A 700-watt gasoline engine and a 300-watt electric motor both do 3 J of work. Which machine can do the work faster? Explain your answer.

Answers

Answer 1

Answer: 700 W gasoline engine do the work faster than electric motor.
What define time of execution of a work is power.
If gasoline engine has 700 W so it is faster than electric motor that has only 300 W

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A failure of the red-sensitive nerves in the eye to respond to light properly causes A. astigmatism.
B. farsightedness.
C. chromatic aberration.
D. color blindness.

Answers

D. colour blindness

"red-sensitive" just means that the nerves are able to see red light

D. Color blindness is your correct answer

The earth has a circumference of about 25000 miles. How is it expressed in scientific notation?

Answers

2.5 * 10^4 miles

Yay!!!

A hot air balloon is moving vertically upwards at a velocity of 3m/s. A sandbag is dropped when the balloon reaches 150m. How long does the sandbag take to reach the ground and what is the maximum height it reaches?

Answers

This is a perfect opportunity to stuff all that data into the general equation for the height of an object that has some initial height, and some initial velocity, when it is dropped into free fall.

   H(t) = (H₀) + (v₀ T) + (1/2 a T²)

Height at any time 'T' after the drop =

(initial height) +
      (initial velocity) x (T) +
   (1/2) x (acceleration) x (T²) .

For the balloon problem ...

-- We have both directions involved here, so we have to define them:

   Upward = the positive direction

   Initial height = +150 m
   Initial velocity = + 3 m/s

   Downward = the negative direction

   Acceleration (of gravity) = -9.8 m/s²

Height when the bag hits the ground = 0 .

   H(t) = (H₀) + (v₀ T) + (1/2 a T²)

   0 = (150m) + (3m/s T) + (1/2 x -9.8 m/s² x T²)

   -4.9 T² + 3T + 150 = 0

Use the quadratic equation:

   T = (-1/9.8) [ -3 plus or minus √(9 + 2940) ]

   = (-1/9.8) [ -3 plus or minus 54.305 ]

   = (-1/9.8) [ 51.305 or -57.305 ]

      T = -5.235 seconds or 5.847 seconds .

(The first solution means that the path of the sandbag is part of
the same path that it would have had if it were launched from the
ground 5.235 seconds before it was actually dropped from balloon
while ascending.)

Concerning the maximum height ... I don't know right now any other
easy way to do that part without differentiating the big equation.
So I hope you've been introduced to a little bit of calculus.

H(t) = (H₀) + (v₀ T) + (1/2 a T²)

   H'(t) = v₀ + a T

The extremes of 'H' (height) correspond to points where h'(t) = 0 .

Set    v₀ + a T = 0

+3 - 9.8 T = 0

Add 9.8 to each side: 3   = 9.8 T

Divide each side by 9.8 :   T = 0.306 second

That's the time after the drop when the bag reaches its max altitude.

Oh gosh ! I could have found that without differentiating.

- The bag is released while moving UP at 3 m/s .

- Gravity adds 9.8 m/s of downward speed to that every second.
So the bag reaches the top of its arc, runs out of gas, and starts
falling, after
   (3 / 9.8) = 0.306 second .

At the beginning of that time, it's moving up at 3 m/s.
At the end of that time, it's moving with zero vertical speed).
Average speed during that 0.306 second = (1/2) (3 + 0) = 1.5 m/s .

Distance climbed during that time = (average speed) x (time)

   = (1.5 m/s) x (0.306 sec)

   = 0.459 meter (hardly any at all)

   But it was already up there at 150 m when it was released.

It climbs an additional 0.459 meter, topping out at 150.459 m,
then turns and begins to plummet earthward, where it plummets
to its ultimate final 'plop' precisely 5.847 seconds after its release.

We can only hope and pray that there's nobody standing at
Ground Zero at the instant of the plop.

I would indeed be remiss if were to neglect, in conclusion,
to express my profound gratitude for the bounty of 5 points
that I shall reap from this work. The moldy crust and tepid
cloudy water have been delicious, and will not soon be forgotten.

Oceanic electric fields produce electric currents. True False

Answers

Wow ! That's wild !
I never heard of an "oceanic" electric field before.

Electric currents are the flow of great numbers of electrons
inside a conducting material. What makes them move is
an electrostatic or magnetic field inside the conductor.

Things that happen in the ocean produce ocean currents,
not electric ones.

In 1780, in what is now referred to as "Brady's Leap," Captain Sam Brady of the U.S. Continental Army escaped certain death from his enemies by running over the edge of the cliff above Ohio's Cuyahoga River in (Figure 1) , which is confined at that spot to a gorge. He landed safely on the far side of the river. It was reported that he leapt 22 ft (≈ 6.7 m) across while falling 20 ft (≈ 6.1 m).What is the minimum speed with which he’d need to run off the edge of the cliff to make it safely to the far side of the river?

Express your answer to two significant figures and include the appropriate units.

Answers

The minimum speed with which the captain Sam Brady of the US continental army had to run off the edge of the cliff to make it safely to the far side of the river is $\boxed{19.667\text{ ft/s}}$ or $\boxed{5.998\text{ m/s}}$ or $\boxed{6\text{ m/s}}$ or $\boxed{599.8\text{ cm/s}}$ .

