Hello there.
<span>Which of the following is an example of the concepts of growth and development functioning together? <span>
</span>
</span><span> C. Children learning to print their names
</span>
Answer:
Yes the ramp can be safely used
Explanation:
Here, we have
Length of longest ramp = 5 ft
Height of wall = 2 ft
Therefore, the sine of the angle adjacent to the ramp which is equal to the angle of elevation is given by;

Where:
The opposite side to angle = 2 ft wall and
Hypotenuse side = Ramp = 5 ft
Therefore,
and θ = sin⁻¹0.4 = 23.55 °
The ramp can be safely used as the angle it is adjacent to is less than the specified 30°.
Answer:
Kinetic energy = 44.1 J
Explanation:
Given:
Velocity of ball = 14 m/s
Mass of ball = 0.45 kg
Find:
Kinetic energy
Computation:
An item's kinetic energy is force it has as a result of its motion. It is the amount of work necessary to move a body of a known volume from rest to a specified velocity.
Ke = (1/2)(m)(v²)
Ke = (1/2)(0.45)(14²)
Ke = (1/2)(0.45)(196)
Kinetic energy = 44.1 J
Answer:
823.46 kgm/s
Explanation:
At 9 m above the water before he jumps, Henri LaMothe has a potential energy change, mgh which equals his kinetic energy 1/2mv² just as he reaches the surface of the water.
So, mgh = 1/2mv²
From here, his velocity just as he reaches the surface of the water is
v = √2gh
h = 9 m and g = 9.8 m/s²
v = √(2 × 9 × 9.8) m/s
v = √176.4 m/s
v₁ = 13.28 m/s
So his velocity just as he reaches the surface of the water is 13.28 m/s.
Now he dives into 32 cm = 0.32 m of water and stops so his final velocity v₂ = 0.
So, if we take the upward direction as positive, his initial momentum at the surface of the water is p₁ = -mv₁. His final momentum is p₂ = mv₂.
His momentum change or impulse, J = p₂ - p₁ = mv₂ - (-mv₁) = mv₂ + mv₁. Since m = Henri LaMothe's mass = 62 kg,
J = (62 × 0 + 62 × 13.28) kgm/s = 0 + 823.46 kgm/s = 823.46 kgm/s
So the magnitude of the impulse J of the water on him is 823.46 kgm/s
There are some missing data in the problem. The full text is the following:
"<span>A </span>real<span> (</span>non-Carnot<span>) </span>heat engine<span>, </span>operating between heat reservoirs<span> at </span>temperatures<span> of 710 K and 270 K </span>performs 4.1 kJ<span> of </span>net work<span>, and </span>rejects<span> 9.7 </span>kJ<span> of </span>heat<span>, in a </span>single cycle<span>. The </span>thermal efficiency<span> of a </span>Carnot heat<span> engine, operating between the same </span>heat<span> reservoirs, in percent, is closest to.."
Solution:
The efficiency of a Carnot cycle working between cold temperature </span>

and hot temperature

is given by

and it represents the maximum efficiency that can be reached by a machine operating between these temperatures. If we use the temperatures of the problem,

and

, the efficiency is

Therefore, the correct answer is D) 62 %.