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3 years ago

Heat likes to remain ONMFRIU ← Whats that unscrambled?

1 answer:

5 0

<span>So the question is how does heat prefer to remain and to unscramble the letters ONMFRIU. The unscrambled letters mean: UNIFORM. The heat likes to remain uniform because thermodynamic systems always tend to reach thermal equilybrium after some period of time that is specific for each system. </span>

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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

8_murik_8 [283]

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.

4 0

3 years ago

Estimate the magnitude of the electric field due to the proton in a hydrogen atom at a distance of

Lana71 [14]

Electric field due to a point charge is given as

$E = \frac{kq}{r^2}$

here we know that

$q = 1.6 \times 10^{-19} C$

also the distance is given as

$r = 5.29 \times 10^{-11} m$

now we will have

$E = \frac{(9\times 10^9)(1.6 \times 10^{-19})}{(5.29 \times 10^{-11})^2}$

so we will have

$E = 5.14 \times 10^{11} N/C$

so above is the electric field due to proton

5 0

3 years ago

A 3 kg penguin is pushed by his penguin friends who give him an initial speed vo at the top of a 30 m hill. The penguin is hopin

Strike441 [17]

Answer:

This question can be answered by using conversation of energy.

$K_1 + U_1 = K_2 + U_2$

$\frac{1}{2}mv_{0}^2 + mgh_1 = 0 + mgh_2$

$\frac{1}{2}(3)v_0^2 + (3)(9.8)(30) = (3)(9.8)(45)\\\frac{1}{2}(3)v_0^2 = 441\\v_0^2 = 294\\v_0 = 17.14 m/s$

Explanation:

Note that we take $K_2 = 0$ because we are looking for the minimum initial speed for the penguin to reach the top of the second hill. Any other speed more than this will already be enough for him.