The specific gravity is how the density of the object compares to the density of water. Water's density is 1gram per milliliter. We just need to figure out the density of the object.
The object is .8 kg and it displaces 500mL of water, so the density is the mass divided by the volume. Since the density of water is given in grams, we have to convert the objects mass from kg to g and then we can get the density.
.8kg * 1000g/kg = 800 grams
So
800g/500ml = 1.6grams/mL this is the density.
So divide the density of your object by the density of water, which is 1g/mL, you get 1.6 as the specific gravity. This means the object is 1.6 times more dense than water.
Answer:
Mintu should have chosen material with lower specific heat.
Hope this is useful!! <3
Answer:-50.005 kJ
Explanation:
Given
mass of system =10 kg
work done=0.147 kJ/kg
Change in elevation![(\Delta h)=-50 m](https://tex.z-dn.net/?f=%28%5CDelta%20h%29%3D-50%20m)
initial velocity ![(v_1)=15 m/s](https://tex.z-dn.net/?f=%28v_1%29%3D15%20m%2Fs)
Final Velocity![(v_2)=30 m/s](https://tex.z-dn.net/?f=%28v_2%29%3D30%20m%2Fs)
Specific internal Energy![(\Delta U)=-5 kJ/kg](https://tex.z-dn.net/?f=%28%5CDelta%20U%29%3D-5%20kJ%2Fkg)
from first Law of thermodynamics
![Q-W=\Delta H](https://tex.z-dn.net/?f=Q-W%3D%5CDelta%20H)
![\Delta H=\Delta KE+ \Delta PE +\Delta U](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5CDelta%20KE%2B%20%5CDelta%20PE%20%2B%5CDelta%20U)
where KE= kinetic energy
PE=potential energy
U=internal Energy
![\Delta H=\frac{m(v_2^2-v_1^2)}{2}+mg(\Delta h)+\Delta U](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Cfrac%7Bm%28v_2%5E2-v_1%5E2%29%7D%7B2%7D%2Bmg%28%5CDelta%20h%29%2B%5CDelta%20U)
![Q=W+\Delta KE+ \Delta PE +\Delta U](https://tex.z-dn.net/?f=Q%3DW%2B%5CDelta%20KE%2B%20%5CDelta%20PE%20%2B%5CDelta%20U)
![Q=0.147\times 10+\frac{10\cdot (30^2-15^2)}{2}+10\cdot 9.81\cdot (-5)-5\times 10](https://tex.z-dn.net/?f=Q%3D0.147%5Ctimes%2010%2B%5Cfrac%7B10%5Ccdot%20%2830%5E2-15%5E2%29%7D%7B2%7D%2B10%5Ccdot%209.81%5Ccdot%20%28-5%29-5%5Ctimes%2010)
Q=1.47+3.375-4.850-50
Q=-50.005 kJ
<h2><u>Question</u><u>:</u><u>-</u></h2>
Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the work done by Ryan?
<h2><u>Answer:</u><u>-</u></h2>
<h3>Given,</h3>
=> Force applied by Ryan = 10N
=> Distance covered by the book after applying force = 30 cm
<h3>And,</h3>
30 cm = 0.3 m (distance)
<h3>So,</h3>
=> Work done = Force × Distance
=> 10 × 0.3
=> 3 Joules
![\small \boxed{work \: done \: by \: Ryan \: = 3 \: Joules}](https://tex.z-dn.net/?f=%20%5Csmall%20%5Cboxed%7Bwork%20%5C%3A%20done%20%5C%3A%20%20by%20%5C%3A%20Ryan%20%5C%3A%20%20%3D%203%20%5C%3A%20Joules%7D)
Some factors that caused the shape and structure of the earth are earthquakes and erosion.