Answer:
The found acceleration in terms of h and t is:
![a=\frac{h}{5(t_1)^2}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bh%7D%7B5%28t_1%29%5E2%7D)
Explanation:
(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)
We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.
<h3>
</h3><h3>
Stage 1</h3>
Constant acceleration, starts from rest.
Distance = ![y = \frac{1}{2}a(t_1)^2](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B1%7D%7B2%7Da%28t_1%29%5E2)
Velocity = ![v_1=at_1](https://tex.z-dn.net/?f=v_1%3Dat_1)
<h3>Stage 2</h3>
Constant velocity where
Velocity = ![v_o=v_1=at_1](https://tex.z-dn.net/?f=v_o%3Dv_1%3Dat_1)
Distance =
<h3>
![y_2=v_2(t_2)\\\text{Where~}t_2=4t_1 ~\text{and}~ v_2=v_1=at_1\\y_2=(at_1)(4t_1)\\y_2=4a(t_1)^2\\](https://tex.z-dn.net/?f=y_2%3Dv_2%28t_2%29%5C%5C%5Ctext%7BWhere~%7Dt_2%3D4t_1%20~%5Ctext%7Band%7D~%20v_2%3Dv_1%3Dat_1%5C%5Cy_2%3D%28at_1%29%284t_1%29%5C%5Cy_2%3D4a%28t_1%29%5E2%5C%5C)
</h3><h3 /><h3>Stage 3</h3>
Constant deceleration where
Velocity = ![v_0=v_1=at_1](https://tex.z-dn.net/?f=v_0%3Dv_1%3Dat_1)
Distance =
![y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2](https://tex.z-dn.net/?f=y_3%3Dv_1t_3-%5Cfrac%7B1%7D%7B2%7Da%28t_3%29%5E2%5C%5C%5Ctext%7BWhere%7D~t_3%3Dt_1%5C%5Cy_3%3Dv_1t_1-%5Cfrac%7B1%7D%7B2%7Da%28t_1%29%5E2%5C%5C%5Ctext%7BWhere%7D~%20v_1t_1%3Da%28t_1%29%5E2%5C%5Cy_3%3Da%28t_1%29%5E2-%5Cfrac%7B1%7D%7B2%7Da%28t_1%29%5E2%5C%5C%5Ctext%7BSubtracting%20both%20terms%3A%7D%5C%5Cy_3%3D%5Cfrac%7B1%7D%7B2%7Da%28t_1%29%5E2)
<h3>Total Height</h3>
Total height = y₁ + y₂ + y₃
Total height = ![\frac{1}{2}a(t_1)^2+4a(t_1)^2+\frac{1}{2}a(t_1)^2 = 5a(t_1)^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Da%28t_1%29%5E2%2B4a%28t_1%29%5E2%2B%5Cfrac%7B1%7D%7B2%7Da%28t_1%29%5E2%20%3D%205a%28t_1%29%5E2)
<h3 /><h3>Acceleration</h3>
Find acceleration by rearranging the found equation of total height.
Total Height = h
h = 5a(t₁)²
![a=\frac{h}{5(t_1)^2}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bh%7D%7B5%28t_1%29%5E2%7D)
Answer:
It will take the plant
days or 4.44 days to grow to a height of 200 inches tall.
Explanation:
From the question, the rate at which the species of the bamboo tree grows is 36 inches per day.
To determine how long it would take a plant 40 inches tall initially to grow at this rate (that is, 36 inches per day) to a height of 200 inches.
This means we will calculate the number of days it will take the plant to grow additional 160 inches ( 200 inches - 40 inches) at this rate.
Now,
If the plant grows 36 inches in 1 day
then it will grow 160 inches in x days
x = (160 inches × 1 day) / 36 inches
x = 160 / 36
x =
days or 4.44 days
Hence, it will take the plant
days or 4.44 days to grow to a height of 200 inches tall.
Answer:east
Explanation:Earth rotates or spins toward the east, and that's why the Sun, Moon, planets, and stars all rise in the east and make their way westward across the sky.
A. B. D. C. D, A, A, C, B, B, D, D
Answer:
![E = 2.84 * 10^5 N/C](https://tex.z-dn.net/?f=E%20%3D%202.84%20%2A%2010%5E5%20N%2FC)
Explanation:
The speed increased from 2.0 * 10^7 m/s to 4.0 * 10^7 m/s over a 1.2 cm distance.
Let us find the acceleration:
![v^2 = u^2 + 2as](https://tex.z-dn.net/?f=v%5E2%20%3D%20u%5E2%20%2B%202as)
![(4.0 * 10^7)^2 = (2.0 * 10^7)^2 + 2 * a * 0.012\\\\(4.0 * 10^7)^2 - (2.0 * 10^7)^2 = 0.024a\\\\1.2 * 10^{15}= 0.024a\\\\a = 1.2 * 10^{15} / 0.024\\\\a = 5 * 10^{16} m/s^2](https://tex.z-dn.net/?f=%284.0%20%2A%2010%5E7%29%5E2%20%3D%20%282.0%20%2A%2010%5E7%29%5E2%20%2B%202%20%2A%20a%20%2A%200.012%5C%5C%5C%5C%284.0%20%2A%2010%5E7%29%5E2%20-%20%282.0%20%2A%2010%5E7%29%5E2%20%3D%200.024a%5C%5C%5C%5C1.2%20%2A%2010%5E%7B15%7D%3D%200.024a%5C%5C%5C%5Ca%20%3D%201.2%20%2A%2010%5E%7B15%7D%20%2F%200.024%5C%5C%5C%5Ca%20%3D%205%20%2A%2010%5E%7B16%7D%20m%2Fs%5E2)
Electric force is given as the product of charge and electric field strength:
F = qE
where q = electric charge
E = Electric field strength
Force is generally given as:
F = ma
where m = mass
a = acceleration
Equating both:
ma = qE
E = ma / q
For an electron:
m = 9.11 × 10^{-31} kg
q = 1.602 × 10^{-19} C
Therefore, the electric field strength of the electron is:
![E = \frac{9.11 * 10^{-31} * 5 * 10^{16}}{1.602 * 10^{-19}} \\\\E = 2.84 * 10^5 N/C](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B9.11%20%2A%2010%5E%7B-31%7D%20%2A%205%20%2A%2010%5E%7B16%7D%7D%7B1.602%20%2A%2010%5E%7B-19%7D%7D%20%5C%5C%5C%5CE%20%3D%202.84%20%2A%2010%5E5%20N%2FC)