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Lesechka [4]
3 years ago
6

Charlie drove around the block at constant velocity, is it true or false?​

Physics
1 answer:
Paha777 [63]3 years ago
7 0

Answer:

FALSE I THINK

Explanation:

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Which of the following is an example of the concepts of growth and development functioning together?   
DedPeter [7]
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 
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7 0
3 years ago
A ramp is needed to allow vehicles to climb a 2 foot wall. The angle of elevation in order for the vehicles to safely go up must
stepan [7]

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;

Sin\theta = \frac{Opposite \, side \, to\,  angle}{Hypothenus\, side \, of\,  triangle}

Where:

The opposite side to angle = 2 ft wall and

Hypotenuse side = Ramp = 5 ft

Therefore,

Sin\theta = \frac{2}{5} = 0.4 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°.

5 0
4 years ago
I was playing goalie in a soccer league and made a save. The ball was traveling at 14 m/s and has a mass of 0.45 kg. How muck ki
Zigmanuir [339]

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

6 0
3 years ago
Until he was in his seventies, Henri LaMothe excited audiences by belly-flopping from a height of 9 m into 32 cm. of water. Assu
baherus [9]

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

6 0
3 years ago
A real (non-Carnot) heat engine, operating between heat reservoirs at temperatures of and performs 4.1 kJ of net work, and rejec
Sati [7]
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>T_C and  hot temperature T_H is given by
\eta = 1 - \frac{T_C}{T_H}
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, T_C=270 K and T_H=710 K, the efficiency is
\eta = 1 - \frac{270 K}{710 K}=0.62 = 62%

Therefore, the correct answer is D) 62 %.
6 0
3 years ago
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