Answer:
open cluster. globular cluster. eclipsing binary
Answer:
1500 milliradians
Explanation:
Data provided in the question:
1.5 radians
Now,
1 radians consists of 1000 milliradians
1 milli = 1000
thus for the 1.5 radians, we have
1.5 radians = 1.5 multiplied by 1000 milliradians
or
1.5 radians = 1500 milliradians
Hence, after the conversion
1.5 radians equals to the value 1500 milliradians
Answer: Symbol A
Explanation:
The four symbols described here represent:
- Symbol A shows two dots and a line draw from one not connected to the other. --> this is an open switch. A switch is component of a circuit that is used to open/close the circuit in order to interrupt/allow the flow of current through the circuit. In this case, the switch is open, since the line does not connect the second dot.
- Symbol B shows two dots and a line draw from one connected to the other. --> this is the symbol used to represent the switch when it is closed, so it is a closed switch.
- Symbol C shows vertical lines in the pattern long, short, long, and short with a plus and minus symbol on it. --> this symbol represents a battery, which consists of two or more cells and provides the electromotive force that pushes the electrons along the circuit.
Therefore, the correct symbol representing the open switch is
Symbol A
We would just get three times heavier.
weight = mass x gravity (9.8 m/s/s)
Answer:
ee that the lens with the shortest focal length has a smaller object
Explanation:
For this exercise we use the constructor equation or Gaussian equation
where f is the focal length, p and q are the distance to the object and the image respectively.
Magnification a lens system is
m = 
= - 
h ’= -\frac{h q}{p}
In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞
Let's calculate the distance to the image for each lens
f = 6.0 cm
as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf
q = f = 6.0 cm
for the lens of f = 12.0 cm q = 12.0 cn
to find the size of the image we use
h ’= h q / p
where p has a high value and is the same for all systems
h ’= h / p q
Thus
f = 6 cm h ’= fo 6 cm
f = 12 cm h ’= fo 12 cm
therefore we see that the lens with the shortest focal length has a smaller object
