PHYSICS HIGH SCHOOL

A bicyclist steadily speeds up from rest to 11.0m/s in 3.40s. How far did she travel during this time?

Answers

Answer 1
Answer:

Answer:

To find the distance traveled by the bicyclist during the given time, we can use the formula:

Distance = (Initial Velocity * Time) + (0.5 * Acceleration * Time^2)

Since the bicyclist starts from rest, the initial velocity is 0 m/s.

Given:

Initial velocity (u) = 0 m/s

Final velocity (v) = 11.0 m/s

Time (t) = 3.40 s

Using the formula, we can calculate the distance traveled:

Distance = (0 * 3.40) + (0.5 * Acceleration * 3.40^2)

To find the acceleration, we can use the equation:

Acceleration = (Final Velocity - Initial Velocity) / Time

Acceleration = (11.0 - 0) / 3.40

Acceleration = 11.0 / 3.40

Now, we substitute the value of acceleration into the distance formula:

Distance = (0 * 3.40) + (0.5 * (11.0 / 3.40) * 3.40^2)

Simplifying further:

Distance = 0 + (0.5 * (11.0 / 3.40) * 11.56)

Distance = (0.5 * (11.0 / 3.40) * 11.56)

Distance = (0.5 * 11.0 * 3.40)

Distance = 0.5 * 37.4

Distance = 18.7 meters

Therefore, the bicyclist traveled a distance of 18.7 meters during the given time of 3.40 seconds.

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Charge Q = +4.00 mC is distributed uniformly over the volume of an insulating sphere that has radius R = 5.00 cm. What is the potential difference between the center of the sphere and the surface of the sphere?

Answers

The potential difference between the center of the sphere and the surface of the sphere will be PD=3.6 x 10^8 V

What will be the potential difference?

It is given that:-

Charge Q = +4.00 mC  or  4*10^(-3) c

insulating sphereradius R = 5.00 cm = 0.05 m

The formula for the potentialdifference at the surface of the sphere is given by

V_(s) =(K* Q)/(r) =(9*10^(9) * 4*10^(-3) )/(0.05)=7.2*10^(8) V  

The formula for the potentialdifference at the center of the sphere  

Vc_{} =(3)/(2) * V_(s)

V_(c) = 1.5*7.2*10^(8) v = 10.8*10^(8 ) V      

The difference between the potential difference will be  

V=V_(c) -V_(s) = 10.8*10^(8) =3.6*10^(8) V

Thus the potential difference between the center of the sphere and the surface of the sphere will be PD=3.6 x 10^8 V

To know more about Charge follow

brainly.com/question/14372859

Answer:

3.6 x 10^8 V

Explanation:

Q = 4 m C = 4 x 10^-3 C

r = 5 cm = 0.05 m

The formula for the potential at the surface is

Vs = K Q / r = (9 x 10^9 x 4 x 10^-3) / 0.05 = 7.2 x 10^8 V

The formula for the potential at the centre is

Vc = 3/2 Vs

Vc = 1.5 x 7.2 x 10^8 V = 10.8 x 10^8 V

The difference in potential is

V = Vc - Vs = 10.8 x 10^8 - 7.2 x 10^8 = 3.6 x 10^8 V

A charge of 8.5 × 10–6 C is in an electric field that has a strength of 3.2 × 105 N/C. What is the electric force acting on the charge?

Answers

ok
here is your anwer
O hope it is useful for you

Answer:

2.7N

Explanation:

( Pennfoster plz help ) Which of the following does not accurately describe the forces that exist with an atom? A. Electrons closer to the nucleus and electrons farther from the nucleus repel due to like charges. B. Electrons positioned farther away from the nucleus are less attracted to the nucleus than are electrons positioned closer to the nucleus. C. Electrons positioned closer to the nucleus have a greater attraction to the protons and are more likely to be discharged from the atom than electrons farther away. D. Electrons in an atom are attracted to the protons within the nucleus due to opposite charges.

Answers

I would say C i'm not 100% sure

Choice-C is nonsense.

Electrons positioned closer to the nucleus are closer to the protons in the nucleus and more strongly attracted to them.  Therefore these electrons are LESS likely to be discharged from the atom than electrons farther away from the nucleus are.

a 5 v battery is connected to a resistor which draws a 2 a current. What is the power used by this circuit

Answers


Electrical power = (voltage) x (current)

                         = (5 volts) x (2 amperes)  =  10 watts.

When electrons are removed from the outermost shell of a calcium atom, the atom becomes

Answers

A cation has a smaller radius than the atom.
a cation
Cations are positive ions that have lost an electron

Which one of the following parenting factors is associated with improved IQ scores in children? A. Strict discipline
B. Lenient discipline and fewer rules
C. A stable home and varied activities
D. Early instruction in reading

Answers

Answer: Option (C) is the correct answer.

Explanation:

The term IQ stands form intelligent quotient. When a child is provided with stable home environment in which there are less or rare conflicts, happy environment, understanding between individuals etc will lead to a stable growth of child.

Also, varied activities will help child to learn quickly and develop various skills.

Thus, a stable home and varied activities is the one parenting factor which is associated with improved IQ scores in children.
The correct answer is C. A stable home and varied activities.

Hope this helps! xoxo
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