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atroni [7]
3 years ago
10

A scientist is observing a eukarotic cell and a prokaryotic cell. Which structure could she only observe in the eukaryotic cell?

Physics
2 answers:
Ksenya-84 [330]3 years ago
7 0
The answer is nucleus
tangare [24]3 years ago
4 0

Answer:

It's definitely Nucleus

Hope this Helps!

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B.) Write a clear, English sentence which describes a general relationship between
Lynna [10]

If a student took seven science classes, his/her at score would be 1200.

<h3>What is a graph?</h3>

A graph is a relationship in which data is presented on the axes of cartesian coordinates. The presentation of the graph must be such that it is easy to read off the graph.

Now we can see that the graph is a plot of SAT scores versus the number of science classes that students take. In this case, the slope would tell us the score per subject of a student in the SAT.

a) The mathematical model that re[resents this graph is;

y =100x + 500

b) The slope of the graph must be obtained by the use of two points thus;

m = y2 - y1/x2 - x1

m = 600 - 500/1 - 0

m = 100

c) Using the relation;

y = mx + c

m = 100

c = 500

y = 100(7) + 500

y = 1200

Thus if a student took seven science classes, his/her at score would be 1200.

Learn more about graph:brainly.com/question/16608196

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4 0
1 year ago
For an object in uniform circular motion, what can you say about the directions of the velocity, acceleration, and net force vec
zavuch27 [327]

Answer: b) The velocity vector is perpendicular to the acceleration vector; the acceleration vector is parallel to the net force vector.

Explanation: A change in velocity creates an acceleration. As the object rotates through the circular path it is constantly changing direction, and hence accelerating, which causes a constant force to act upon the object. This Force acts towards the center of curvature, directly toward the axis of rotation in a direction parallel to the acceleration of the body along the path. Because the object is moving perpendicular to the force, the path followed by the object is a circular one. Hence the velocity of the object is perpendicular to the acceleration.

3 0
3 years ago
An electron experiences a magnetic force with a magnitude of 4.90×10−15 n when moving at an angle of 60.0 ∘ with respect to a ma
Leya [2.2K]
Missing questions: "find the speed of the electron".

Solution:
the magnetic force experienced by a charged particle in a magnetic field is given by
F=qvB \sin \theta
where
q is the particle charge
v its velocity
B the magnitude of the magnetic field
\theta the angle between the directions of v and B.

Re-arranging the formula, we find:
v= \frac{F}{qB \sin \theta}
and by substituting the data of the problem (the charge of the electron is q=1.6 \cdot 10^{-19} C), we find the velocity of the electron:
v= \frac{F}{qB \sin \theta}= \frac{4.90 \cdot 10^{-15}N}{(1.6 \cdot 10^{-19}C)(3.70 \cdot 10^{-3} T)(\sin 60^{\circ})}=9.56 \cdot 10^6 m/s
4 0
3 years ago
Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 9.0 km/h due east. Runner B is initia
sergejj [24]

Answer:

0.176m from the flagpole, westward.

Explanation:

Let the Eastward be the positive direction. So initially runner A is at position -6km, running with velocity of 9km/h while runner B is at position 5km running at a velocity of -8km/h. We can conduct the following equation for their distances over the same time t

s_A = -6 + 9t

s_B = 5 - 8t

When A an B meets, they are at the same position and at the same time. So

s_A = s_B

-6 +9t = 5 - 8t

17t = 5 + 6 = 11

t = 11/17 = 0.647 s

s_A = -6 + 9*0.647 = -0.176 m

So where they meet is 0.176m from the flagpole, westward.

5 0
3 years ago
The law of ellipse simple explanation
nirvana33 [79]

Kepler's first law - sometimes referred to as the law of ellipses - explains that planets are orbiting the sun in a path described as an ellipse. ... The resulting shape will be an ellipse. An ellipse is a special curve in which the sum of the distances from every point on the curve to two other points is a constan

4 0
2 years ago
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