It is 4.0 km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10 km/h (which uses up energy at the rate of 700 W), or you could walk it leisurely at 3.0 km/h (which uses energy at 290 W).A. Which choice would burn up more energy? running or walkingB. How much energy (in joules) would it burn?C. Why is it that the more intense exercise actually burns up less energy than the less intense one?

Explanation:

To solve this problem we have to take into account that the energy consumed per second is the power. Hence, by multipling the power and the time spent to arrive to the lab we obtain the total energy consumed.

But first we have to calculate the time

Now we use E=W*t for both times

A. Hence, by running the energy consumed is lower.

B.

E1=1008000J

E2=1392000J

C. Because the more intense exercise is made in a lower time in comparison with the less intense exercise, and higher the time, more energy is consumed.

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1

chemical properties of thing