If you drop a <span>6.0x10^-2 kg ball from height of 1.0m above hard flat surface, and a</span>fter the ball had bounce off the flat surface, the kinetic energy of the ball would be mgh - 0.14 = 0.45.
1110 atm
Let's start by calculating how many cm deep is 36,000 feet.
36000 ft * 12 in/ft * 2.54 cm/in = 1097280 cm
Now calculate how much a column of water 1 cm square and that tall would mass.
1097280 cm * 1.04 g/cm^3 = 1141171.2 g/cm^2
We now have a number using g/cm^2 as it's unit and we desire a unit of Pascals ( kg/(m*s^2) ).
It's pretty obvious how to convert from g to kg. But going from cm^2 to m is problematical. Additionally, the s^2 value is also a problem since nothing in the value has seconds as an unit. This indicates that a value has been omitted. We need something with a s^2 term and an additional length term. And what pops into mind is gravitational acceleration which is m/s^2. So let's multiply that in after getting that cm^2 term into m^2 and the g term into kg.
1141171.2 g/cm^2 / 1000 g/kg * 100 cm/m * 100 cm/m = 11411712 kg/m^2
11411712 kg/m^2 * 9.8 m/s^2 = 111834777.6 kg/(m*s^2) = 111834777.6 Pascals
Now to convert to atm
111834777.6 Pa / 1.01x10^5 Pa/atm = 1107.2750 atm
Now we gotta add in the 1 atm that the atmosphere actually provides (but if you look closely, you'll realize that it won't affect the final result).
1107.274 atm + 1 atm = 1108.274 atm
And finally, round to 3 significant figures since that's the accuracy of our data, giving 1110 atm.
Answer:
D. Calculate the area under the graph.
Explanation:
The distance made during a particular period of time is calculated as (distance in m) = (velocity in m/s) * (time in s)
You can think of such a calculation as determining the area of a rectangle whose sides are velocity and time period. If you make the time period very very small, the rectangle will become a narrow "bar" - a bar with height determined by the average velocity during that corresponding short period of time. The area is, again, the distance made during that time. Now, you can cover the entire area under the curve using such narrow bars. Their areas adds up, approximately, to the total distance made over the entire span of motion. From this you can already see why the answer D is the correct one.
Going even further, one can make the rectangular bars arbitrarily narrow and cover the area under the curve with more and more of these. In fact, in the limit, this is something called a Riemann sum and leads to the definition of the Riemann integral. Using calculus, the area under a curve (hence the distance in this case) can be calculated precisely, under certain existence criteria.
The electric field E of a charge is defined as E=F/Q where F is the Coulomb force and Q is the test charge.
E=(1/Q)*k*(q*Q)/r², where k=9*10^9 N*m²/C², q is the point charge, Q is the test charge and r is the distance between the charges.
So E=(k*q)/r²
When we input the numbers we get that electric field E of a point chage q is:
E=(9*10^9)*(5.4*10^-8)/0.2²=486/0.04=12150 N/C.
This is roughly E=12000 N/C =1.2*10^4 N/C
The correct answer is B.
The space between the cars closes at (70+65)=135 km/hr. They collide in (405/135)=3 hours after they start moving. In 3 hours, the canary covers (100km/hr x 3)= 300 km.
