ANS : 313℃
You need to use K in this.
To convert ℃ to Kelvin (K), add 273.15 to ℃.
Missing question: "What is the spring's constant?"
Solution:
The object of mass m=6.89 kg exerts a force on the spring equal to its weight:
![F=mg=(6.89 kg)(9.81 m/s^2)=67.6 N](https://tex.z-dn.net/?f=F%3Dmg%3D%286.89%20kg%29%289.81%20m%2Fs%5E2%29%3D67.6%20N)
When the object is attached to the spring, the displacement of the spring with respect to its equilibrium position is
![\Delta x=43.2 cm-33.6 cm=9.6 cm=0.096 m](https://tex.z-dn.net/?f=%5CDelta%20x%3D43.2%20cm-33.6%20cm%3D9.6%20cm%3D0.096%20m)
And by using Hook's law, we can find the constant of the spring:
Answer:
This question is asking to identify the following variables:
Independent variable (IV): TYPE OF SOIL
Dependent variable (DV): HEIGHT AND NUMBER OF LEAVES
Control group: None in this experiment
Constant: SAME ROSE PLANT, SAME TIME INTERVAL (1 WEEK)
Explanation:
Independent variable in an experiment is the variable that is manipulated or changed by the experimenter in order to effect a measurable outcome. In this case, the independent variable is the TYPE OF SOIL used.
Dependent variable is the measurable variable that responds to changes made to the independent variable. In this experiment, the dependent variable is the HEIGHT AND NUMBER OF LEAVES of each rose.
Constants or control variable is the variable that is kept unchanged or constant for all groups throughout the experiment. In this experiment, the constants are SAME ROSE PLANT, SAME TIME INTERVAL (1 WEEK).
Control group are the groups that does not receive the experimental treatment. In this case, all the groups received the experimental treatment (different soil types). Hence, there is no control
The answer is 1,600 J.
A work (W) can be expressed as a product of a force (F) and a
distance (d):
W = F · d<span>
We have:
W = ?
F = 20 N = 20 kg*m/s</span>²
d = 80 m
_____
W = 20 kg*m/s² * 80 m
W = 20 * 80 kg*m/s² * m
W = 1600 kg*m²/s²
W = 1600 J
Answer:
m = 14*26 = 364
Explanation:
overall magnification is given as m
![m = m_{o}* m_{e}](https://tex.z-dn.net/?f=m%20%3D%20m_%7Bo%7D%2A%20m_%7Be%7D)
mo magnification of objective lens
me magnification of EYE lens
where mo is given as
![m_{o} = \frac{v_{o}}{-u _{0}}](https://tex.z-dn.net/?f=m_%7Bo%7D%20%3D%20%5Cfrac%7Bv_%7Bo%7D%7D%7B-u%20_%7B0%7D%7D)
and me as
![m_{e} = 1+\frac{D}{f_{e}}](https://tex.z-dn.net/?f=m_%7Be%7D%20%3D%201%2B%5Cfrac%7BD%7D%7Bf_%7Be%7D%7D)
d is distant of distinct vision = 25.0 cm for normal eye
fe = focal length of eye piece
focal length of objective lense is 0.140 cm
we know that
![\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bv_%7B0%7D%7D-%5Cfrac%7B1%7D%7Bu_%7B0%7D%7D%3D%5Cfrac%7B1%7D%7Bf_%7B0%7D%7D)
![\frac{1}{v_{0}} = \frac{1}{u_{0}} + \frac{1}{f_{0}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bv_%7B0%7D%7D%20%3D%20%5Cfrac%7B1%7D%7Bu_%7B0%7D%7D%20%2B%20%5Cfrac%7B1%7D%7Bf_%7B0%7D%7D)
![\frac{1}{v_{0}} = \frac{1}{0.150} + \frac{1}{0.14}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bv_%7B0%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B0.150%7D%20%2B%20%5Cfrac%7B1%7D%7B0.14%7D)
![\frac{1}{v_{o}} = 2.1 cm](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bv_%7Bo%7D%7D%20%3D%202.1%20cm)
![m_{o} = \frac{2.1}{0.150} = 14](https://tex.z-dn.net/?f=m_%7Bo%7D%20%3D%20%5Cfrac%7B2.1%7D%7B0.150%7D%20%3D%2014)
![m_{e} = 1+\frac{25}{1}](https://tex.z-dn.net/?f=m_%7Be%7D%20%3D%201%2B%5Cfrac%7B25%7D%7B1%7D)
![m_{e} =26](https://tex.z-dn.net/?f=m_%7Be%7D%20%3D26)
![m = m_{o}* m_{e}](https://tex.z-dn.net/?f=m%20%3D%20m_%7Bo%7D%2A%20m_%7Be%7D)
m = 14*26 = 364