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Oliga [24]
2 years ago
9

Use the drop-down menu to complete the statement. One difference between prokaryotic cells and eukaryotic cells is that a eukary

otic cell contains , while a prokaryotic cell does not.
Physics
1 answer:
Ahat [919]2 years ago
8 0

Answer:

Nucleus

Explanation: I did the quiz and got it right. Have an amazing day!

You might be interested in
A projectile of mass 2.0 kg is fired in the air at an angle of 40.0 ° to the horizon at a speed of 50.0 m/s. At the highest poin
tekilochka [14]

Answer:

a) The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

                                         0.7 kg is 109.4 ms on the x axis

b)  Y = 109.3 m

Explanation:

This is a moment and projectile launch exercise.

a) Let's start by finding the initial velocity of the projectile

       sin 40 = voy / v₀

       v_{oy} = v₀ sin 40

       v_{oy} = 50.0 sin40

       v_{oy} = 32.14 m / s

       cos 40 = v₀ₓ / V₀

       v₀ₓ = v₀ cos 40

       v₀ₓ = 50.0 cos 40

       v₀ₓ = 38.3 m / s

Let us define the system as the projectile formed t all fragments, for this system the moment is conserved in each axis

Let's write the amounts

Initial mass of the projectile M = 2.0 kg

Fragment mass 1 m₁ = 1.0 kg and its velocity is vₓ = 0 and v_{y} = -10.0 m / s

Fragment mass 2 m₂ = 0.7 kg moves in the x direction

Fragment mass 3 m₃ = 0.3 kg moves up (y axis)

Moment before the break

X axis

     p₀ₓ = m v₀ₓ

Y Axis y

    p_{oy} = 0

After the break

X axis

   p_{fx} = m₂ v₂

Axis y

     p_{fy} = m₁ v₁ + m₃ v₃

Let's write the conservation of the moment and calculate

Y Axis  

     0 = m₁ v₁ + m₃ v₃

Let's clear the speed of fragment 3

     v₃ = - m₁ v₁ / m₃

     v₃ = - (-10) 1 / 0.3

     v₃ = 33.3 m / s

X axis

     M v₀ₓ = m₂ v₂

     v₂ = v₀ₓ M / m₂

     v₂ = 38.3  2 / 0.7

     v₂ = 109.4 m / s

The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

                                         0.7 kg is 109.4 ms on the x axis

b) The speed of the fragment is 33.3 m / s and has a starting height of where the fragmentation occurred, let's calculate with kinematics

       v_{fy}² = v_{oy}² - 2 gy

       0 =  v_{oy}²-2gy

       y =  v_{oy}² / 2g

       y = 32.14² / 2 9.8

       y = 52.7 m

This is the height where the break occurs, which is the initial height for body movement of 0.3 kg

      v_{f}² =  v_{y}² - 2 g y₂

      0 =  v_{y}² - 2 g y₂

     y₂ =  v_{y}² / 2g

     y₂ = 33.3²/2 9.8

     y₂ = 56.58 m

Total body height is

      Y = y + y₂

      Y = 52.7 + 56.58

     Y = 109.3 m

8 0
2 years ago
Sinkholes can form when
kherson [118]
Land surfaces change, erosion happens, the ground collapses, etc.
7 0
3 years ago
Read 2 more answers
Two objects, C & D, have the same momentum. Object C has ½ the mass of object D. Find the value of the ratio of velocity C t
Savatey [412]
These are two questions and two answers.

Part 1. Fin the value of the ration of velocity C to velocity D.


Answer: 2

Explanation:

1) Formula: momentum = mass * velocity

2) momentum C = mass C * velocity C

3) momentum D = mass D * velocity D.

4) C and D have the same momentum =>

mass C * velocity C = mass D * velocity D

5) mass C = (1/2) mass D => mass C / mass C = 1/2

6) use in the equation stated in the point 4)

velocit C / velocity D = mass D / mass C

using the equation stated in point 5:

mass D / mass C = 1 / [ mass C / mass D] = 1 / [1/2] = 2

=>

7) velocity C / velocity D = mass D / mass C = 2

Part 2: <span>ratio of kinetic energy C to kinetic energy D.
</span>
Answer: 2

Explanation:

1) formula: kinetic energy KE = (1/2) mass * (velocity)^2

2) KE C = (1/2) mass C * (velocity C)^2

3) KE D = (1/2) mass D * (velocity D)^2

4) KE C / KE D =

(1/2) mass C * (velocity C)^2        mass C        (velocity C)^2
--------------------------------------- = --------------- * ---------------------- = (1/2) * (2)^2
(1/2) mass D *( velocity D)^2        mass D        v(velocity D)^2

= 4 / 2 = 2
3 0
3 years ago
A plane accelerates from rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long. Assume that the plane reaches
tiny-mole [99]

Answer:

26.8 seconds

Explanation:

To solve this problem we have to use 2 kinematics equations: *I can't use subscripts for some reason on here so I am going to use these variables:

v = final velocity

z = initial velocity

x = distance

t = time

a = acceleration

{v}^{2}  =  {z}^{2}  + 2ax

v = z + at

First let's find the final velocity the plane will have at the end of the runway using the first equation:

{v}^{2}  =  {0}^{2}  + 2(5)(1800)

v = 60 \sqrt{5}

Now we can plug this into the second equation to find t:

60 \sqrt{5}  = 0 + 5t

t = 12 \sqrt{5}

Then using 3 significant figures we round to 26.8 seconds

3 0
2 years ago
What is the mass of an object that is accelerated at 25 m/s2 by a force of 135 N?
yKpoI14uk [10]

Answer:

<h3>The answer is 5.4 kg</h3>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{135}{25}   =  \frac{27}{5} \\

We have the final answer as

<h3>5.4 kg</h3>

Hope this helps you

4 0
3 years ago
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