Answer:
To find the circumference (orbit) of an object, you use Pi x Diameter.
As you have the circumference of B, you divide it by Pi to get the Diameter.
So 120 divided by 3.141592654 = 38.2 minutes for the Diameter.
As' radius and Diameter will be 3x greater than B.
38.2 x 3 = 114.6
To get back to the orbital period, times 114.6 by Pi, and you will get 360 minutes
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Answer:
To convert m/sec into km/hr, multiply the number by 18 and then divide it by 5.
Explanation:
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Answer:
Explanation:
Let the charge particle have charge equal to +q .
force due to electric field will be along the field that is along y - axis . To balance it force by magnetic force must be along - y axis. ( negative of y axis )
force due to magnetic field = q ( v x B ) , v is velocity and B is magnetic field.
F = q ( v i x B k ) , ( velocity is along x direction and magnetic field is along z axis. )
= (Bqv) - j
= - Bqv j
The force will be along - ve y - direction .
If we take charge as negative or - q
force due to electric field will be along - y axis .
magnetic force = F = -q ( v i x B k )
= + Bqv j
magnetic force will be along + y axis
So it is difficult to find out the nature of charge on the particle from this experiment.
Answer:
the answer choice is B
Explanation:
as we move from left to right , the atomic size decreases due to higher number of protons in the nucleus, which are able to attract the electrons more strongly. and so the electronegativity and electron affinity increases for the same reason. the nuclear charge increase due to more protons , and without an increase in inner electrons , there is less shielding effect. so effective nuclear charge increases.
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