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3 years ago

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Which of the following is an example of an irregular meter? A. Four beats per measure B. Five beats per measure C. Two beats per

measure D. Three beats per measure

1 answer:

Komok [63]3 years ago

7 0

B. 5 beats per measure.

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A mass spectrometer was used in the discovery of the electron. In the velocity selector, the electric and magnetic fields are se

Mama L [17]

Answer:

Explanation:

Radius of dee, r = 8 mm = 0.008 m

Electric field, e = 400 V/m

Magnetic field, B = 4.7 x 10^-4 T

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

(a) Let v is the speed of electrons.

$v = \frac{Bqr}{m}$

$v = \frac{4.7\times 10^{-4}\times 1.6\times 10^{-19}\times 0.008}{9.1 \times 10^{-31}}$

v = 661098.9 = 661099 m/s

(b)

$\frac{e}{m}=\frac{1.6 \times 10^{-19}}{9.1\times 10^{-31}}$

e / m = 1.76 x 10^14 C / kg

(c) Let K be the kinetic energy

K = 0.5 x mv²

K = 0.5 x 9.1 x 10^-31 x 661099 x 661099

K = 1.99 x 10^-19 J

K = 1.24 eV

So, the potential difference is

V = 1.24 V

(d) if the acceleration voltage is doubled

V = 2 x 1.24 = 2.48 V

So, Kinetic energy

K = 2.48 eV

K = 2.48 x 1.6 x 10^-19 = 3.968 x 10^-19 J

Let v is the speed

K = 0.5 x mv²

3.968 x 10^-19 = 0.5 x 9.1 x 10^-31 x v²

v = 933856.5 m/s

Let the new radius is r.

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

$r=\frac{9.1\times 10^{-31}\times 933856.5}{4.7\times 10^{-4}\times 1.6\times 10^{-19}}$

r = 0.0113 m = 1.13 cm

7 0

3 years ago

If grade 7 students learn six periods and if one period is 40 minutes then how many seconds they learn per day

Ludmilka [50]

Answer:

C. 14400

Explanation:

40min × 6 periods = 240mins total.

1 minute = 60 seconds.

240 minutes × 60 = 14400 seconds.

6 0

3 years ago

Read 2 more answers

A runner traveling with an initial velocity of 1.1 m/s accelerates at a constant rate of 0.8 m/s2fora time of 2.0 s.(a).What is

pychu [463]

Answer:

The final velocity of the runner at the end of the given time is 2.7 m/s.

Explanation:

Given;

initial velocity of the runner, u = 1.1 m/s

constant acceleration, a = 0.8 m/s²

time of motion, t = 2.0 s

The velocity of the runner at the end of the given time is calculate as;

$v = u + at$

where;

v is the final velocity of the runner at the end of the given time;

v = 1.1 + (0.8)(2)

v = 2.7 m/s

Therefore, the final velocity of the runner at the end of the given time is 2.7 m/s.

7 0

3 years ago

Find the speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth. (Radiu

Gnesinka [82]

The orbiting velocity of the satellite is 4.2km/s.

To find the answer, we need to know about the orbital velocity of a satellite.

<h3>What's the expression of orbital velocity of a satellite?</h3>

Mathematically, orbital velocity= √(GM/r)
r = radius of the orbital, M = mass of earth

<h3>What's the orbital velocity of a satellite orbiting earth with a radius 3.57 times the earth radius?</h3>

M= 5.98×10²⁴ kg, r= 3.57× 6.37×10³ km = 22.7×10⁶m
Orbital velocity= √(6.67×10^(-11)×5.98×10²⁴/22.7×10⁶)

=4.2km/s

Thus, we can conclude that the orbiting velocity of the satellite is 4.2km/s.

Learn more about the orbital velocity here:

brainly.com/question/22247460

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8 0

2 years ago

What term best describes the regular path of a spacecraft or other object around a planetary body?

Law Incorporation [45]

B. Orbit. The planets orbit the sun, the moon orbits earth, etc.

7 0

3 years ago