Ferns are located in which zone of deciduous forests?a. tree stratum
b. small tree and sapling
c. shrub
d. herb

Answers

Answer 1

Answer: A fern is a vascular plant that have neither seeds nor flowers. Fern spieces live in the variety of habitats: shady forests, acid wetlands and tropical areas. Tree stratum zone contains the tallest trees. The small tree zone is lower than tree stratum zone. The shurb zone contains schurbs which grow the food for animals. Finally, the herb layer is made up of wildflowers and ferns. Answer: D ) Herb zone.

Answer 2

Answer: D. Herb
I hope that helps! :)

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The power associated with one horsepower is 735.5 watts

Answers

The only vehicle capable of outpacing the Ninja H2R out of the box was the severely tuned, 1,350 horsepower SPE Nissan GT-R, which caught and passed the superbike at the.

Which is faster, the Kawasaki Ninja H2R or the Bugatti?

Slowly coming off the line, the biker eventually catches up with and surpasses the Veyron.When the H2R triggers the speed camera at 194.5 mph—nearly 15 mph faster than that of the Bugatti—it is still harsh in acceleration.

A bike is faster than a Lamborghini, right?

The bike easily overtakes the Lamborghini after another strong start from the latter.In comparison, the Huracan needs 9.6 seconds to finish the quarter mile.

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A golf ball is dropped from rest from a height of 9.5m. It hits the pavement then bounces back up rising just 5.7 m before falling back down again. A boy then catches the ball on the way down when it is 1.20 m above the pavement. Ignoring air resistance, calculate the total amount of time the ball is in the air, from drop to catch.

Answers

Answer: 3.4s

Explanation:

There are three stages in the motion of the ball, so you have to calculate the times for every stage.

1) Ball dropping from 9.5m: free fall

d = Vo + gt² / 2

Vo = 0 ⇒ d = gt² / 2 ⇒ t² = 2d / g = 2 × 9.5 m / 9.81 m/s² = 1.94 s²

⇒ t = √ (1.94 s²) = 1.39s

2) Ball rising 5.7m (vertical rise)

i) Determine the initial speed:

Vf² = Vo² - 2gd

Vf² = 0 ⇒ Vo² = 2gd = 2 × 9.81 m/s² × 5.7m = 111.8 m²/s²

⇒ Vo = 10.6 m/s

ii) time rising

Vf = Vo - gt

Vf = 0 ⇒ Vo = gt ⇒

t = Vo / g = 10.6 m/s / 9.81 m/s² = 1.08 s

3) Ball dropping from 5.7 m to 1.20m above the pavement (free fall)

i) d = 5.7m - 1.20m = 4.5m

ii) d = gt² / 2 ⇒ t² = 2d / g = 2 × 4.5 m / 9.81 m/s² = 0.92 s²

⇒ t = √ (0.92 s²) = 0.96s

4) Total time

t = 1.39s + 1.08s + 0.96s = 3.43s ≈ 3.4s

The time the ball takes to fall 9.5 meters is the square root of (19/g), where g is gravitational acceleration.
The time it takes to rise to 5.7 meters is the square root of (11.4/g), for the same value of g.
The time it takes to fall from 5.7 meters to 1.2 is the square root of (9/g).
So the answer is [sqrt(19)+sqrt(11.4)+sqrt(9)]/sqrt(g). If g=10, the answer is 3.39 seconds; if g=9.8, the answer is 3.43 seconds.

The boy on the tower throws a ball 20 meters downrange as shown.What is his pitching speed? (Take g = 9.8 m/s²)
Show work clearly and completely (including formulas, calculations, units, etc. as done in class.)

Answers

The problem states that the ball is thrown 20 meters downrange. Since the ball is thrown horizontally, we can assume that the initial vertical velocity is zero. The only force acting on the ball is gravity, which causes it to accelerate downwards at a rate of 9.8 m/s² 1.

To calculate the pitching speed, we need to find the time it takes for the ball to travel 20 meters horizontally. Unfortunately, this information is not provided in the problem statement. Therefore, we cannot calculate the pitching speed.

The equation E = mc2 relates energy andA. force
B. work
C. gravity
D. mass

Answers

The equation E = mc² relates energy and mass and the correct option is option D.

The equation was formulated by Albert Einstein as part of his theory of special relativity. In this equation, "E" represents energy, "m" represents mass, and "c" represents the speed of light in a vacuum, which is approximately 3 x 10⁸ meters per second. The equation states that energy is equal to the mass of an object multiplied by the square of the speed of light.

This equation demonstrates the equivalence between mass and energy, suggesting that mass can be converted into energy and vice versa. It is a foundational equation in physics and has significant implications in understanding the relationship between matter and energy.

Thus, the ideal selection is option D.

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E=mc2 is when energy is set equal to mass then it is multiplied by the velocity of light squared. The equation relates to energy and D.Mass

If you can throw a ball at 20m/s what is the maximum distance you can throw it?

Answers

it depends on how long the ball stays in the air
1 s=20m
10 s=200m

Final answer:

The maximum distance that a ball can be thrown at a speed of 20 m/s depends on the angle at which the ball is thrown. In ideal conditions and at a 45-degree angle, the theoretical maximum distance is approximately 40.57 meters.

Explanation:

The question is asking about the maximum distance you can throw a ball given an initial speed, which is a topic in Physics known as projectile motion. In an ideal condition, where air resistance is ignored, the maximum distance a projectile can travel is achieved when it is launched at an angle of 45 degrees.

However, we are missing a piece of information in this situation, which is the launch angle. Without knowing the angle at which the ball is thrown, we cannot accurately calculate the maximum distance. Theoretically, if the ball is thrown at an angle of 45 degrees, the distance (d) can be obtained using the formula for the range of a projectile: d = (v^2)/g, where v is the initial speed and g is the acceleration due to gravity. Substituting the value, d = (20^2)/9.81 = 40.57 meters. But this is an estimation and the value could change according to the actual circumstances when the ball is thrown.

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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 $\boxed{599.8\text{ cm/s}}$ .

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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.