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4 years ago

One major difference between plant and animal cells is that plant cells are the only ones with ?

Physics

2 answers:

inysia [295]4 years ago

8 0

they are the only ones with a cell wall.

xz_007 [3.2K]4 years ago

3 0

Plants are the only ones with a CELL WALL , plus they have photosynthesis.

dose this help you?

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The compressor in a certain carnot refrigerator performs 480 j of work to remove 150 j of heat from the interior of the refriger

Liono4ka [1.6K]

630 J is the answer :)

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4 years ago

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Alona [7]

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3 years ago

We investigated a jet landing on an aircraft carrier. In a later maneuver, the jet comes in for a landing on solid ground with a

AlexFokin [52]

Answer:

Time is 14.8 s and cannot landing

Explanation:

This is a kinematic exercise with constant acceleration, we assume that the acceleration of the jet to take off and landing are the same

Calculate the time to stop, where it has zero speed

Vf² = Vo² + a t

t = - Vo² / a

t = - 110²/(-7.42)

t = 14.8 s

This is the time it takes to stop the jet

Let's analyze the case of the landing at the small airport, let's look for the distance traveled to land, where the speed is zero

Vf² = Vo² + 2 to X

X = -Vo² / 2 a

X = -110² / 2 (-7.42)

X = 815.4 m

Since this distance is greater than the length of the runway, the jet cannot stop

Let's calculate the speed you should have to stop on a track of this size

Vo² = 2 a X

Vo = √ (2 7.42 800)

Vo = 109 m / s

It is conclusion the jet must lose some speed to land on this track

4 0

3 years ago

Plants use sunlight to fuel photosynthesis (the process they use to produce their food). Which of the following is the best desc

Montano1993 [528]

I think it’s “light energy to chemical energy” let me know if i’m wrong

6 0

3 years ago

What is the ideal banking angle (in degrees) for a gentle turn of 2.00 km radius on a highway with a 125 km/h speed limit (about

nikitadnepr [17]

An "ideal" banking angle assumes no friction is required to keep a car on the road as it turns. Let θ denote the banking angle, and consult the attached free-body diagram for a car making the turn. There are only 2 relevant forces acting on the car,

• the normal force with magnitude n

• the car's weight with magnitude w

and the net force points toward the center of the circle made by the turn, with centripetal acceleration

a = (125 km/h)² / (2.00 km) = 7812.5 km/h² ≈ 0.603 m/s²

Split up the forces into components acting perpendicular (⟂) and parallel (//) to the banked curve, so that by Newton's second law,

∑ F (⟂) = N + W (⟂) = m a (⟂)

and

∑ F (//) = W (//) = m a (//)

Let the direction of N be the positive perpendicular axis, and down the incline and toward the center of the circle the positive parallel axis. The net force vector and acceleration both make an angle θ with the banked curve, and W makes the same angle with the negative perpendicular axis, so that the equations above reduce to

N - m g cos(θ) = m a sin(θ)

and

m g sin(θ) = m a cos(θ)

The second equation is all we need at this point to find the ideal θ. The mass m cancels out, and we can solve for θ to get

tan(θ) = a/g ≈ (0.603 m/s²) / (9.80 m/s²) ≈ 0.0615

→ θ ≈ 3.52°

8 0

4 years ago