5.06 x 10⁻³ m/s²

Explanation:

Given;

initial velocity of the airplane. u = 34.5 m/s

distance traveled by the airplane, s = 46,100 m

final velocity of the airplane, v = 40.7 m/s

The acceleration of the airplane is calculated from the following kinematic equation;

v² = u² + 2as

$2as= v^2 - u^2\\\\a = \frac{v^2 - u^2}{2s} \\\\a = \frac{(40.7)^2 -(34.5)^2}{2 \times 46,100} \\\\a = 5.06 \ \times \ 10^{-3} \ m/s^2$

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3 years ago

Why is gravity an example of a scientific law?

valina [46]

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3 years ago

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If you swim with the current in a river, your speed is increased by the speed of the water; if you swim against the current, you

Blababa [14]

Answer:

11.23%

Explanation:

Lets take

Speed of man in still water =u= 1.73 m/s

Speed of flow of water = v=0.52 m/s

When swims in downward direction then speed of man = u + v

When swims in upward direction then speed of man = u - v

Lets time taken by man when he swims in downward direction is $t_1$ and when he swims in downward direction is $t_2$

Lets distance is d and it will be remain constant in both the case

$d=(u+v)t_1$

$d=(u-v)t_2$

$(1.73+0.52)t_1=(1.73-0.52)t_2$

$t_2=1.85t_1$

Time taken in still water

2 d= t x 1.73

t=1.15 x d sec

$t_1=0.44d\ sec$

$t_2=0.82d\ sec$

total time in current = 0.82 +0.44 d=1.26 d sec

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$percentage\ time =\dfrac{1.28-1.15}{1.15}$

Percentage time =11.32%

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3 years ago

Sayid made a chart listing data of two colliding objects. A 5-column table titled Collision: Two Objects Stick Together with 2 r

Alborosie

Answer:

6 m/s is the missing final velocity

Explanation:

From the data table we extract that there were two objects (X and Y) that underwent an inelastic collision, moving together after the collision as a new object with mass equal the addition of the two original masses, and a new velocity which is the unknown in the problem).

Object X had a mass of 300 kg, while object Y had a mass of 100 kg.

Object's X initial velocity was positive (let's imagine it on a horizontal axis pointing to the right) of 10 m/s. Object Y had a negative velocity (imagine it as pointing to the left on the horizontal axis) of -6 m/s.

We can solve for the unknown, using conservation of momentum in the collision: Initial total momentum = Final total momentum (where momentum is defined as the product of the mass of the object times its velocity.

In numbers, and calling $P_{xi}$ the initial momentum of object X and $P_{yi}$ the initial momentum of object Y, we can derive the total initial momentum of the system: $P_{total}_i=P_{xi}+P_{yi}= 300*10 \frac{kg*m}{s} -100*6\frac{kg*m}{s} =\\=(3000-600 )\frac{kg*m}{s} =2400 \frac{kg*m}{s}$

Since in the collision there is conservation of the total momentum, this initial quantity should equal the quantity for the final mometum of the stack together system (that has a total mass of 400 kg):

Final momentum of the system: $M * v_f=400kg * v_f$

We then set the equality of the momenta (total initial equals final) and proceed to solve the equation for the unknown(final velocity of the system):

$2400 \frac{kg*m}{s} =400kg*v_f\\\frac{2400}{400} \frac{m}{s} =v_f\\v_f=6 \frac{m}{s}$

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Aleksandr-060686 [28]

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