Further explanation:

As Captain Sam Brady jumps from the cliff, he moves in two dimension under the action of gravity.

Given:

The height of free fall of the captain Brady is $20\text{ ft}$ or $6.1\text{ m}$ .

The horizontal distance moved by the captain Brady is $22\text{ ft}$ or $6.7\text{ m}$ .

Concept:

The time required to free fall of a body can be calculated by using the expression given below.

$\left( { - s}\right)=ut-(1)/(2)g{t^2}$ ……. (1)

The displacement is considered negative because the captain is moving in vertically downward direction.

Here, $s$ is the distance covered by the body in free fall, $u$ is the initial velocity of the object, $g$ is the acceleration due to gravity and $t$ is the time taken in free fall of a body.

As the Caption jumps off the cliff, he has his velocity in the horizontal direction. The velocity of the captain in vertical direction is zero.

Substitute $0$ for $u$ in the equation (1) .

$s=(1)/(2)g{t^2}$

Rearrange the above expression for $t$ .

$\boxed{t=\sqrt {\frac{{2s}}{g}}}$ …… (2)

Converting acceleration due to gravity in $\text{ft}/\text{s}^2$ .

$\begin{aligned}g&=\left( {9.81\,{\text{m/}}{{\text{s}}^{\text{2}}}} \right)\left( {\frac{{1.0\,{\text{ft/}}{{\text{s}}^{\text{2}}}}}{{0.305\,{\text{m/}}{{\text{s}}^{\text{2}}}}}} \right) \n&=32.16\,{\text{ft/}}{{\text{s}}^{\text{2}}} \n \end{aligned}$

Substitute $20\text{ ft}$ for $s$ and $32.16\,{\text{ft/}}{{\text{s}}^{\text{2}}}$ for $g$ in equation (2) .

$\begin{aligned}t&=\sqrt {\frac{{2\left( {20\,{\text{ft}}} \right)}}{{\left( {32.16\,{\text{ft/}}{{\text{s}}^{\text{2}}}} \right)}}} \n&=1.116\,{\text{s}} \n \end{aligned}$

Therefore, the time taken by captain to free fall a height $20\text{ ft}$ is $1.116\text{ s}$ .

In the same time interval captain has to move $22\text{ ft}$ in horizontal direction. The acceleration is zero in horizontal direction. So, the velocity will be constant throughout the motion in the horizontal direction.

The distance travelled by captain in the horizontal direction is given by,

$x=v\cdot t$

Rearrange the above expression for $v$ .

$\boxed{v=(x)/(t)}$ …… (3)

Here, $x$ is the distance travelled in horizontal direction, $v$ is the velocity of the captain and $t$ is the time.

Substitute $22\text{ ft}$ for $x$ and $1.116\text{ s}$ for $t$ in equation (3) .

$\begin{aligned}v&=\frac{{22\,{\text{ft}}}}{{1.116\,{\text{s}}}} \n&=19.71\,{\text{ft/s}} \n \end{aligned}$

Thus, the minimum speed with which the captain Sam Brady of the US continental army had to run off the edge of the cliff to make it safely to the far side of the river is $\boxed{19.667\text{ ft/s}}$ or $\boxed{5.998\text{ m/s}}$ or $\boxed{6\text{ m/s}}$ or $\boxed{599.8\text{ cm/s}}$ .

Learn more:

1. Energy density stored in capacitor brainly.com/question/9617400

2. Kinetic energy of the electrons brainly.com/question/9059731

3. Force applied by the car on truck brainly.com/question/2235246

Keywords:

Free fall, projectile, gravity, 1780, Brady’s, leap, Captain, Sam Brady, US, continental army, enemies, Ohio’s, Cuyahoga river, 22 ft, 6.7 m, 20 ft, 6.1 m, minimum speed, run off, edge, cliff, safely, far side, river, 19.71 ft/s, 6 m/s, 6 meter/s, 5.99 m/s, 599.8 cm/s.

Final answer:

Using the principles of projectile motion from Physics, Captain Sam Brady would need to run with an initial horizontal speed of approximately 19.64 ft/s to reach the far side of the river.

Explanation:

This problem can be solved using basic Physics, specifically projectile motion. Here, Captain Sam Brady had to run off the edge of the cliff to make it safely to the far side of the river which is 22 ft away while falling 20 ft down. We assume that he jumps horizontally (i.e., his initial vertical velocity is 0).

Firstly, we calculate the time for the vertical fall. Using the equation t = sqrt (2h/g) where h is height and g is the acceleration due to gravity (32.2 ft/s²), we get time t ≈ 1.12s (rounded to two significant figures).

Next, we can use this time to figure out his initial horizontal velocity needed. The equation v = d/t where v is velocity, d is distance, and t is time gives us v ≈ 19.64 ft/s (rounded to two significant figures).

So, Captain Sam Brady would need to run with an initial horizontal speed of approximately 19.64 ft/s to make it safely across the river.

Learn more about Projectile Motion here:

brainly.com/question/20627626

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Covalent bonds tend to make atoms more stable by helping them

Answers

by helping them achieve a full outer shell

Covalent bonds are where two atoms SHARE electrons, that way they can be stable :D

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