If I can't open the lid of a jelly jar, I'd keep trying and if I can't open the lid of a jelly jar after the MANY tries I took, I'd ask for help.
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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Yes, you're correct. If you put all of them on the K scale, then 0C is at 273, 0F is at 255, and 0K is at zero.
Answer:
a) 200m, 100m/s
b) 710.20m
c) -117.98 m/s
d) 26.24 s
Explanation:
To solve this we have to use the formulas corresponding to a uniformly accelerated motion problem:
(1)
(2)
(3)
where:
Vo is initial velocity
Xo=intial position
V=final velocity
X=displacement
a)

the intial position is zero because is lauched from the ground and the intial velocitiy is zero because it started from rest.



b)
The intial velocity is 100m/s we know that because question (a) the acceleration is -9.8
because it is going downward.

c)
In order to find the velocity when it crashes, we can use the formula (3).
the initial velocity is 0 because in that moment is starting to fall.

the minus sign means that the object is going down.
d)
We can find the total amount of time adding the first 4 second and the time it takes to going down.
to calculate the time we can use the formula (2) setting the reference at 200m:

solving this we have: time taken= 22.24 seconds
total time is:
total=22.24+4=26.24 seconds.
Answer:
there is not enough information