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SVETLANKA909090 [29]
3 years ago
9

Gus likes to create pictures or clusters to show the information he has learned. What kind of learner is Gus likely to be?

Physics
2 answers:
viva [34]3 years ago
4 0
D. Visual because he draws what he learns and this helps him visualize this. Kinesthetic learners like to feel things and touch them while they learn. Auditory learners like to listen to the lessons.
beks73 [17]3 years ago
3 0

Answer:

B) prioritizer

Explanation:

i thing

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An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 m in t
Liono4ka [1.6K]

(a) 5.69 N/C, vertically downward

We can calculate the acceleration of the electron by using the SUVAT equation:

d=ut+\frac{1}{2}at^2

where

d = 4.50 m is the distance travelled by the electron

u = 0 is the initial velocity of the electron

t=3.00 \mu s = 3.0 \cdot 10^{-6} s is the time of travelling

a is the acceleration

Solving for a,

a=\frac{2d}{t^2}=\frac{2(4.50)}{(3.0\cdot 10^{-6})^2}=1.0\cdot 10^{12} m/s^2

Given the mass of the electron,

m=9.11\cdot 10^{-31} kg

We can find the electric force acting on the electron:

F=ma=(9.11\cdot 10^{-31})(1.0\cdot 10^{12})=9.11\cdot 10^{-19}N

And the electric force can be written as

F=qE

where

q=-1.6\cdot 10^{-19}C is the charge of the electron

E is the magnitude of the electric field

Solving for E,

E=\frac{F}{q}=\frac{9.11\cdot 10^{-19}}{-1.6\cdot 10^{-19}}=-5.69 N/C

The negative sign means that the direction of the electric field is opposite to the direction of the force (because the charge is negative): since the force has same direction of the acceleration (vertically upward), the electric field must point vertically downward.

(b) Yes

We can answer the question by calculating the magnitude of the gravitational force acting on the electron, to check if it is relevant or not. The gravitational force on the electron is:

F=mg

where

m=9.11\cdot 10^{-31} kg is the mass of the electron

g=9.81 m/s^2 is the acceleration due to gravity

Substituting,

F=(9.11\cdot 10^{-31})(9.81)=8.93\cdot 10^{-30}N

We see that the gravitational force is basically negligible compared to the electric force calculated in part (a), therefore we can say it is justified to ignore the effect of gravity in the problem.

7 0
3 years ago
A(n) _______ wave carries energy through space.<br><br> Fill in the blank.
Gwar [14]

Answer:

electromagnetic.

Explanation:

Electrons can jump from one energy level to another.

4 0
3 years ago
Do physical changes include changes in the states of matter?
VikaD [51]
I think so yeah if I’m not right sorry
8 0
3 years ago
Read 2 more answers
The blood pressure of a<br>person is 140/100 mm of Hg, what does it mean?​
KonstantinChe [14]

Answer:

If the blood pressure of a person is more than 120/80 mm of Hg, then it is especially called as high blood pressure.

Explanation:

But regarding answer to this question, The blood pressure of a

person is 140/100 mm of Hg means that the systolic blood pressure of a person is 140 mm of Hg and diastolic blood pressure of a person is 100 mm of Hg.

I hope this answer will help you

3 0
3 years ago
An uncharged capacitor is connected to the terminals of a 4.0 V battery, and 9.0 μC flows to the positive plate. The 4.0 V batte
Lelechka [254]

Answer:

2.25\mu C

Explanation:

At the beginning, we have:

V = 4.0 V potential difference across the capacitor

Q=9.0 \mu C=9.0\cdot 10^{-6}C charge stored on the capacitor

Therefore, we can calculate the capacitance of the capacitor:

C=\frac{Q}{V}=\frac{9.0 \cdot 10^{-6} C}{4.0 V}=2.25\cdot 10^{-6} F

Later, the battery is replaced with another battery whose voltage is

V = 5.0 V

Since the capacitance of the capacitor does not change, we can calculate the new charge stored:

Q=CV=(2.25\cdot 10^{-6} F)(5.0 V)=11.25 \cdot 10^{-6} C=11.25 \mu C

Since the capacitor has been connected exactly as before, we have that the charge on the positive plate has increased from 9.0 \mu C to 11.25 \mu C. Therefore, the additional charge that moved to the positive plate is

\Delta Q = 11.25 \mu C-9.0 \mu C=2.25 \mu C

5 0
3 years ago
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