Answer:
1.61 second
Explanation:
Angle of projection, θ = 53°
maximum height, H = 7.8 m
Let T be the time taken by the ball to travel into air. It is called time of flight.
Let u be the velocity of projection.
The formula for maximum height is given by
![H = \frac{u^{2}Sin^{2}\theta }{2g}](https://tex.z-dn.net/?f=H%20%3D%20%5Cfrac%7Bu%5E%7B2%7DSin%5E%7B2%7D%5Ctheta%20%7D%7B2g%7D)
By substituting the values, we get
![7.8= \frac{u^{2}Sin^{2}53 }{2\times 9.8}](https://tex.z-dn.net/?f=7.8%3D%20%5Cfrac%7Bu%5E%7B2%7DSin%5E%7B2%7D53%20%7D%7B2%5Ctimes%209.8%7D)
u = 9.88 m/s
Use the formula for time of flight
![T = \frac{2uSin\theta }{g}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2uSin%5Ctheta%20%7D%7Bg%7D)
![T = \frac{2\times 9.88\times Sin53 }{9.8}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2%5Ctimes%209.88%5Ctimes%20Sin53%20%7D%7B9.8%7D)
T = 1.61 second
Think of it this way: If you test only one variable then you know that the difference in the experimental and control setup is that one independent variable. If you test more than one you will not know which one made the difference. Hope this helps :)
Answer: No
Can the change in cyclin concentration during mitosis be explained by the fact that the cell divides in two and thus divides the material in the cell into two smaller volumes?
Explanation: The cyclin concentration is not halved but degraded during mitosis.
There is an increase in cyclin concentrations at interphase. These changes are caused by the presence of Cyclin Dependent Kinase (CDK) complexes. CDK being a substrate of cyclin catalyses cyclin, thereby increasing its concentration. During mitosis, cyclins are destroyed, signifying the end of mitosis and cytokinesis.
Without this process, it will be impossible for the cell to exit mitosis.
Answer
given,
diameter of planet = 1.8 x 10⁷ m
radius of planet = 0.9 x 10⁷ m
time period = 22.3 hours
the planet orbits 2.2 x 10¹¹ m period of 402 earth days.
acceleration= 12.2 m/s²
we know
![g = \dfrac{GM_{p}}{r^2}](https://tex.z-dn.net/?f=g%20%3D%20%5Cdfrac%7BGM_%7Bp%7D%7D%7Br%5E2%7D)
![M_{p} = \dfrac{gr^2}{G}](https://tex.z-dn.net/?f=M_%7Bp%7D%20%3D%20%5Cdfrac%7Bgr%5E2%7D%7BG%7D)
![M_{p} = \dfrac{12.2 \times (0.9 \times 10^7)^2}{6.67 \times 10^{-11}}](https://tex.z-dn.net/?f=M_%7Bp%7D%20%3D%20%5Cdfrac%7B12.2%20%5Ctimes%20%280.9%20%5Ctimes%2010%5E7%29%5E2%7D%7B6.67%20%5Ctimes%2010%5E%7B-11%7D%7D)
M_p = 1.48 x 10²⁵ Kg
b) Formula to calculate the mass of star
![M_{s} = \dfrac{4\pi^2}{GT_s^2}(R^3)](https://tex.z-dn.net/?f=M_%7Bs%7D%20%3D%20%5Cdfrac%7B4%5Cpi%5E2%7D%7BGT_s%5E2%7D%28R%5E3%29)
![M_{s} = \dfrac{4\pi^2}{6.67 \times 10^{-11} \times (402 \times 86400)^2}((2.2 \times 10^{11})^3)](https://tex.z-dn.net/?f=M_%7Bs%7D%20%3D%20%5Cdfrac%7B4%5Cpi%5E2%7D%7B6.67%20%5Ctimes%2010%5E%7B-11%7D%20%5Ctimes%20%28402%20%5Ctimes%2086400%29%5E2%7D%28%282.2%20%5Ctimes%2010%5E%7B11%7D%29%5E3%29)
M_s = 5.22 x 10³³ Kg