Answer:
Half as large.
Explanation:
Using Newton's law of universal gravitation, if the mass of the planet is <em>M</em> and of the Moons 1 and 2 is <em>m</em>, them the force exerted by the planet on them will be:
Which clearly shows that the force that the planet exerts on the Moon 2 is half the force it exerts on the Moon 1.
Lets say sphere 1 has a charge of 12 + and sphere 2 has a charge of 0 +. After they are touched Sphere 1 becomes 6 + and sphere 2 6 +. So 6 - 12 = a change of 6 -, while 6 - 0 = a change of 6 + Therfore,
Answer: The sign of the charge change / transfered are opposites.
Answer:
3.036×10⁻¹⁰ N
Explanation:
From newton's law of universal gravitation,
F = Gm1m2/r² .............................. Equation 1
Where F = Gravitational force between the balls, m1 = mass of the first ball, m2 = mass of the second ball, r = distance between their centers.
G = gravitational constant
Given: m1 = 7.9 kg, m2 = 6.1 kg, r = 2.0 m, G = 6.67×10⁻¹¹ Nm²/C²
Substituting into equation 1
F = 6.67×10⁻¹¹×7.9×6.1/2²
F = 321.427×10⁻¹¹/4
F = 30.36×10⁻¹¹
F = 3.036×10⁻¹⁰ N
Hence the force between the balls = 3.036×10⁻¹⁰ N
Answer:
a = 2.72 [m/s2]
Explanation:
To solve this problem we must use the following kinematics equation:
where:
Vf = final velocity = 1200 [km/h]
Vo = initial velocity = 25 [km/h]
t = time = 2 [min] = 2/60 = 0.0333 [h]
1200 = 25 + (a*0.0333)
a = 35250.35 [km/h2]
if we convert these units to units of meters per second squared
![35250.35[\frac{km}{h^{2} }]*(\frac{1}{3600^{2} })*[\frac{h^{2} }{s^{2} } ]*(\frac{1000}{1} )*[\frac{m}{km} ] = 2.72 [\frac{m}{s^{2} } ]](https://tex.z-dn.net/?f=35250.35%5B%5Cfrac%7Bkm%7D%7Bh%5E%7B2%7D%20%7D%5D%2A%28%5Cfrac%7B1%7D%7B3600%5E%7B2%7D%20%7D%29%2A%5B%5Cfrac%7Bh%5E%7B2%7D%20%7D%7Bs%5E%7B2%7D%20%7D%20%5D%2A%28%5Cfrac%7B1000%7D%7B1%7D%20%29%2A%5B%5Cfrac%7Bm%7D%7Bkm%7D%20%5D%20%3D%202.72%20%5B%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D%20%5D)
B4 the tackle:
<span>The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north </span>
<span>and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east </span>
<span>After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle </span>
<span>The vector triangle is right angled: </span>
<span>magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s </span>
<span>so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] </span>
<span>v(f) = 5.6 m/s (to 2 sig figs) </span>
<span>direction of v(f) is the same as the direction of the final momentum </span>
<span>so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) </span>
<span>so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E </span>
<span>btw ... The direction can be given heaps of different ways ... N 31° E is probably the easiest way to express it when using the vector triangle to find it</span>
