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timama [110]
3 years ago
8

An electron moves with a constant horizontal velocity of 3.0 × 106 m/s and no initial vertical velocity as it enters a deflector

inside a TV tube. The electron strikes the screen after traveling 11 cm horizontally and 34 cm vertically upward with no horizontal acceleration. What is the constant vertical acceleration provided by the deflector? (The effects of gravity can be ignored.)
Physics
1 answer:
Ghella [55]3 years ago
7 0

Answer:

a = 5.05 x 10¹⁴ m/s²

Explanation:

Consider the motion along the horizontal direction

v_{x} = velocity along the horizontal direction = 3.0 x 10⁶ m/s

t = time of travel

X = horizontal distance traveled = 11 cm = 0.11 m

Time of travel can be given as

t = \frac{X}{v_{x}}

inserting the values

t = 0.11/(3.0 x 10⁶)

t = 3.67 x 10⁻⁸ sec

Consider the motion along the vertical direction

Y = vertical distance traveled = 34 cm = 0.34 m

a = acceleration = ?

t = time of travel  = 3.67 x 10⁻⁸ sec

v_{y} = initial velocity along the vertical direction = 0 m/s

Using the kinematics equation

Y = v_{y} t + (0.5) a t²

0.34 = (0) (3.67 x 10⁻⁸) + (0.5) a (3.67 x 10⁻⁸)²

a = 5.05 x 10¹⁴ m/s²

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a van moves with a constant speed of 79 km/h how long will it take to travel a distance of 502 kilometers
Mazyrski [523]

Answer:

6.35hours

Explanation:

s=vt

t=s/v=502/79=6.35hours

5 0
2 years ago
2. Kevin works as a janitor, and he is pushing a fully-
dybincka [34]

The time taken for him to move the bin 6.5 m is 2.30 s.

The given parameters;

  • <em>weight of the load, w = 557 N</em>
  • <em>force applied , F = 410 N</em>
  • <em>angle of force, =  15°</em>
  • <em>coefficient of kinetic friction  = 0.46</em>
  • <em>distance moved, d = 6.5 m</em>

The net horizontal force on the recycling bin is calculated as follows;

Fcos\theta - F_k = ma

where;

  • <em>m is the mass of the recycling bin</em>
  • <em />F_k<em> is the frictional force </em>

W = mg

557 = 9.8m\\\\m = \frac{557}{9.8} \\\\m = 56.84 \ kg

The net horizontal force on the recycling bin is calculated as;

Fcos \theta - F_k = ma\\\\Fcos\theta - \mu_kF_n  = ma\\\\410\times cos(15) \ - \ 0.46(557) = 56.84 a\\\\139.8 = 56.84a\\\\a = \frac{139.8}{56.84} \\\\a = 2.46 \ m/s^2

The time taken for him to move the bin 6.5 m is calculated as follows;

s = v_0t + \frac{1}{2} at^2\\\\6.5 = 0 + \frac{1}{2}  \times 2.46\times t^2\\\\6.5 = 1.23 t^2\\\\t^2 = \frac{6.5 }{1.23} \\\\t^2 = 5.285\\\\t = \sqrt{5.285} \\\\t = 2.30 \ s

Thus, the time taken for him to move the bin 6.5 m is 2.30 s.

Learn more here:brainly.com/question/21684583

7 0
2 years ago
It takes 105 j of work to move 2.7 c of charge from the negative plate to the positive plate of a parallel plate capacitor. what
wolverine [178]
Se necesita 105 j de trabajo para mover 2,7 c de carga de la placa negativa a la placa positiva de un condensador de placa paralela. Qué diferencia de tensión existe entre las placas

5 0
3 years ago
. A 1.50kg mass on a spring has a displacement as a function of time given by the equation: x(t) = (7.40cm)cos[(4.16s-1)t – 2.42
rusak2 [61]

Answer:

Solution:

we have given the equation of motion is x(t)=8sint [where t in seconds and x in centimeter]

Position, velocity and acceleration are all based on the equation of motion.

The equation represents the position.  The first derivative gives the velocity and the 2nd derivative gives the acceleration.

x(t)=8sint

x'(t)=8cost

x"(t)=-8sint

now at time t=2pi/3,

position, x(t)=8sin(2pi/3)=4*squart(3)cm.

velocity, x'(t)=8cos(2pi/3)==4cm/s

acceleration, x"(t)==8sin(2pi/3)=-4cm/s^2

so at present the direction is in y-axis.

5 0
3 years ago
Pe=___
Nesterboy [21]

Answer:

H = start height (v = 0)

h = present height

v = present speed

assuming no friction

total energy = PE + KE

mgH = mgh + .5mv^2

if PE = KE then

mgH = mgh + mgh

h = H/2

potential energy = kinetic energy when object is at half its start height.

Explanation:

8 0
3 years ago
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