<span>10.3 cm
The wavelength will be the distance that light travels in 1 second divided by the frequency of the radiation. Since the over operates at 2.60 ghz, the frequence is 2.6 billion times per second, or 2.60 x 10^9. The speed of light is defined as 299792458 m/s exactly. So
299792458 m/s / 2.60 x 10^9 1/s = 0.10337671 m = 10.337671 cm
Since we only have 3 significant digits, the answer rounds to 10.3 cm</span>
Answer: They are in the same group because they have similar chemical properties, but they are in different periods because they have very different atomic numbers.
Explanation: On Edgenuity!!
<span>5.98 x 10^-2 ohms.
Resistance is defined as:
R = rl/A
where
R = resistance in ohms
r = resistivity (given as 1.59x10^-8)
l = length of wire.
A = Cross sectional area of wire.
So plugging into the formula, the known values, including the area of a circle being pi*r^2, gives:
R = 1.59x10^-8 * 3.00 / (pi * (5.04 x 10^-4)^2)
R = (4.77 x 10^-8) / (pi * 2.54016 x 10 ^-7)
R = (4.77 x 10^-8) / (7.98015 x 10^-7)
R = 5.98 x 10^-2 ohms
So that wire has a resistance of 5.98 x 10^-2 ohms.</span>
Energy is calculated as power*time, so give the wattage of 1200 W (equivalent to 1200 Joules/second) and time of 30 seconds, multiplying these gives 36000 J or 36 kJ of electrical energy.
If electrical charge or current is needed: Power = voltage * current, so given the power of 1200 watts and voltage of 120 V, current is 1200 W / 120 V = 10 Amperes. Charge is calculated by multiplying 10 A*30 s = 300 C.
Answer:
83%
Explanation:
On the surface, the weight is:
W = GMm / R²
where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.
In orbit, the weight is:
w = GMm / (R+h)²
where h is the height of the shuttle above the surface of the Earth.
The ratio is:
w/W = R² / (R+h)²
w/W = (R / (R+h))²
Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:
w/W = (6.4×10⁶ / 7.03×10⁶)²
w/W = 0.83
The shuttle in orbit retains 83% of its weight on Earth.
