What does a physical map show?
the names of countries, states, and cities
the history of an area
the geographical features of an area
the rest stops and restaurants in an area
15:) using more force in your muscle will increase the force used to bounce the basketball
16:) the pulling of gravity livitation does not allow the ball to go back up with the hieght it was dropped from on the scientifical drop point
14:) <span>a weight hung from a fixed point so that it can swing freely backward and forward, especially a rod with a weight at the end that regulates the mechanism of a clock that is the deffinition of to which of the word pendulum read it do not plagarize and i hope ii helped and have a great day bye.)::</span>
Answer:
r = 4.24x10⁴ km.
Explanation:
To find the radius of such an orbit we need to use Kepler's third law:
<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km. </em>
From equation (1), r₁ is:
![r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}](https://tex.z-dn.net/?f=%20r_%7B1%7D%20%3D%20r_%7B2%7D%20%5Csqrt%5B3%5D%20%7B%28%5Cfrac%7BT_%7B1%7D%7D%7BT_%7B2%7D%7D%29%5E%7B2%7D%7D%20)
![r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}](https://tex.z-dn.net/?f=%20r_%7B1%7D%20%3D%203.84%5Ccdot%2010%5E%7B5%7D%20km%20%5Csqrt%5B3%5D%20%7B%28%5Cfrac%7B1%20d%7D%7B0.07481%20y%20%5Ccdot%20%5Cfrac%7B365%20d%7D%7B1%20y%7D%7D%29%5E%7B2%7D%7D%20)
Therefore, the radius of such an orbit is 4.24x10⁴ km.
I hope it helps you!
Explanation:
12) q = mCΔT
125,600 J = (500 g) (4.184 J/g/K) (T − 22°C)
T = 82.0°C
13) Solving for ΔT:
ΔT = q / (mC)
a) ΔT = 1 kJ / (0.4 kg × 0.45 kJ/kg/K) = 5.56°C
b) ΔT = 2 kJ / (0.4 kg × 0.45 kJ/kg/K) = 11.1°C
c) ΔT = 2 kJ / (0.8 kg × 0.45 kJ/kg/K) = 5.56°C
d) ΔT = 1 kJ / (0.4 kg × 0.90 kJ/kg/K) = 2.78°C
e) ΔT = 2 kJ / (0.4 kg × 0.90 kJ/kg/K) = 5.56°C
f) ΔT = 2 kJ / (0.8 kg × 0.90 kJ/kg/K) = 2.78°C
14) q = mCΔT
q = (2000 mL × 1 g/mL) (4.184 J/g/K) (80°C − 20°C)
q = 502,000 J
20) q = mCΔT
q = (2000 g) (4.184 J/g/K) (100°C − 15°C) + (400 g) (0.9 J/g/K) (100°C − 15°C)
q = 742,000 J
24) q = mCΔT
q = (0.10 g) (0.14 J/g/K) (8.5°C − 15°C)
q = -0.091 J
