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Vinil7 [7]
4 years ago
12

It is 5.00 km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10.0 k

m/hr (which uses up energy at the rate of 700 W ), or you could walk it leisurely at 3.00 km/hr (which uses energy at 290 W ).A. Which choice would burn up more energy? 1. running.2. walking.B. 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?
Physics
1 answer:
8_murik_8 [283]4 years ago
4 0

Answer:

a. Walking burns up more energy.

b. 1740 kJ

c. This is because more intense exercise releases a lot of energy in a short period of time, whereas, less intense energy releases it energy gradually over a long period of time.

Explanation:

a. We know energy W = Pt where P = power and t = time.

Now for walking, t = d/v where d = distance = 5.00 km and v = speed = 3.00 km/hr and P = 290 W

So, t = d/v = 5.00 km/3.00 km/hr = 5/3 hr = 5/3 × 3600 s = 6000 s

W = Pt = 290 W × 6000 s = 1740000 = 1740 kJ

Now for running, t = d/v where d = distance = 5.00 km and v = speed = 10.00 km/hr

So, t = d/v = 5.00 km/10.00 km/hr = 0.5 hr = 0.5 × 3600 s = 1800 s and P = 700 W

W = Pt = 700 W × 1800 s = 1260000 = 1260 kJ

Since walking burns up 1740 kJ and running burns up 1260 kJ, walking burns up more energy.

b. It burns up 1740 kJ

c. This is because more intense exercise releases a lot of energy in a short period of time, whereas, less intense energy releases it energy gradually over a long period of time.

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Answer:

Option C is the correct answer.

Explanation:

Absolute  pressure is sum of gauge pressure and atmospheric pressure.

That is

               P_{abs}=P_{gauge}+P_{atm}

We have

          P_{abs}=550.280 kPa\\\\P_{atm}=1atm=101325Pa=101.325kPa

Substituting

         P_{abs}=P_{gauge}+P_{atm}\\\\550.280=P_{gauge}+101.325\\\\P_{gauge}=448.955kPa

Option C is the correct answer.

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a moving billiard ball collides with an identical stationary billiard ball in an elastic collision. after the collision, the sec
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A billiard ball collides with a stationary identical billiard ball to make it move. If the collision is perfectly elastic, the first ball comes to rest after collision.

<h3>Why does the first ball comes to rest after collision ?</h3>

Let m be the mass of the two identical balls.  

u1 = velocity before the collision of ball 1

u2 = 0 = velocity of second ball that is at rest

v1 and v2 are the velocities of the balls after the collision.

From the conservation of momentum,

∴ mu1 + mu2 = mv1 + mv2

∴ mu1 = mv1 + mv2

∴ u1 = v1 + v2

In an elastic collision, the kinetic energy of the system before and after collision remains same.

\frac{1}{2}  mu_1^2+0=\frac{1}{2}  mv_1^2+\frac{1}{2}  mv_2^2

∴  \frac{1}{2}  m(v_1+v_2 )^2=\frac{1}{2} mv_1^2+\frac{1}{2}mv_2^2

∴ \frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2+mv_1 v_2=\frac{1}{2}  mv_1^2+\frac{1}{2} mv_2^2

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<h3>What is collision ?</h3>

An elastic collision is a collision between two bodies in which the total kinetic energy of the two bodies remains constant. There is no net transfer of kinetic energy into other forms such as heat, noise, or potential energy in an ideal, fully elastic collision.

Can learn more about elastic collision from brainly.com/question/12644900

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