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2 years ago

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

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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 $2.55\times 10^5m$