Answer:
t = 6.09 seconds
Explanation:
Given that,
Speed, v = 44.1 cm/s
Distance, d = 269 cm
We need to find the time interval of the marble. Speed is distance per unit time.
Hence, the time interval of the marble is 6.09 seconds.
The net force is 270 N
Explanation:
We can solve this problem by using Newton's second law, which states that the net force on an object is equal to the product between its mass and its acceleration:
where
F is the force
m is the mass
a is the acceleration
In this problem, we have
m = 90.0 kg
Substituting, we find the net force on the object:
Learn more about Newton's second law:
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Answer:
The decrease is due to the bulge at the equator (putting more distance between the rest of the planet and the surface
Explanation:
The parcel will undergo projectile motion, which means that it will have motion in both the horizontal and vertical direction.
First, we determine how long the parcel will fall using:
s = ut + 1/2 at²
where s will be the height, u is the initial vertical velocity of the parcel (0), t is the time of fall and a is the acceleration due to gravity.
5.5 = (0)(t) + 1/2 (9.81)(t)²
t = 1.06 seconds
A
The solution would be like
this for this specific problem:
<span>
The force on m is:</span>
<span>
GMm / x^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] ->
1
The force on 2m is:</span>
<span>
GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2]
-> 2
From (1), you’ll get M = 2mx^2 / L^2 and from
(2) you get M = m(L - x)^2 / L^2
Since the Ms are the same, then
2mx^2 / L^2 = m(L - x)^2 / L^2
2x^2 = (L - x)^2
xsqrt2 = L - x
x(1 + sqrt2) = L
x = L / (sqrt2 + 1) From here, we rationalize.
x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1)
x = L(sqrt2 - 1) / (2 - 1)
x = L(sqrt2 - 1) </span>
= 0.414L
<span>Therefore, the third particle should be located the 0.414L x
axis so that the magnitude of the gravitational force on both particle 1 and
particle 2 doubles.</span>
