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Helen [10]
2 years ago
7

II.

Physics
1 answer:
Alekssandra [29.7K]2 years ago
3 0

Answer:

I just need to get points soorry

Explanation:

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The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.4cm apart with a 28kV potential d
yulyashka [42]

Answer:

a)   F = 3.2 10⁻¹⁰ N , b)       v = 9.9 10⁷ m / s

Explanation:

a) The electric force is

       F = q E

The electric field is related to the potential reference

     V = E d

     E = V / d

Let's replace

    F = e V / d

Let's calculate

    F = 1.6 10⁻¹⁹ 28 10³ / 1.4 10⁻²

    F = 3.2 10⁻¹⁰ N

b) For this part we can use kinematics

          v² = v₀ + 2 a d

          v = √ 2 ad

Acceleration can be found with Newton's second law

        e V / d = m a

        a = e / m V / d

        a = 1.6 10⁻¹⁹ / 9.1 10⁻³¹ 28 10³ / 1.4 10⁻²

        a = 3,516 10⁻¹⁷ m / s²

Let's calculate the speed

       v = √ (2 3,516 10¹⁷ 1.4 10⁻²)

       v = √ (98,448 10¹⁴)

       v = 9.9 10⁷ m / s

3 0
2 years ago
Yoyoyoyoyoyoyoyoyoyoyoyoyoyoyoy
Vikentia [17]

Answer:

Yoyoyoyoyooyoyy

Explanation:

Yoyoyoyoyoyoyoyoyooy

4 0
2 years ago
Read 2 more answers
A steel spur pinion has a module of 2 mm, 17 teeth cut on the 20° full-depth system, and a face width of 20 mm. At a speed of 16
erastovalidia [21]

Answer:

The value of bending stress on the pinion 35.38 M pa

Explanation:

Given data

m = 2 mm

Pressure angle \phi = 20°

No. of teeth T = 17

Face width (b) = 20 mm

Speed N = 1650 rpm

Power = 1200 W

Diameter of the pinion gear

D = m T

D = 2 × 17

D = 34 mm

Velocity of the pinion gear

V =\pi D( \frac{N}{60} )

V = 3.14 (0.034) \frac{(1650)}{60}

V = 2.93 \frac{m}{s}

Form factor for the pinion gear is

Y = 0.303

Now

K_{v} = \frac{6.1 +0.303}{6.1} = 1.049

Force on gear tooth

F = \frac{P}{V}

F = \frac{1200}{2.93}

F = 408.73 N

Now the bending stress is given by the formula

\sigma = \frac{K_{v} F}{m b y}

\sigma = \frac{(1.049)(408.73)}{(0.002)(0.02)(0.303)}

\sigma = 35.38 M pa

This is the value of bending stress on the pinion

8 0
3 years ago
If the kinetic energy of an electron is 4.1e-18 j, what is the speed of the electron? (you can use the approximate (nonrelativis
arlik [135]
The kinetic energy of the electron is
K= \frac{1}{2}mv^2
where m=9.1 \cdot 10^{-31} kg is the mass of the electron and v its speed. Since we know the value of the kinetic energy, K=4.1 \cdot 10^{-18} J, we can find the value of the speed v:
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2\cdot 4.1 \cdot 10^{-18}J}{9.1 \cdot 10^{-31}kg} }  = 3\cdot 10^6 m/s
3 0
3 years ago
A shell is fired from the ground with an initial speed of 1.60 × 103 m/s (approximately five times the speed of sound) at an ini
Volgvan

Answer:

The horizontal range will be 2.55\times 10^5m

Explanation:

We have given initial speed of the shell u = 1.6\times 10^3m/sec

Angle of projection = 51°

Acceleration due to gravity g=9.8m/sec^2

We have to find maximum range

Horizontal range in projectile motion is given by

R=\frac{u^2sin2\Theta }{g}=\frac{(1.60\times 10^3)^2sin(2\times 51^{\circ})}{9.81}=2.55\times 10^5m

So the horizontal range will be 2.55\times 10^5m

6 0
2 years ago
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