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4 years ago

Could anyone help with this

Physics

1 answer:

pishuonlain [190]4 years ago

4 0

The answer is "B) Most species names are written in Swedish because Linnaeus was Swedish." Actually most if not all names in the Modern System of Taxonomic Classification are in Latin.

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Bryan Allen pedaled a human-powered aircraft across the English channel from the cliffs of Dover to Cap Gris-Nez on June 12, 197

tigry1 [53]

Answer:

The answer is " $5.53 \ \frac{m}{s}$ "

Explanation:

apply the formula for calculating the average velocity to the relative air

$V_{PG} =V_{PA}+V_{AG}$

$\Rightarrow V_{PA} = V_{PG} -V_{AG}$

Given value:

$V_{AG} = -2 \ \frac{m}{s}$

$V_{PG} =3.53$

$\Rightarrow V_{PA} = 3.53 - (-2) \\\\\Rightarrow V_{PA} = 3.53 +2 \\\\\Rightarrow V_{PA} = 5.53 \\\\$

The final answer is " $5.53 \ \frac{m}{s}$ " in the south-east direction.

4 0

3 years ago

Which of these statements explains the difference between nuclear binding energy and the strong nuclear force? Check all that ap

Maksim231197 [3]

B, C, F are the answers

6 0

3 years ago

Read 2 more answers

PLEASE HELP 15 POINTS!!!!

goldfiish [28.3K]

Answer:

15 degrees Celsius

Explanation:

8 0

3 years ago

Whats the measure of how much energy a sound wave carries, the loudness of a sound?

musickatia [10]

Decibels I believe? I’m not 100% sure

6 0

3 years ago

A car of mass 1600 kg traveling at 27.0 m/s is at the foot of a hill that rises vertically 135 m after travelling a distance of

Zarrin [17]

Answer:

Neglecting any frictional losses, the average power delivered by the car's engine is 10565 W

Explanation:

The energy conservation law indicates that the energy must be the same at the bottom of the hill and at the top of the hill.

The energy at the bottom is only the Kinect energy (K_1) of the car in motion, but in the top, the energy is the sum of its Kinect energy (K_2), potential energy (P) and the work (W) done by the engine.

$K_1 = K_2 + P + W$

then, the work done by the engine is:

$W = K_1 - K_2 - P$

The formulas for the Kinetic and potential energy are:

$K=\frac{1}{2}mV^2\\P=mgh$

where, m is the mass of the car, V the velocity, g the gravity and h is the elevation of the hill.

Using the formulas:

$W=\frac{1}{2}mV_1^2-\frac{1}{2}mV_2^2-mgh$

Replacing the values:

$W=\frac{1}{2}(1600Kg)(27m/s)^2-\frac{1}{2}(1600Kg)(14m/s)^2-(1600Kg)(9.8m/s^2)(135m)\\W=-1690400 J$

The negative of this value indicates the direction of the work done, but for the problem, you only care about the magnitude, so the power is W=1690400 J. Now, the power is equal to work/time so you need to find the time the car took to get to the top of the hill.

The average speed of the car is (27+14)/2=20m/s, and t=d/v so the time is:

$t=\frac{3200m}{20m/s}=160s$

the power delivered by the car's engine was:

$power=\frac{work}{time}=\frac{1690400J}{160s}=10565W$

8 0

4 years ago