Answer:
A) 1.88 * 10^17 m
B) 1.22 * 10^34 J
C) 1.95 * 10^34 J
Explanation:
Parameters given:
Mass of planet = 7.00 * 10^25 kg
Radius of orbit = 6.00 * 10^11 m
Force exerted on planet = 6.51 * 10^22 N
Velocity of planet = 2.36 * 10^4 m/s
A) The distance traveled by the planet is half of the circumference of the orbit (which is circular).
The circumference of the orbit is
C = 2 * pi * R
R = radius of orbit
C = 2 * 3.142 * 6.0 * 10¹¹
C = 3.77 * 10¹² m
Hence, distance traveled will be:
D = 0.5 * 3.77 * 10¹²
D = 1.88 * 10 ¹² m/s
B) Work done is given as:
W = F * D
W = 652 * 10²² * 1.88 * 10¹¹
W = 1.22 * 10³⁴ J
C) Change in Kinetic energy is given as:
K. E. = 0.5 * m * v²
K. E. = 0.5 * 7 * 10^25 * (2.36 * 10^4)²
K. E. = 1.95 * 10³⁴ J
Answer:
1 m = 39.37 in = 39.37/12 ft = 3.28 ft
V = 1145 k/hr = 1145k/hr * 6076 ft/k = 6957020 ft / hr
V = 6957020 ft/hr / 3600 s/hr = 1933 ft/sec
V = 1933 ft/sec / (3.28 ft / m) = 589 m/s
Check:
88 ft/sec = 60 mph
(1145 k/hr * 6076 ft / k) 3600 sec/hr = 1933 ft/sec = 589 m/s
1933 ft/sec / (88 ft/sec) * 60 mph = 1318 mph
Also, 1318 / 1145 = 6076 / 5280 as it should
Answer:
F = 2.30 10⁴ N
Explanation:
The force required to link two gates must be equal to or greater than the electrostatic force of repulsion, because the protons have equal charges.
F = k q₁ q₂ / r²
Where k is the Coulomb constant that is worth 8.99 10⁹ N m² / C²
In this case the proton charge is 1.6 10⁻¹⁹ C and the distance between them is approximately the diameter of the core r = 10⁻¹⁵ m
Let's calculate
F = 8.99 10⁹ (1.6 10⁻¹⁹)² / (10⁻¹⁵)²
F = 2.30 10⁴ N
The bond strength must be equal to or greater than this value
the area bounded by the line and the axes of a velocity-time graph is equal to the displacement of an object during that particular time period
Thank you
It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating).
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
