Answer:
= 15.57 N
= 2.60 N
= 16.98 N
The mass of the bag is the same on the three planets. m=1.59 kg
Explanation:
The weight of the sugar bag on Earth is:
g=9.81 m/s²
m=3.50 lb=1.59 kg
=m·g=1.59 kg×9.81 m/s²= 15.57 N
The weight of the sugar bag on the Moon is:
g=9.81 m/s²÷6= 1.635 m/s²
=m·g=1.59 kg× 1.635 m/s²= 2.60 N
The weight of the sugar bag on the Uranus is:
g=9.81 m/s²×1.09=10.69 m/s²
=m·g=1.59 kg×10.69 m/s²= 16.98 N
The mass of the bag is the same on the three planets. m=1.59 kg
Answer:
W = 1.06 MJ
Explanation:
- We will use differential calculus to solve this problem.
- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.
- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.
- Now develop and expression of Force required:
F = p*V*g
F = 1000*(2*0.5*x*8*dx)*g
F = 78480*x*dx
- Now, the work done is given by:
W = F.s
- Where, s is the distance from top of hose to the differential volume:
s = (5 - x)
- We have the work as follows:
dW = 78400*x*(5-x)dx
- Now integrate the following express from 0 to 3 till the tank is empty:
W = 78400*(2.5*x^2 - (1/3)*x^3)
W = 78400*(2.5*3^2 - (1/3)*3^3)
W = 78400*13.5 = 1058400 J
The freezing point of the water is 0 C , and it equals to 273 K
Then, To convert from Kelvins degrees to Celsius degrees we use the relation
Also,
Answer:
Explanation:
The rotated angle is given by:
Since this is a quadratic equation it can be solved using:
Rewriting our equation:
Since 
we discard the negative solution.
Mass: the amount of matter an object contains
Weight: mass•acceleration due to gravity(9.8m/s) / the force of gravity acting on an object
