The tension in the string B) It quadruples.
Explanation:
The ball is in uniform circular motion in a horizontal circle, so the tension in the string is providing the centripetal force that keeps the ball in circular motion. So we can write:
where:
T is the tension in the string
m is the mass of the ball
v is the speed of the ball
r is the radius of the circle (the lenght of the string)
In this problem, we are told that the speed of the ball is doubled, so
v' = 2v
Substituting into the previous equation, we find the new tension in the string:
Therefore, the tension in the string will quadruple.
Learn more about circular motion:
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Answer:
Trial 1 is the largest, trial 3 is the smallest
Explanation:
Given:
<em>Trial 1</em>
M₁ = 6·10²² kg
d₁ = 3 500 km = 3.5·10⁶ м
<em>Trial 2</em>
M₂ = 6·10²² kg
d₂ = 7 000 km = 7·10⁶ м
<em>Trial 3</em>
M₃ = 3·10²² kg
d₃ = 7 000 km = 7·10⁶ м
___________
F - ?
Gravitational force:
F₁ = G·m·M₁ / d₁² = m·6.67·10⁻¹¹·6·10²² / (3.5·10⁶)² = 0.37·m (N)
F₂ = G·m·M₂ / d₂² = m·6.67·10⁻¹¹·6·10²² / (7·10⁶)² = 0.08·m (N)
F₃ = G·m·M₃ / d₃² = m·6.67·10⁻¹¹·3·10²² / (7·10⁶)² = 0.04·m (N)
Trial 1 is the largest, trial 3 is the smallest
Density = mass/volume
= 4300/10000
= 0.43 g/mL.
Therefore density of liquid propane is 0.43 g/mL.
Hope this helps!
Answer:
R = 36.885 km
Explanation:
In order to distinguish the two planes we must use the Rayleigh criterion that establishes two distinguishable objects if in their diffraction the central maximum of one coincides with the first minimum of the other
The diffraction equation for slits is
a sin θ = m λ
the first minimum occurs for m = 1
sin θ = λ a
as the diffraction experiments the angles are very small, we approximate
sin θ = θ
θ = λ / a
This expression is for a slit, in the case of circular objects, when solving the system in polar coordinates, a numerical constant appears, leaving the expression of the form
θ = 1.22 λ / a
In this problem they give us the frequency, let's find the wavelength with the relation
c = λ f
λ = c / f
θ = 1.22 c/ f a
since they ask us for the distance between the planes, we can use the definition of radians
θ = s / R
if we assume that the distance is large, we can approximate the arc to the horizontal distance
s = x
we substitute
x / R = 1.22 c / fa
R = x f a / 1.22c
Let's reduce the magnitudes to the SI system
f = 9000 MHz = 9 109 Hz
a = 15 m
x = 100 m
let's calculate
R = 100 10⁹ 15 / (1.22 3 108)
R = 3.6885 10⁴ m
let's reduce to km
R = 3.6885 10¹ km
R = 36.885 km
