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3 years ago

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If a negatively charged particle enters a region of uniform magnetic field which is perpendicular to the particle's velocity, wi

ll the kinetic energy of the particle increase, decrease, or stay the same?

1 answer:

lys-0071 [83]3 years ago

3 0

Answer:

Same

Explanation:

While moving through a magnetic field in a direction perpendicular to a B-field, a continuous force experienced by a charged particle. If this magnetic field remains uniform, the force exerted also remains same and hence the velocity with which the particle is moving remains same. However, the particle is forced to move on a curved path until it forms a complete circle.

Hence, the kinetic energy remains the same because the speed is same

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One hundred jumping beans are placed along the center line of a gymnasium floor at six-inch intervals. Twelve hours later, the d

stira [4]

Answer:

a) Diffusion coefficient, D = 1.5 in/hr

b) Mean jump frequency, f = 0.0833 Hz

Explanation:

a) The relationship between the diffusion coefficient, time and mean displacement and can be given by the expression:

$^{2} = 2Dt$ ..........(1)

Where <r> = mean displacement

D = Diffusion coefficient

t = time = 12 hrs

sum of the squares of the distance divided by 100 is 36 in2.

<r>²= 36 in²

Substituting these values into equation (1) above

$36 = 2 * D *12\\36 = 24 D\\D = 36/24\\D = 1.5 in/hr$

b) Mean jumping distance, <r> = 0.1 inches

Applying equation (1) again

Where D = 1.5 in/hr

$^{2} = 2Dt$

$0.1^{2} = 2 * 1.5t\\0.01 = 3t\\t = 0.01/3\\t = 0.0033 hrs\\t = 0.0033 * 3600\\t = 12 seconds$

The mean jump frequency, f = 1/t

f = 1/12

f = 0.0833 Hz

8 0

3 years ago

Describe the shape of the light waves that would be created from a single, uncovered light bulb

lilavasa [31]

Curvy lines with light glooming

6 0

3 years ago

A cat is shot horizontally out of a cannon that is on top of a 125 meter tall cliff. If the cat lands 286 m away from the base o

Elden [556K]

The initial velocity of the cat is 56.6 m/s

Explanation:

The motion of the cat is a projectile motion, consisting of:

- a uniform horizontal motion with constant velocity

- a vertical accelerated motion with constant acceleration (free fall)

We start by analyzing the vertical motion, to find the time it takes for the cat to reach the ground. We use the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where we have (choosing downard as positive direction)

s = 125 m is the vertical displacement of the cat

u = 0 is the initial vertical velocity

t is the time taken to reach the ground

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t, we find

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(125)}{9.8}}=5.05 s$

Now we can analzye the horizontal motion. Since the motion is uniform and the horizontal velocity is constant, it is given by

$v_x =\frac{d}{t}$

where

d = 286 m is the horizontal distance travelled

t = 5.05 s is the time taken

Solving the equation,

$v_x = \frac{286}{5.05}=56.6 m/s$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

3 0

3 years ago

Let's say that we have a pilot that is dropping a package from a plane that is flying horizontally at a constant speed. If we ne

antoniya [11.8K]

Answer:

The package will be directly below the location of the plane.

Explanation:

Look up projectile motion for more information. The horizontal speed of the package is separate from the vertical speed of the package. The vertical speed of the falling package will be based on the rate of acceleration and the height of the package when dropped. The horizontal speed of the package will be the same as the plane so the package will remain directly below the plane the entire time until the package hits the ground.

6 0

3 years ago

If a ball is thrown vertically upward from the roof of 32 foot building with a velocity of 64 ft/sec, its height after t seconds

stepan [7]

Answer:

a) $s_{max} = 96\,ft$ , b) $v(4.449\,s) = -78.368\,\frac{ft}{s}$

Explanation:

a) The maximum height is obtained with the help of the First and Second Derivative Tests:

First Derivative

$v(t) = 64 - 32\cdot t$

$64 - 32\cdot t = 0$

$t = 2\,s$

Second Derivative

$a(t) = -32$ (absolute maximum)

The maximum height reached by the ball is:

$s (2\,s) = 32 + 64\cdot (2\,s) - 16\cdot (2\,s)^{2}$

$s_{max} = 96\,ft$

b) The time required by the ball to hit the ground is:

$32+64\cdot t - 16\cdot t^{2} = 0$

$-16\cdot (t^{2}-4\cdot t - 2) = 0$

$t^{2}-4\cdot t - 2 = 0$

$(t -4.449)\cdot (t+0.449)\approx 0$

Just one root offers a solution that is physically reasonable:

$t = 4.449\,s$

The velocity of the ball when it hits the ground is:

$v(4.449\,s) = 64 - 32\cdot (4.449\,s)$

$v(4.449\,s) = -78.368\,\frac{ft}{s}$

6 0

3 years ago

Read 2 more answers