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motikmotik
3 years ago
5

The voltage between the cathode and the screen of a television set is 22 kV. If we assume a speed of zero for an electron as it

leaves the cathode, what is its speed just before it hits the screen?
Physics
1 answer:
il63 [147K]3 years ago
4 0

Answer:

v= 8.8*10⁷ m/s

Explanation:

  • Assuming no friction present, the change in electrical potential energy, must be equal in magnitude, to the change in kinetic energy of the electron.
  • The change in the electrical potential energy, can be expressed as follows:

       \Delta U = (-e)*\Delta V

  • The change in kinetic energy, assuming that the electron started from rest, can be written as follows:

       \Delta K = \frac{1}{2} *m*v^{2}

       ⇒\Delta K = -\Delta U

  • From the equation above, replacing by ΔK and ΔU, we have:

       -\Delta U =- (-e)*\Delta V =\Delta K = \frac{1}{2} *m*v^{2}

  • Solving for v:

        v=\sqrt{\frac{2*e*\Delta V}{m_{e}} } =\sqrt{\frac{2*1.6e-19C*22e3V}{9.1e-31kg}} }\\ v= 8.8e7 m/s

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A child sleds down a hill with an acceleration of 2.94 m/s2. If her initial speed is 0.0 m/s and her final speed is 17.5 m/s, ho
hram777 [196]

Answer:

The child will take 5.952 seconds to travel from the top of the hill to the bottom.

Explanation:

Given that the child accelerates uniformly and that both initial (v_{o}) and final speeds (v_{f}), measured in meters per second, and acceleration (a), measured in meters per square second, are known, we proceed to use the following kinematic equation to determine the time taken to travel from the top of the hill to the bottom (t), measured in seconds, is:

t = \frac{v_{f}-v_{o}}{a} (1)

If we know that v_{o} = 0\,\frac{m}{s}, v_{f} = 17.5\,\frac{m}{s} and a = 2.94\,\frac{m}{s^{2}}, then the time taken is:

t = \frac{17.5\,\frac{m}{s}-0\,\frac{m}{s} }{2.94\,\frac{m}{s^{2}} }

t = 5.952\,s

The child will take 5.952 seconds to travel from the top of the hill to the bottom.

8 0
3 years ago
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eduard
Uhh? Do you have any questions or need help?
7 0
3 years ago
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A large helium filled balloon is used as the center piece for a graduation party. The balloon alone has a mass of 222 kg and it
inn [45]

Answer:

The buoyant force is 3778.8 N in upward.

Explanation:

Given that,

Mass of balloon = 222 Kg

Volume = 328 m³

Density of air = 1.20 kg/m³

Density of helium = 0.179 kg/m³

We need to calculate the buoyant force acting

Using formula of buoyant force

F_{b}=\rho_{air}\times V_{b}\times g

Where, \rho_{air} = density of air

V = Volume of balloon

g = acceleration due to gravity

Put the value into the formula

F_{b}=1.20\times321\times9.81

F_{b}=3778.8\ N

This buoyant force is in upward direction.

Hence, The buoyant force is 3778.8 N in upward.

4 0
3 years ago
Sophia is planning on going down an 8-m high water slide. Her weight is 50 N. What is Sophia's Gravitational Potential Energy at
pogonyaev
Gpe = 50 x 8 = 400 Joules
6 0
3 years ago
what equastion do you use to solve Riders in a carnival ride stand with their backs against the wall of a circular room of diame
Hitman42 [59]

Answer:

μsmín = 0.1

Explanation:

  • There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
  • This friction force has a maximum value, that can be written as follows:

       F_{frmax} = \mu_{s} *F_{n} (1)

       where  μs is the coefficient of static friction, and Fn is the normal force,

       perpendicular to the wall and aiming to the center of rotation.

  • This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
  • This force has the following general expression:

       F_{c} =  m* \omega^{2} * r (2)

       where ω is the angular velocity of the riders, and r the distance to the

      center of rotation (the  radius of the circle), and m the mass of the

      riders.

      Since Fc is actually Fn, we can replace the right side of (2) in (1), as

      follows:

     F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)

  • When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

       m* g = m* \mu_{smin} * \omega^{2} * r (4)

  • (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
  • Cancelling the masses on both sides of (4), we get:

       g = \mu_{smin} * \omega^{2} * r (5)

  • Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

      60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)

  • Replacing by the givens in (5), we can solve for μsmín, as follows:

       \mu_{smin} = \frac{g}{\omega^{2} *r}  = \frac{9.8m/s2}{(6.28rad/sec)^{2} *2.5 m} =0.1 (7)

5 0
2 years ago
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