There is no "why", because that's not what happens. The truth is
exactly the opposite.
Whatever the weight of a solid object is in air, that weight will appear
to be LESS when the object is immersed in water.
The object is lifted by a force equal to the weight of the fluid it displaces.
It displaces the same amount of air or water, and any amount of water
weighs more than the same amount of air. So the force that lifts the
object in water is greater than the force that lifts it in air, and the object
appears to weigh less in the water.
Answer:
No. 67
Peter Street
12th Road
Chennai
24th June 201_
Dear Amrish
I have come to know that since your school has closed for the Autumn Break you have plenty of free time at your disposal at the moment. I would like to tell you that even I am having holidays now.
It has been a long time since we have spent some time together. If you are free, I would welcome to have your company this weekend. Why don’t you come over to my house and spend a day or so with me?
I am anxiously waiting for your reply.
Yours affectionately
your name
Answer:
Option A
You need a Angle C congruent to angle F
Explanation:
EX) Side angle Side = sas
Answer:
(a) ![a_s=0.62\frac{m}{s^2}](https://tex.z-dn.net/?f=a_s%3D0.62%5Cfrac%7Bm%7D%7Bs%5E2%7D)
(b) ![a_s=0.13\frac{m}{s^2}](https://tex.z-dn.net/?f=a_s%3D0.13%5Cfrac%7Bm%7D%7Bs%5E2%7D)
(c) ![x_f=2.6m](https://tex.z-dn.net/?f=x_f%3D2.6m)
Explanation:
(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.
![a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}](https://tex.z-dn.net/?f=a_s%3D%5Cfrac%7BF%7D%7Bm_s%7D%5C%5Ca_s%3D%5Cfrac%7B5.2N%7D%7B8.4kg%7D%5C%5Ca_s%3D0.62%5Cfrac%7Bm%7D%7Bs%5E2%7D)
(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:
![a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}](https://tex.z-dn.net/?f=a_g%3D%5Cfrac%7BF%7D%7Bm_g%7D%5C%5Ca_g%3D%5Cfrac%7B5.2N%7D%7B40kg%7D%5C%5Ca_g%3D0.13%5Cfrac%7Bm%7D%7Bs%5E2%7D)
(c) Using the kinematics equation:
![x_f=x_0+v_0t \pm \frac{at^2}{2}](https://tex.z-dn.net/?f=x_f%3Dx_0%2Bv_0t%20%5Cpm%20%20%5Cfrac%7Bat%5E2%7D%7B2%7D)
For the girl, we have
and
. So:
![x_f_g=\frac{a_gt^2}{2}(1)](https://tex.z-dn.net/?f=x_f_g%3D%5Cfrac%7Ba_gt%5E2%7D%7B2%7D%281%29)
For the sled, we have
. So:
![x_f_s=x_0_s-\frac{a_st^2}{2}(2)](https://tex.z-dn.net/?f=x_f_s%3Dx_0_s-%5Cfrac%7Ba_st%5E2%7D%7B2%7D%282%29)
When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:
![x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s](https://tex.z-dn.net/?f=x_0_s-%5Cfrac%7Ba_st%5E2%7D%7B2%7D%3D%5Cfrac%7Ba_st%5E2%7D%7B2%7D%5C%5Ct%5E2%28a_g%2Ba_s%29%3D2x_0_s%5C%5Ct%3D%5Csqrt%7B%5Cfrac%7B2x_s_0%7D%7Ba_g%2Ba_s%7D%7D%5C%5Ct%3D%5Csqrt%7B%5Cfrac%7B2%2815m%29%7D%7B0.13%5Cfrac%7Bm%7D%7Bs%5E2%7D%2B0.62%5Cfrac%7Bm%7D%7Bs%5E2%7D%7D%7D%5C%5Ct%3D6.32s)
Now, we solve (1) for ![x_f_g](https://tex.z-dn.net/?f=x_f_g)
![x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m](https://tex.z-dn.net/?f=x_f_g%3D%5Cfrac%7B0.13%5Cfrac%7Bm%7D%7Bs%5E2%7D%286.32s%29%5E2%7D%7B2%7D%5C%5Cx_f_g%3D2.6m%5C%5Cx_f%3D2.6m)