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Ahat [919]
3 years ago
12

Which of the following terms is given to a pair of stars that appear to change position in the sky, indicating that they are orb

iting one another?A) visual binaryB) eclipsing binaryC) spectroscopic binaryD) double starE) none of the above
Physics
1 answer:
Ad libitum [116K]3 years ago
5 0

Answer:

option A

Explanation:

The correct answer is option A

The binary star system is the system in which two stars are continuously orbiting each other,

In the eclipsing binary system, two stars revolve about there center of mass and in this system one one-star eclipse another star.

Spectroscopic binary stars are found from the observation of radial velocity and the brighter member of such binary can be seen to have continuously changed the wavelength and periodic velocity.

When the pairs of stars appear to change position in the sky then it is known as visual binary.

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An object with a charge of −2.1 μC and a mass of 0.0044 kg experiences an upward electric force, due to a uniform electric field
zheka24 [161]

Answer:

(1) 2.05 x 10^4 N/C

(2) Downward

(3) upward, 9.8 m/s^2

(4) upward, 9.8 m/s^2

Explanation:

q = - 2.1 micro coulomb, m = 0.0044 kg, g = 9.8 m/s^2

(1) The electric force is given by F = - q x E

The magnitude of electric force is balanced by the weight of the charged particle

q x E = m x g

E = mg / q

E = \frac{0.0044 \times 9.8}{2.1 \times 10^{-6}}

E = 2.05 x 10^4 N/C

(2) As the electric force is acting upward and the weight is downward so the elecric field is in downward direction.

(3) The charge is doubled,

then the electric field becomes half.

E = 2.05 x 10^4 / 2 = 1.025 x 10^4 N/C

The direction is same that is in downward direction.

Acceleration = Force / mass

a = mg / m = 9.8 m/s^2 upward

(4) The charge is doubled,

then the electric field becomes half.

E = 2.05 x 10^4 / 2 = 1.025 x 10^4 N/C

The direction is same that is in downward direction.

Acceleration = Force / mass

a = mg / m = 9.8 m/s^2 upward

7 0
3 years ago
Global winds move warm air toward the _____.
Bezzdna [24]

global winds move warm air toward the poles

5 0
3 years ago
Read 2 more answers
A bicyclist starting from rest applies a force of F = 454 N to ride his bicycle across flat ground for a distance of d = 250 m b
frutty [35]

Answer:

1.) 113500J

2.) 237m

Explanation:

Hello!

To solve this exercise follow the following steps, the description and complete process is in the attached image

1. Draw the full sketch of the problem.

2. The work is defined as the product of the trajectory by the force that is parallel to this direction, for this reason to find the work done we multiply the horizontal distance (250m) by the applied force (454N)

3. The potential energy is equal to the product of mass, gravity and height and is equal to the work done by the force applied by the cyclist, of this relationship and using algebra we can find the height that the cyclist climbed

4. We use the sine function to find the diagonal distance using the height and angle of the slope

7 0
3 years ago
A 22.0 nF capacitor is connected across an AC generator that produces a peak voltage of 5.80 V. part a
mariarad [96]

Answer:

Explanation:

Impedence of the circuit = peak voltage / peak current

= 5.8 / 51 x 10⁻³

= 113.725 ohm.

1 / wC =113.725

w = 1 / (113.725 x 22 x 10⁻⁹ )

= 10⁹ / 2.5 x 10³

=10⁶ / 2.5

40 x 10⁴

frequency n = 40 x 10⁴ / 2 x 3.14

6.37 x 10⁴ Hz.

b ) charge on the capacitor = 1 C

V = Q / C

= Charge / capacitor

= 1 / 22 x 10⁻⁹

4.54 x 10⁷ V.

4 0
3 years ago
You are a detective investigating why someone was hit on the head by a falling flowerpot. One piece of evidence is a smartphone
umka21 [38]

Answer:

0.37 m

Explanation:

Given :

Window height, h_1 = 1.27 m

The flowerpot falls 0.84 m off the window height, i.e.

h_2 = (1.27 x 0.84 ) m in a time span of $t=\frac{8}{30}$   seconds.

Assuming that the speed of the pot just above the window is v then,

h_2=ut+\frac{1}{2}gt^2

$(1.27 \times 0.84) = v \times \left( \frac{8}{30} \right) + \frac{1}{2} \times 9.81 \times \left( \frac{8}{30} \right)^2$

$v=\left(\frac{30}{8}\right) \left[ (1.27 \times 0.84) - \left( \frac{1}{2} \times 9.81 \times \left( \frac{8}{30 \right)^2 \right) \right]}$

$v= 2.69$ m/s

Initially the pot was dropped from rest. So,  u = 0.

If it has fallen from a height of h above the window then,

$h = \frac{v^2}{2g}$

$h = \frac{(2.69)^2}{2 \times 9.81}$

h = 0.37 m

3 0
3 years ago
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