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3 years ago

A pendulum is in a spacecraft to measure

Physics

1 answer:

Tcecarenko [31]3 years ago

4 0

Answer:

For small angles (✓ < 30 degrees), the period of a simple pendulum4can be approximated by: 3A radian is an angle measure based upon the circumference of a circle C =2⇡r where r is the radius of the circle. A complete circle (360 ) is said to have 2⇡ radians. Therefore, a 1/4 circle (90 )is⇡/2 radians.

Explanation:

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Which sentence correctly uses a comma to separate coordinate adjectives?

lubasha [3.4K]

Answer:

Explanation:

Coordinate adjectives are adjectives that describe the same noun equally. They should be separated with commas. In this sentence, both red and round are describing the ball. The correct answer is, "The red, round ball is Adley's favorite toy to play with."

6 0

2 years ago

Uranus and neptune have methane clouds but jupiter and saturn do not. Which factor explains why?.

mr Goodwill [35]

Answer:

Temperature on Jupiter and Saturn are too high for methane to condense.

Explanation:

However, methane can condense on Uranus and Neptune because they are farther from the sun and hence colder.

4 0

2 years ago

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

morpeh [17]

Answer:

63.750KeV

Explanation:

We are given that

Initial velocity of second electron, $u_2=0$

Radius, $r_1=0$

$r_2=2.3 cm=\frac{2.3}{100}=0.023 m$

1 m=100 cm

Magnetic field,B=0.0370 T

We have to determine the energy of the incident electron.

Mass of electron, $m=9.1\times 10^{-31} kg$

Charge on an electron, $q=-1.6\times 10^{-19} C$

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

Using the formula

Speed of electron, $v_1=\frac{Bqr_1}{m}=\frac{0.0370\times 1.6\times 10^{-19}\times 0}{9.1\times 10^{-31}}=0$

Speed of second electron, $v_2=\frac{0.0370\times 1.6\times 10^{-19}\times 0.023}{9.1\times 10^{-31}}$

$v_2=1.5\times 10^8 m/s$

Kinetic energy of incident electron= $\frac{1}{2}mv^2_1+\frac{1}{2}mv^2_2$

Kinetic energy of incident electron= $0+\frac{1}{2}(9.1\times 10^{-31})(1.5\times 10^8)^2=1.02\times 10^{-14} J$

Kinetic energy of incident electron= $\frac{1.02\times 10^{-14}}{1.6\times 10^{-19}}=63750eV=\frac{63750}{1000}=63.750KeV$

1KeV=1000eV

3 0

3 years ago

PLEASE HELP ME

Snowcat [4.5K]

Answer:

Compound.

Explanation:

6 0

4 years ago

Read 2 more answers

An electron is moving east in a uniform electric field of 1.47 {\rm N/C} directed to the west. At point A, the velocity of the e

zimovet [89]

Answer:

a) $v = 6.25\times 10^{5} m/s$

b) $v = 1.73\times 10^{4} m/s$

Explanation:

Given data:

Electric field = 1.47 N/C

velocity of electron is $4.55\times 10^5 m/s$

distance of point b from point A is 0.55 m

we know that acceleration of particle is given as

a) for electron

$a =\frac{q E}{m}$

$a = \frac{1.6\times 10^{-19} \times 1.47}{9.1\times 10^{-31}}$

$a = 2.58\times 10^{11} m/s^2$

from equation of motion we have

$v^2 = u^2 + 2aS$

$= 20.7025 \times 10^{10} + 2\times 2.58\times 10^{11} \times 0.355$

$v = 6.25\times 10^{5} m/s$

b) for proton

$a = \frac{1.6\times 10^{-19} \times -1.47}{1.6\times 10^{-27}}$

$a = -1.41\times 10^{8} m/s^2$

from equation of motion we have

$v^2 = u^2 + 2aS$

$= 3.8025 \times 10^{8} - 2\times 1.41\times 10^{8} \times 0.355$

$v = 1.73\times 10^{4} m/s$

3 0

3 years ago