The tension in the string holding the tassel and the vertical will the tension in the string
<h3>What is the tension in the string holding the tassel. ?</h3>
Generally, the equation for Tension is mathematically given as
Therefore
T = 0.1953 N
b).
Where
a = 1.13 m/s^2
In conclusion
T* sinФ = ma
2msinФ = ma
2sinФ = a
Ф = 34.4 °
In conclusion, The tension in the string holding the tassel and the vertical will the tension in the string
T = 0.1953 N
Ф = 34.4 °
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Let us say that:
1 = 1st player notation
2 = 2nd player notation (the opponent)
a. First let us establish the distance travelled by the 2nd
player:
d2 = 13 m/s * (t + 1.5)
d2 = 13 t + 19.5
Then the distance of the 1st player:
d1 = v0 t + 0.5 a t^2 (v0
initial velocity = 0 since he started from rest)
d1 = 0.5 * 4 m/s^2 * t^2
d1 = 2 t^2
The two distances must be equal, d1 = d2:
2 t^2 = 13 t + 19.5
t^2 – 6.5 t = 9.75
Completing the square:
(t – 3.25)^2 = 9.75 + (- 3.25)^2
t – 3.25 = ±4.5
t = -1.25, 7.75
Since time cannot be negative, therefore:
t = 7.75 seconds
So he catches his opponent after 7.75 seconds.
b. Using the equation:
d1 = 2 t^2
d1 = 2 * (7.75)^2
d1 = 120.125 m
So he travelled about 120.125 meters when he catches up
to his opponent.
Answer:
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Explanation:
Using Kepler's 3rd law which is: T² = 4π²r³ / GM
Solved for r :
r = [GMT² / 4π²]⅓
Where G is the universal gravitational constant,M is the mass of the sun,T is the asteroid's period in seconds, andr is the radius of the orbit.
Change 5.00 years to seconds :
5.00years = 5.00years(365days/year)(24.0hours/day)(6... = 1.58 x 10^8s
The radius of the orbit then is computed:
r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.58 x 10^8s)² / 4π²]⅓ = 4.38 x 10^11m
They are all units of measure of length
Explanation:
Length is a scalar quantity representing a distance between two points, and it can be expressed in different units.
The SI units of the length is the metre (m), which is defined as the length of the path travelled by light in vacuum during a time interval of 1/299,792,458 of a second.
A unit which is common used is a multiple of the meter, the kilometre (km) which corresponds to 1000 metres:
1 km = 1000 m
Another unit used in the UK system is the mile (mi), where the conversion factor between miles and metres is
1 mi = 1609.34 m
Finally, these units are not suitable to be used to measure astronomical distances - such as those between stars and galaxies. For this, another unit is used, which is the light-year (ly), which corresponds to the distance travelled by the light in a vacuum in one year, and its conversion factor to the metre is:
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