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3 years ago

Calculate the potential energy of a five KG object sitting at the top of a 2 m ramp

Physics

1 answer:

docker41 [41]3 years ago

5 0

5 x 10 x 200 = 10000

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3 years ago

Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce

lutik1710 [3]

Answer:

speed of electrons = 3.25 × $10^{7}$ m/s

acceleration in term g is 3.9 × $10^{17}$ g.

radius of circular orbit is 2.76 × $10^{-4}$ m

Explanation:

given data

voltage = 3 kV

magnetic field = 0.66 T

solution

law of conservation of energy

PE = KE

qV = 0.5 × m × v²

v = $\sqrt{\frac{2qV}{m}}$

v = $\sqrt{\frac{2\times 1.6 \times 10^{-19}\times 3}{9.1\times 10^{-31}}$

v = 3.25 × $10^{7}$ m/s

and

magnetic force on particle movie in magnetic field

F = Bqv

ma = Bqv

a = $\frac{Bqv}{m}$

a = $\frac{0.67\times 1.6\times 10^{-19}\times 3.25\times 10^7}{9.1\times 10^{-31}}$

a = 3.82 × $10^{18}$ m/s²

and acceleration in term g

a = $\frac{3.82\times 10^{18}}{9.81}$

a = 3.9 × $10^{17}$ g

acceleration in term g is 3.9 × $10^{17}$ g.

and

electron moving in circular orbit has centripetal force

F = $\frac{mv^2}{r}$

Bqv = $\frac{mv^2}{r}$

r = $\frac{mv}{Bq}$

r = $\frac{9.1\times 10^{-31}\times 3.25\times 10^7}{0.67\times 1.6\times 10^{-19}}$

r = 2.76 × $10^{-4}$ m

radius of circular orbit is 2.76 × $10^{-4}$ m

8 0

3 years ago

Your teacher (175kg) and the lab (15kg) are 2m apart. What is the gravitational force between them? Draw a FBD, and label

Pachacha [2.7K]

The gravitational force between two objects is given by
$F=G \frac{m_1 m_2 }{r^2}$
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is the separation between the two objects

In this problem, $m_1 = 175 kg$ , $m_2 =15 kg$ and $r=2 m$ , therefore the gravitational force between the two objects is
$F=(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2}) \frac{(175 kg)(15 kg)}{(2m)^2}=4.38 \cdot10^{-8} N$

7 0

2 years ago