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irga5000 [103]
3 years ago
9

Which instrument is launched into the atmosphere to collect pressure, temperature, humidity, wind speed, and other data?

Physics
1 answer:
koban [17]3 years ago
5 0
C. Radiosonde is the answer

the above mentioned is not correct
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A 2N and 6N force pull on an object to the right and a 4N force pulls to the left a 0.5kg object. What is the net force on the o
Angelina_Jolie [31]

Explanation:

F net = 2+6-4 ( 2 and 6 N are in same direction so they get added, 4N in opposite direction so it will be subtracted)

F net=4 N

3 0
2 years ago
A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points
mr Goodwill [35]

Answer:

Approximately 11.0\; \rm m \cdot s^{-1}. (Assuming that g = 9.81 \; \rm N \cdot kg^{-1}, and that the tabletop is level.)

Explanation:

Weight of the book:

W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N.

If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, F(\text{normal force}) \approx 10.202\; \rm N.

As a side note, the F_N and W on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction F(\text{kinetic friction}) on it will be equal to

  • \mu_{\rm k}, the coefficient of kinetic friction, times
  • F(\text{normal force}), the normal force that's acting on it.

That is:

\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}.

Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that 15.0\; \rm N applied force. The net force on the book shall be:

\begin{aligned}& F(\text{net force})  \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}.

Apply Newton's Second Law to find the acceleration of this book:

\displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}.

6 0
2 years ago
Ethan made a diagram to compare examples of the first and second laws of thermodynamics. What belongs in the areas marked X and
bazaltina [42]

Answer:

The answer is X: Thermal energy is converted to light energy

Y: A cold spoon placed in hot liquid gets warmer

Explanation:

I took the quiz

4 0
3 years ago
A car of mass 800kg travels a distance of 40m at constant speed in a duration of 2.0s. The car exerts a forward force of 15kN.
Alex17521 [72]

W = F × s

W = 15kN × 40 m

W = 15.000 N × 40 m

W = 600.000 J

P = W/t

P = 600.000 J/2 s

P = 300.000 Watt

P = 300kWatt

#LearnWithEXO

6 0
2 years ago
Can you help me with this??
s2008m [1.1K]

Answer:

i want to say flip the coins but im not really sure sry

Explanation:

3 0
3 years ago
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