7. solar flare: f.
8. core: h.
9. chromosphere: b.
10. sunspot: d.
11. corona: c.
12. nuclear fusion: j.
13. photosphere: a.
14. solar wind: g.
15. prominence: e.
16. radiation zone: k.
17. convection zone: i.
Answer:
Because winding roads have a gentle slope on hills, so it's easy to climb it than a steepy.
The ball only accelerates during the brief time that the club is in contact
with it. After it leaves the club face, it takes off at a constant speed.
If it accelerates at 20 m/s² during the hit, then
Force = (mass) x (acceleration) = (0.2kg) x (20 m/s²) = <em>4 newtons</em> .
Answer:
a = 1 m/s² and
Explanation:
The first two parts can be seen in attachment
We use Newton's second law on each axis
Y axis
Ty - W = 0
Ty = w
X axis
Tx = m a
With trigonometry we find the components of tension
Sin θ = Ty / T
Ty = T sin θ
Cos θ = Tx / T
Tx = T cos θ
We calculate the acceleration with kinematics
Vf = Vo + a t
a = (Vf -Vo) / t
a = (20 -10) / 10
a = 1 m/s²
We substitute in Newton's equations
T Sin θ = mg
T cos θ = ma
We divide the two equations
Tan θ = g / a
θ = tan⁻¹ (g / a)
θ = tan⁻¹ (9.8 / 1)
θ = 84º
We see that in the expression of the angle the mass does not appear therefore you should not change the angle
Explanation:
The given data is as follows.
radius (r) = 3.25 cm, ![\alpha = 11.6 rad/s^{2}](https://tex.z-dn.net/?f=%5Calpha%20%3D%2011.6%20rad%2Fs%5E%7B2%7D)
Now, we will calculate the tangential acceleration as follows.
![a_{tangential} = \alpha \times r](https://tex.z-dn.net/?f=a_%7Btangential%7D%20%3D%20%5Calpha%20%5Ctimes%20r)
Putting the given values into the above formula as follows.
![a_{tangential} = \alpha \times r](https://tex.z-dn.net/?f=a_%7Btangential%7D%20%3D%20%5Calpha%20%5Ctimes%20r)
= ![11.6 rad/s^{2} \times 3.25 cm](https://tex.z-dn.net/?f=11.6%20rad%2Fs%5E%7B2%7D%20%5Ctimes%203.25%20cm)
= 37.7 ![rad cm/s^{2}](https://tex.z-dn.net/?f=rad%20cm%2Fs%5E%7B2%7D)
Thus, we can conclude that the tangential acceleration of a point on the rim of the flywheel during this spin-up process is 37.7
.