What is the rate at which work is done?

Explanation: Roughly 180 - 200 million years ago, just before the first dinosaurs evolved. Mammals themselves evolved from a group or reptiles which exhibited mammal-like traits. One of them was specialized teeth. Reptiles tend to have teeth all the same shape. The mammal-like reptiles evolved tiny teeth in front of the jaw and two pairs of over sized fangs along the the sides. Like modern mammals, the head was large in proportion to the rest of the body. The jaws were also evolving another mammal trait, the ability to move sideways. Despite the lack of specialized teeth, acute hearing and the ability to chew, the dinosaurs evolved an adaptation which made them far more successful than mammals--modified leg bones. These limbs could be articulated directly under their bodies. This enabled the legs to support more weight, since the limbs were now under the body instead of at the sides. Then dinosaurs did something which secured their dominance for the next 120 million years - they began to stand on two legs. Although the back was still parallel to the ground, running on two legs greatly increased the dinosaur's speed. Mammals could simply not compete with swift, giant predators and were forced to remain small, and most became nocturnal to evade dinosaurs which were probably active during the day. Despite that they managed to survive which allowed the further evolution of mammals into us, humans.

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The following three hot samples have the same temperature. The same amount of heat is removed from each sample. Which one experi

melomori [17]

Complete Question:

The following three hot samples have the same temperature. The same amount of heat is removed from each sample. Which one experiences the smallest drop in temperature, and which one experiences the largest drop? Sample A: 4.0 kg of water [c = 4186 J/(kg·C°)] Sample B: 2.0 kg of oil [c = 2700 J/(kg·C°)] Sample C: 9.0 kg of dirt [c = 1050 J/(kg·C°)]

Answer:

A. Smallest B. Largest.

Explanation:

Assuming no heat exchange except for the heat removed from any sample (which we know is the same for the three ones), and that the process is done using only conduction, we can use the equation that relates the heat lost or gained by one object, with the mass of the object and the consequent change in temperature, as follows:

Q = c*m*ΔT, where c, is a proportionality constant called specific heat, which is different for each material.

As we know that the heat removed is the same for the three samples, we can equate the right sides of the equation for each sample, as follows:

cw*mw*ΔTw = co*mo*ΔTo = cd*md*ΔTd

Replacing by the givens, we have:

4.0 kg. 4,186 J/kgºC*ΔT(ºC) = 2.0 kg*2,700 J/kgºC*ΔT(ºC) =9.0kg*1,050J/kgºC*ΔT(ºC)

As the three expressions must be equal each other, it's clear that the unknown term (the drop in temperature) must compensate the product of the mass times the specific heat.

This product is the following for the three samples:

Water: 4.0 kg*4,186 J/kgºC = 16,744 J/ºC

Oil : 2.0 kg*2,700 J/kgºC = 5,400 J/ºC

Dirt: 9.0 * 1,050 J/kgºC = 9,450 J/ºC

Clearly, we see that in order to keep the heat exchange equations equal each other, the water must suffer the smallest drop in temperature, and the oil must experience the largest one.

So, the sample A experiencies the smallest drop in temperature, and sample B does the largest one.

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3 years ago

Galactic Alliance Junior Mission Officer (GAJMO) Bundit Nermalloy is predicting the kinetic energy of a supply spacecraft, which

antiseptic1488 [7]

Answer:

the ship's energy is greater than this and the crew member does not meet the requirement

Explanation:

In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship

W =∫ F dx = ΔK

Let's replace

∫ (α x³ + β) dx = ΔK

α x⁴ / 4 + β x = ΔK

Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J

x (α x³ + β) = $K_{f}$ - K₀

$K_{f}$ = K₀ + x (α x³ + β)

Assuming that the low limit is x = 0, measured from the cargo hangar

Let's calculate

$K_{f}$ = 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)

Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)

Kf = 2.7 10¹¹ - 1.1475 10¹¹

Kf = 1.55 10¹¹ J

In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement

We evaluate the kinetic energy if the System is well calibrated

W = x F₀ = $K_{f}$ –K₀

$K_{f}$ = K₀ + x F₀

We calculate

$K_{f}$ = 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶

$K_{f}$ = (2.7 -2.625) 10¹¹

$K_{f}$ = 7.5 10⁹ J

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3 years ago