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zhannawk [14.2K]
3 years ago
13

It is 5.5 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).
Part A. Which choice would burn up more energy?
Part B. How much energy (in joules) would it burn?
Part C. Why is it that the more intense exercise actually burns up less energy than the less intense one?
Physics
1 answer:
stealth61 [152]3 years ago
6 0

Explanation:

Displacement = 5 km

A.

Converting km/h to m/s,

10 km/h * 1000 m/1 km * 1 h/3600 s

= 25/9 m/s

Remember,

700 watt = 700 J/s

Velocity = displacement/time

Time = 5000/(25/9)

= 1800 s

Energy = power * time

= 700 * 1800

= 1,260,000

= 1260 kJ

B.

Converting km/h to m/s,

3 km/h * 1000 m/1 km * 1 h/3600 s

= 5/6 m/s

290 watt = 290 J/s

Velocity = displacement/time

Time = 5000/(5/6)

= 6000 s

Energy = power * time

= 290 * 6000

= 1,740,000

= 1740 kJ

C.

Walking burns more energy; 1,740,000 joules. It burns more because you walk for a greater period of time.

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Gelneren [198K]

Answer:

response

Explanation:

Acceleration is your changing Velocity. An object that is ACCELERATING is experiencing a change in velocity. usually positive. if an object such as a car reduces velocity, it is called deceleration

3 0
3 years ago
Based on Newton's law of motion, which combination of rocket bodies and engine will result in the acceleration of 40 m/s ^2 at t
ad-work [718]

The question is incomplete. The complete question is :

The Rocket Club is planning to launch a pair of model rockets. To build the rocket, the club needs a rocket body paired with an engine. The table lists the mass of three possible rocket bodies and the force generated by three possible engines.

A 4-column table with 3 rows. The first column labeled Body has entries 1, 2, 3. The second column labeled Mass (grams) has entries 500, 1500, 750. The third column labeled Engine has entries 1, 2, 3. The fourth column labeled Force (Newtons) has entries 25, 20, 30.

Based on Newton’s laws of motion, which combination of rocket bodies and engines will result in the acceleration of 40 m/s2 at the start of the launch?

Body 3 + Engine 1

Body 2 + Engine 2

Body 1 + Engine 2

Body 1 + Engine 1

Solution :

Given :

Body       Mass (gram)     Engine      Force (newtons)

1                   500                 1                     25

2                  1500                2                    20

3                  750                  3                    30

The body 1 has a mass of 500 gram which is equal to 0.5 kg

And engine 2 has a force of 20 newtons.

We know that according to Newton's laws of motion,

Force = mass x acceleration

 20    = 0.5 x acceleration

Acceleration $=\frac{20}{0.5}$

                      $=\frac{200}{5}$

                      $= 40 \ m/s^2$

Therefore, based on laws of motion of Newton, the Body 1 + Engine 2 combination of the rocket bodies and engines will result in an acceleration of $ 40 \ m/s^2$ at the start of the launch.

8 0
3 years ago
Earth that are large enough to be detected by radar
Novosadov [1.4K]

Answer:

I don't now sorry HHHAHAH GOOD LUCK

8 0
3 years ago
M (b) Calculate the speed of an electron that is accelerated through the same electric potential difference.
MaRussiya [10]

The speed of a electron that is accelerated from rest through an electric potential difference of 120 V  is 6.49\times10^6m/s

<h3>How to calculate the speed of the electron?</h3>

We know, that the  energy of the system is always conserved.

Using the Law of Conservation of energy,

\triangle K+\triangle U=0

Here, K is the kinetic energy and U is the potential energy.

Now, substituting the formula of U and K, we get:

(\frac{1}{2} mv^2)+(-q\triangle V) =0------(1)

Here,

m is the mass of the electron

v is the speed of the electron

q is the charge on the electron

V is the potential difference

Let v_f and v_i represent the final and initial speed.

Here, v_i =0

Solving for v_f, we get:

v_f=\sqrt{\frac{-2q\triangle V}{m} }

=\sqrt{\frac{2\times1.602\times10^{-19}\times 120}{9.109\times10^{-31}} }

=6.49\times10^6m/s

To learn more about the conservation of energy, refer to:

brainly.com/question/2137260

#SPJ4

4 0
1 year ago
How long would it take for Sofia to walk 300 meters if she is walking at a velocity of 2.5 m/s?
Vaselesa [24]

Answer:

Time=120seconds

Explanation:

S=300m

V=2.5m/s

t=?

V=S/t

t=S/V

t=300/2.5

t=120 second

7 0
3 years ago
Read 2 more answers
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