Answer:
0.05 m
Explanation:
From the question given above, the following data were obtained:
Mass of first object (M1) = 9900 kg
Gravitational force (F) = 12 N
Mass of second object (M2) = 52000 kg
Distance apart (r) =?
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Thus, we can obtain the distance between the two objects as shown below:
F = GM1M2/r²
12 = 6.67×10¯¹¹ × 9900 × 52000 /r²
Cross multiply
12 × r² = 6.67×10¯¹¹ × 9900 × 52000
Divide both side by 12
r² = (6.67×10¯¹¹ × 9900 × 52000)/12
Take the square root of both side
r = √[(6.67×10¯¹¹ × 9900 × 52000)/12]
r = 0.05 m
Therefore, the distance between the two objects is 0.05 m
Answer:
Answer is C. Both technicians A and B.
Refer below.
Explanation:
Two technicians are discussing the testing of a catalytic converter. Technician A says that a vacuum gauge can be used and observed to see if the vacuum drops with the engine at 2500 RPM for 30 seconds. Technician B says that a pressure gauge can be used to check for backpressure. The following technician is correct:
Both technicians A and B
Answer:
* The first thing we observe is that the frequency response does not change
* The current that circulates in the circuit decreases due to the new resistance at the resonance point,
Z = R + R₂
Explanation:
The impedance of a series circuit is
Z₀² = R² + (X_L-X_C) ²
when we place another resistor in series the initial resistance impedance changes to
Z² = (R + R₂) ² + (X_L - X_C) ²
let's analyze this expression
* The first thing we observe is that the frequency response does not change
* The current that circulates in the circuit decreases due to the new resistance at the resonance point,
Z = R + R₂
There are lots of variables that directly and indirectly contribute to the presence of gas on a surface
if the size of a planet is relatively small it will in turn be that of a smaller area which results in the less area to be covered for gas which basically means higher presence
I can go in depth more but I don't think that would be necessary all you need to know is this ...based on the size and gas will in turn be parallel to it's conformity
<h2>
Answer: U-238</h2>
Explanation:
Let's begin by explaining that for radioactive geological dating (also called radioisotope dating) in which radioactive impurities were selectively incorporated when the fossil materials were formed, it is very useful to compare it with a naturally occurring radioisotope having a known half-life.
Now, taking into account that the <u>fossils are millions and millions of years old, radioisotopes are needed that exceed this measure.
</u>
To understand it better:
The longer the half-life of a radioisotope, the greater its utility for estimating fossil ages or geological formations.
In this sense, uranium-238 (U238) has a half-life of 4,470 million years, therefore, it is among the most commonly used radioisotopes for fossil and geological dating.
